4.2: Visualize Fractions (Part 2)
Model Equivalent Fractions
Let’s think about Andy and Bobby and their favorite food again. If Andy eats \(\dfrac{1}{2}\) of a pizza and Bobby eats \(\dfrac{2}{4}\) of the pizza, have they eaten the same amount of pizza? In other words, does \(\dfrac{1}{2} = \dfrac{2}{4}\)? We can use fraction tiles to find out whether Andy and Bobby have eaten equivalent parts of the pizza.
Equivalent fractions are fractions that have the same value.
Fraction tiles serve as a useful model of equivalent fractions. You may want to use fraction tiles to do the following activity. Or you might make a copy of Figure 4.3 and extend it to include eighths, tenths, and twelfths.
Start with a \(\dfrac{1}{2}\) tile. How many fourths equal one-half? How many of the \(\dfrac{1}{4}\) tiles exactly cover the \(\dfrac{1}{2}\) tile?
Figure \(\PageIndex{7}\)
Since two \(\dfrac{1}{4}\) tiles cover the \(\dfrac{1}{2}\) tile, we see that \(\dfrac{2}{4}\) is the same as \(\dfrac{1}{2}\), or \(\dfrac{2}{4} = \dfrac{1}{2}\).
How many of the \(\dfrac{1}{6}\) tiles cover the \(\dfrac{1}{2}\) tile?
Figure \(\PageIndex{8}\)
Since three \(\dfrac{1}{6}\) tiles cover the \(\dfrac{1}{2}\) tile, we see that \(\dfrac{3}{6}\) is the same as \(\dfrac{1}{2}\). So, \(\dfrac{3}{6} = \dfrac{1}{2}\). The fractions are equivalent fractions.
Use fraction tiles to find equivalent fractions. Show your result with a figure.
- How many eighths equal one-half?
- How many tenths equal one-half?
- How many twelfths equal one-half?
Solution
- It takes four \(\dfrac{1}{8}\) tiles to exactly cover the \(\dfrac{1}{2}\) tile, so \(\dfrac{4}{8} = \dfrac{1}{2}\).
- It takes five \(\dfrac{1}{10}\) tiles to exactly cover the \(\dfrac{1}{2}\) tile, so \(\dfrac{5}{10} = \dfrac{1}{2}\).
- It takes six \(\dfrac{1}{12}\) tiles to exactly cover the \(\dfrac{1}{2}\) tile, so \(\dfrac{6}{12} = \dfrac{1}{2}\).
Suppose you had tiles marked \(\dfrac{1}{20}\). How many of them would it take to equal \(\dfrac{1}{2}\)? Are you thinking ten tiles? If you are, you’re right, because \(\dfrac{10}{20} = \dfrac{1}{2}\).
We have shown that \(\dfrac{1}{2}, \dfrac{2}{4}, \dfrac{3}{6}, \dfrac{4}{8}, \dfrac{5}{10}, \dfrac{6}{12}\), and \(\dfrac{10}{20}\) are all equivalent fractions.
Use fraction tiles to find equivalent fractions: How many eighths equal one-fourth?
- Answer
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\(2\)
Use fraction tiles to find equivalent fractions: How many twelfths equal one-fourth?
- Answer
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\(3\)
Find Equivalent Fractions
We used fraction tiles to show that there are many fractions equivalent to \(\dfrac{1}{2}\). For example, \(\dfrac{2}{4}, \dfrac{3}{6}\), and \(\dfrac{4}{8}\) are all equivalent to \(\dfrac{1}{2}\). When we lined up the fraction tiles, it took four of the \(\dfrac{1}{8}\) tiles to make the same length as a \(\dfrac{1}{2}\) tile. This showed that \(\dfrac{4}{8} = \dfrac{1}{2}\). See Example \(\PageIndex{13}\).
We can show this with pizzas, too. Figure \(\PageIndex{9a}\) shows a single pizza, cut into two equal pieces with \(\dfrac{1}{2}\) shaded. Figure \(\PageIndex{9b}\) shows a second pizza of the same size, cut into eight pieces with \(\dfrac{4}{8}\) shaded.
Figure \(\PageIndex{9}\)
This is another way to show that \(\dfrac{1}{2}\) is equivalent to \(\dfrac{4}{8}\). How can we use mathematics to change \(\dfrac{1}{2}\) into \(frac{4}{8}\)? How could you take a pizza that is cut into two pieces and cut it into eight pieces? You could cut each of the two larger pieces into four smaller pieces! The whole pizza would then be cut into eight pieces instead of just two. Mathematically, what we’ve described could be written as:
\[\dfrac{1 \cdot \textcolor{blue}{4}}{2 \cdot \textcolor{blue}{4}} = \dfrac{4}{8} \nonumber \]
These models lead to the Equivalent Fractions Property, which states that if we multiply the numerator and denominator of a fraction by the same number, the value of the fraction does not change.
If \(a\), \(b\), and \(c\) are numbers where \(b ≠ 0\) and \(c ≠ 0\), then
\[\dfrac{a}{b} = \dfrac{a \cdot c}{b \cdot c}\]
When working with fractions, it is often necessary to express the same fraction in different forms. To find equivalent forms of a fraction, we can use the Equivalent Fractions Property. For example, consider the fraction one-half.
\[\begin{split} \dfrac{1 \cdot \textcolor{blue}{3}}{2 \cdot \textcolor{blue}{3}} = \dfrac{3}{6} \; & so \; \dfrac{1}{2} = \dfrac{3}{6} \\ \dfrac{1 \cdot \textcolor{blue}{2}}{2 \cdot \textcolor{blue}{2}} = \dfrac{2}{4} \; & so \; \dfrac{1}{2} = \dfrac{2}{4} \\ \dfrac{1 \cdot \textcolor{blue}{10}}{2 \cdot \textcolor{blue}{10}} = \dfrac{10}{20} \; & so \; \dfrac{1}{2} = \dfrac{10}{20} \end{split} \nonumber \]
So, we say that \(\dfrac{1}{2}, \dfrac{2}{4}, \dfrac{3}{6}\), and \(\dfrac{10}{20}\) are equivalent fractions.
Find three fractions equivalent to \(\dfrac{2}{5}\).
Solution
To find a fraction equivalent to \(\dfrac{2}{5}\), we multiply the numerator and denominator by the same number (but not zero). Let us multiply them by \(2\), \(3\), and \(5\).
\[\dfrac{2 \cdot \textcolor{blue}{2}}{5 \cdot \textcolor{blue}{2}} = \dfrac{4}{10} \qquad \dfrac{2 \cdot \textcolor{blue}{3}}{5 \cdot \textcolor{blue}{3}} = \dfrac{6}{15} \qquad \dfrac{2 \cdot \textcolor{blue}{5}}{5 \cdot \textcolor{blue}{5}} = \dfrac{10}{25} \nonumber \]
So, \(\dfrac{4}{10}, \dfrac{6}{15}\), and \(\dfrac{10}{25}\) are equivalent to \(\dfrac{2}{5}\).
Find three fractions equivalent to \(\dfrac{3}{5}\).
- Answer
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Correct answers include \(\dfrac{6}{10}, \dfrac{9}{15}\) and \(\dfrac{12}{20}\)
Find three fractions equivalent to \(\dfrac{4}{5}\).
- Answer
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Correct answers include \(\dfrac{8}{10}, \dfrac{12}{15}\) and \(\dfrac{16}{20}\)
Find a fraction with a denominator of 21 that is equivalent to \(\dfrac{2}{7}\).
Solution
To find equivalent fractions, we multiply the numerator and denominator by the same number. In this case, we need to multiply the denominator by a number that will result in \(21\).
Since we can multiply \(7\) by \(3\) to get \(21\), we can find the equivalent fraction by multiplying both the numerator and denominator by \(3\).
\[\dfrac{2}{7} = \dfrac{2 \cdot \textcolor{blue}{3}}{7 \cdot \textcolor{blue}{3}} = \dfrac{6}{21} \nonumber \]
Find a fraction with a denominator of \(21\) that is equivalent to \(\dfrac{6}{7}\).
- Answer
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\(\dfrac{18}{21}\)
Find a fraction with a denominator of \(100\) that is equivalent to \(\dfrac{3}{10}\).
- Answer
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\(\dfrac{30}{100}\)
Locate Fractions and Mixed Numbers on the Number Line
Now we are ready to plot fractions on a number line. This will help us visualize fractions and understand their values.
Let us locate \(\dfrac{1}{5}, \dfrac{4}{5}, 3, 3 \dfrac{1}{3}, \dfrac{7}{4}, \dfrac{9}{2}, 5\), and \(\dfrac{8}{3}\) on the number line. We will start with the whole numbers \(3\) and \(5\) because they are the easiest to plot.
The proper fractions listed are \(\dfrac{1}{5}\) and \(\dfrac{4}{5}\). We know proper fractions have values less than one, so \(\dfrac{1}{5}\) and \(\dfrac{1}{5}\) are located between the whole numbers \(0\) and \(1\). The denominators are both \(5\), so we need to divide the segment of the number line between \(0\) and \(1\) into five equal parts. We can do this by drawing four equally spaced marks on the number line, which we can then label as \(\dfrac{1}{5}, \dfrac{2}{5}, \dfrac{3}{5}\), and \(\dfrac{4}{5}\). Now plot points at \(\dfrac{1}{5}\) and \(\dfrac{4}{5}\).
The only mixed number to plot is \(3 \dfrac{1}{3}\). Between what two whole numbers is \(3 \dfrac{1}{3}\)? Remember that a mixed number is a whole number plus a proper fraction, so \(3 \dfrac{1}{3} > 3\). Since it is greater than \(3\), but not a whole unit greater, \(3 \dfrac{1}{3}\) is between \(3\) and \(4\). We need to divide the portion of the number line between \(3\) and \(4\) into three equal pieces (thirds) and plot \(3 \dfrac{1}{3}\) at the first mark.
Finally, look at the improper fractions \(\dfrac{7}{4}, \dfrac{9}{2}\), and \(\dfrac{8}{3}\). Locating these points will be easier if you change each of them to a mixed number.
\[\dfrac{7}{4} = 1 \dfrac{3}{4}, \qquad \dfrac{9}{2} = 4 \dfrac{1}{2}, \qquad \dfrac{8}{3} = 2 \dfrac{2}{3} \nonumber \]
Here is the number line with all the points plotted.
Locate and label the following on a number line: \(\dfrac{3}{4}, \dfrac{4}{3}, \dfrac{5}{3}, 4 \dfrac{1}{5}\), and \(\dfrac{7}{2}\).
Solution
Start by locating the proper fraction \(\dfrac{3}{4}\). It is between \(0\) and \(1\). To do this, divide the distance between \(0\) and \(1\) into four equal parts. Then plot \(\dfrac{3}{4}\).
Next, locate the mixed number \(4 \dfrac{1}{5}\). It is between \(4\) and \(5\) on the number line. Divide the number line between \(4\) and \(5\) into five equal parts, and then plot \(4 \dfrac{1}{5}\) one-fifth of the way between \(4\) and \(5\).
Now locate the improper fractions \(\dfrac{4}{3}\) and \(\dfrac{5}{3}\). It is easier to plot them if we convert them to mixed numbers first.
\[\dfrac{4}{3} = 1 \dfrac{1}{3}, \qquad \dfrac{5}{3} = 1 \dfrac{2}{3} \nonumber\]
Divide the distance between \(1\) and \(2\) into thirds.
Next let us plot \(\dfrac{7}{2}\). We write it as a mixed number, \(\dfrac{7}{2} = 3 \dfrac{1}{2}\). Plot it between \(3\) and \(4\).
The number line shows all the numbers located on the number line.
Locate and label the following on a number line: \(\dfrac{1}{3}, \dfrac{5}{4}, \dfrac{7}{4}, 2 \dfrac{3}{5}, \dfrac{9}{2}\).
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Locate and label the following on a number line: \(\dfrac{2}{3}, \dfrac{5}{2}, \dfrac{9}{4}, \dfrac{11}{4}, 3 \dfrac{2}{5}\).
- Answer
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In Introduction to Integers , we defined the opposite of a number. It is the number that is the same distance from zero on the number line but on the opposite side of zero. We saw, for example, that the opposite of \(7\) is \(−7\) and the opposite of \(−\)7 is \(7\).
Fractions have opposites, too. The opposite of \(\dfrac{3}{4}\) is \(− \dfrac{3}{4}\). It is the same distance from \(0\) on the number line, but on the opposite side of \(0\).
Thinking of negative fractions as the opposite of positive fractions will help us locate them on the number line. To locate \(− \dfrac{15}{8}\) on the number line, first think of where \(\dfrac{15}{8}\) is located. It is an improper fraction, so we first convert it to the mixed number \(1 \dfrac{7}{8}\) and see that it will be between \(1\) and \(2\) on the number line. So its opposite, \(− \dfrac{15}{8}\), will be between \(−1\) and \(−2\) on the number line.
Locate and label the following on the number line: \(\dfrac{1}{4}, − \dfrac{1}{4}, 1 \dfrac{1}{3}, −1 \dfrac{1}{3}, \dfrac{5}{2}\), and \(− \dfrac{5}{2}\).
Solution
Draw a number line. Mark \(0\) in the middle and then mark several units to the left and right.
To locate \(\dfrac{1}{4}\), divide the interval between \(0\) and \(1\) into four equal parts. Each part represents one-quarter of the distance. So plot \(\dfrac{1}{4}\) at the first mark.
To locate \(− \dfrac{1}{4}\), divide the interval between \(0\) and \(−1\) into four equal parts. Plot \(− \dfrac{1}{4}\) at the first mark to the left of \(0\).
Since \(1 \dfrac{1}{3}\) is between \(1\) and \(2\), divide the interval between \(1\) and \(2\) into three equal parts. Plot \(1 \dfrac{1}{3}\) at the first mark to the right of \(1\). Then since \(−1 \dfrac{1}{3}\) is the opposite of \(1 \dfrac{1}{3}\) it is between \(−1\) and \(−2\). Divide the interval between \(−1\) and \(−2\) into three equal parts. Plot \(−1 \dfrac{1}{3}\) at the first mark to the left of \(−1\).
To locate \(\dfrac{5}{2}\) and \(− \dfrac{5}{2}\), it may be helpful to rewrite them as the mixed numbers \(2 \dfrac{1}{2}\) and \(−2 \dfrac{1}{2}\). Since \(2 \dfrac{1}{2}\) is between \(2\) and \(3\), divide the interval between \(2\) and \(3\) into two equal parts. Plot \(\dfrac{5}{2}\) at the mark. Then since \(−2 \dfrac{1}{2}\) is between \(−2\) and \(−3\), divide the interval between \(−2\) and \(−3\) into two equal parts. Plot \(− \dfrac{5}{2}\) at the mark.
Locate and label each of the given fractions on a number line: \(\dfrac{2}{3}, − \dfrac{2}{3}, 2 \dfrac{1}{4}, −2 \dfrac{1}{4}, \dfrac{3}{2}, − \dfrac{3}{2}\)
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Locate and label each of the given fractions on a number line: \(\dfrac{3}{4}, − \dfrac{3}{4}, 1 \dfrac{1}{2}, −1 \dfrac{1}{2}, \dfrac{7}{3}, − \dfrac{7}{3}\)
- Answer
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Order Fractions and Mixed Numbers
We can use the inequality symbols to order fractions. Remember that \(a > b\) means that \(a\) is to the right of \(b\) on the number line. As we move from left to right on a number line, the values increase.
Order each of the following pairs of numbers, using \(<\) or \(>\):
- \(− \dfrac{2}{3}\)____\(−1\)
- \(−3 \dfrac{1}{2}\)____\(−3\)
- \(− \dfrac{3}{7}\)____\(− \dfrac{3}{8}\)
- \(−2\)____\(− \dfrac{16}{9}\)
Solution
- \(− \dfrac{2}{3} > −1\)
- \(−3 \dfrac{1}{2} < −3\)
- \(− \dfrac{3}{7} < − \dfrac{3}{8}\)
- \(−2 < − \dfrac{16}{9}\)
Order each of the following pairs of numbers, using \(<\) or \(>\):
- \(− \dfrac{1}{3}\)__\(−1\)
- \(−1 \dfrac{1}{2}\)__\(− 2\)
- \(− \dfrac{2}{3}\)__\(− \dfrac{1}{3}\)
- \(−3\)__\(− \dfrac{7}{3}\)
- Answer a
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\(>\)
- Answer b
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\(>\)
- Answer c
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\(<\)
- Answer d
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\(<\)
Order each of the following pairs of numbers, using \(<\) or \(>\):
- \(−3\)__\(− \dfrac{17}{5}\)
- \(−2 \dfrac{1}{4}\)__\(−2\)
- \(− \dfrac{3}{5}\)__\(− \dfrac{4}{5}\)
- \(−4\)__\(− \dfrac{10}{3}\)
- Answer a
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\(>\)
- Answer b
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\(<\)
- Answer c
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\(>\)
- Answer d
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\(<\)
Access Additional Online Resources
- Introduction to Fractions
- Identify Fractions Using Pattern Blocks
Key Concepts
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Property of One
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Any number, except zero, divided by itself is one.
\(\dfrac{a}{a}=1\), where \(a\neq 0\).
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Any number, except zero, divided by itself is one.
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Mixed Numbers
- A mixed number consists of a whole number \(a\) and a fraction \(\dfrac{b}{c}\) where \(c \neq 0\).
- It is written as follows: \(a\dfrac{b}{c}\) \(c \neq 0\)
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Proper and Improper Fractions
- The fraction \(\frac{a}{b}\) is a proper fraction if \(a<b\) and an improper fraction if \(a \geq b\).
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Convert an improper fraction to a mixed number.
- Divide the denominator into the numerator.
- Identify the quotient, remainder, and divisor.
- Write the mixed number as \(\dfrac{\text{remainder}}{\text{divisor}}\).
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Convert a mixed number to an improper fraction.
- Multiply the whole number by the denominator.
- Add the numerator to the product found in Step 1.
- Write the final sum over the original denominator.
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Equivalent Fractions Property
- If \(a\), \(b\) and \(c\) are numbers where \(b \neq 0\), \(c \neq 0\), then \(\dfrac{a}{b} = \dfrac{a\cdot c}{b\cdot c}\]).
Glossary
- equivalent fractions
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Equivalent fractions are two or more fractions that have the same value.
- fraction
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A fraction is written \(\dfrac{a}{b}\). in a fraction, \(a\) is the numerator and \(b\) is the denominator. A fraction represents parts of a whole. The denominator \(b\) is the number of equal parts the whole has been divided into, and the numerator \(a\) indicates how many parts are included.
- mixed number
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A mixed number consists of a whole number \(a\) and a fraction \(\dfrac{b}{c}\) where \(c \neq 0\). It is written as \(a\dfrac{b}{c}\), where \(c \neq 0\).
- proper and improper fractions
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The fraction \(\dfrac{a}{b}\) is proper if \(a<b\) and improper if \(a>b\).
Practice Makes Perfect
In the following exercises, shade parts of circles or squares to model the following fractions.
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{3}{4}\)
- \(\dfrac{2}{5}\)
- \(\dfrac{5}{6}\)
- \(\dfrac{7}{8}\)
- \(\dfrac{5}{8}\)
- \(\dfrac{7}{10}\)
In the following exercises, use fraction circles to make wholes, if possible, with the following pieces.
- 3 thirds
- 8 eighths
- 7 sixths
- 4 thirds
- 7 fifths
- 7 fourths
In the following exercises, name the improper fractions. Then write each improper fraction as a mixed number.
In the following exercises, draw fraction circles to model the given fraction.
- \(\dfrac{3}{3}\)
- \(\dfrac{4}{4}\)
- \(\dfrac{7}{4}\)
- \(\dfrac{5}{3}\)
- \(\dfrac{11}{6}\)
- \(\dfrac{13}{8}\)
- \(\dfrac{10}{3}\)
- \(\dfrac{9}{4}\)
In the following exercises, rewrite the improper fraction as a mixed number.
- \(\dfrac{3}{2}\)
- \(\dfrac{5}{3}\)
- \(\dfrac{11}{4}\)
- \(\dfrac{13}{5}\)
- \(\dfrac{25}{6}\)
- \(\dfrac{28}{9}\)
- \(\dfrac{42}{13}\)
- \(\dfrac{47}{15}\)
In the following exercises, rewrite the mixed number as an improper fraction.
- \(1 \dfrac{2}{3}\)
- \(1 \dfrac{2}{5}\)
- \(2 \dfrac{1}{4}\)
- \(2 \dfrac{5}{6}\)
- \(2 \dfrac{7}{9}\)
- \(2 \dfrac{5}{7}\)
- \(3 \dfrac{4}{7}\)
- \(3 \dfrac{5}{9}\)
In the following exercises, use fraction tiles or draw a figure to find equivalent fractions.
- How many sixths equal one-third?
- How many twelfths equal one-third?
- How many eighths equal three-fourths?
- How many twelfths equal three-fourths?
- How many fourths equal three-halves?
- How many sixths equal three-halves?
In the following exercises, find three fractions equivalent to the given fraction. Show your work, using figures or algebra.
- \(\dfrac{1}{4}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{3}{8}\)
- \(\dfrac{5}{6}\)
- \(\dfrac{2}{7}\)
- \(\dfrac{5}{9}\)
In the following exercises, plot the numbers on a number line.
- \(\dfrac{2}{3}, \dfrac{5}{4}, \dfrac{12}{5}\)
- \(\dfrac{1}{3}, \dfrac{7}{4}, \dfrac{13}{5}\)
- \(\dfrac{1}{4}, \dfrac{9}{5}, \dfrac{11}{3}\)
- \(\dfrac{7}{10}, \dfrac{5}{2}, \dfrac{13}{8}, 3\)
- \(2 \dfrac{1}{3}, −2 \dfrac{1}{3}\)
- \(1 \dfrac{3}{4}, −1 \dfrac{3}{5}\)
- \(\dfrac{3}{4}, − \dfrac{3}{4}, 1 \dfrac{2}{3}, −1 \dfrac{2}{3}, \dfrac{5}{2}, − \dfrac{5}{2}\)
- \(\dfrac{2}{5}, − \dfrac{2}{5}, 1 \dfrac{3}{4}, −1 \dfrac{3}{4}, \dfrac{8}{3}, − \dfrac{8}{3}\)
In the following exercises, order each of the following pairs of numbers, using < or >.
- −1__\(− \dfrac{1}{4}\)
- −1__\(− \dfrac{1}{3}\)
- \(−2 \dfrac{1}{2}\)__− 3
- \(−1 \dfrac{3}{4}\)__− 2
- \(− \dfrac{5}{12}\)__\(− \dfrac{7}{12}\)
- \(− \dfrac{9}{10}\)__\(− \dfrac{3}{10}\)
- −3__\(− \dfrac{13}{5}\)
- −4__\(− \dfrac{23}{6}\)
Everyday Math
- Music Measures A choreographed dance is broken into counts. A \(\dfrac{1}{1}\) count has one step in a count, a \(\dfrac{1}{2}\) count has two steps in a count and a 1 3 count has three steps in a count. How many steps would be in a \(\dfrac{1}{5}\) count? What type of count has four steps in it?
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Music Measures
Fractions are used often in music. In 4 4 time, there are four quarter notes in one measure.
- How many measures would eight quarter notes make?
- The song “Happy Birthday to You” has 25 quarter notes. How many measures are there in “Happy Birthday to You?”
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Baking
Nina is making five pans of fudge to serve after a music recital. For each pan, she needs 1 2 cup of walnuts.
- How many cups of walnuts does she need for five pans of fudge?
- Do you think it is easier to measure this amount when you use an improper fraction or a mixed number? Why?
Writing Exercises
- Give an example from your life experience (outside of school) where it was important to understand fractions.
- Explain how you locate the improper fraction \(\dfrac{21}{4}\) on a number line on which only the whole numbers from 0 through 10 are marked.
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) If most of your checks were:
…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.
Contributors and Attributions
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Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at<EMAIL_ADDRESS>."


15.3: Christianity and Christian Mysticism
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Although there are many denominations within Christianity , at times generating significant conflict between each other, Christianity as a whole has been a profoundly successful and influential religion. Approximately one-third of the world’s population, over 2.2 billion people, are Christian, and it has spread around the entire world. At its heart, Christianity is a simple religion, and those who live a truly Christian life are guided by one simple philosophy: love your neighbor.
The Foundation of the Christian Faith
Approximately 2,000 years ago a man named Jesus of Nazareth was born. For 30 years he lived a simple life, but then embarked on a religious crusade that lasted for only 3 short years. At the end of those three years, he was crucified for having challenged the right of the religious and political authorities to lead the people, as well as their decency in so doing. However, he never outright challenged anyone, always teaching love, mercy, and peace. Following his death and resurrection he has been called the Christ (meaning Messiah , or anointed), hence the term Christianity.
 They also believe that he was born of a virgin, through the power of the Holy Spirit. Thus, although Christians believe in one God, they view God in a Trinitarian way: Father , Son , and Holy Spirit (Wilkins, 1967). Although Judaic prophecy had foretold of the coming of a Messiah, they do not believe that Jesus was that person. Yet, having arisen out of Judaism, Christians believe much of the Jewish faith, including the importance of the Ten Commandments.
The second is this, ‘You shall love your neighbor as yourself.’ There is no other commandment greater than these.”
from Mark, Chapter 12; Holy Bible
What is surprising about this answer is that Jesus simplified the Ten Commandments even further, emphasizing just two, but those two commandments encompass both the spiritual and the social worlds - love God, and love all people (the famous Good Samaritan parable teaches that all people are one another’s neighbors). Thus, it seems appropriate for us to address the social and psychological aspects of loving and/or caring for other people, as well as for ourselves, as something separate from religious/spiritual pursuits. And yet at the same time, we cannot, and need not, separate our psychological studies from a religious/spiritual context (at least when trying to understand those people for whom religion and spirituality are important daily factors).
discussion question \(\PageIndex{1}\)
Love God, and love your neighbor as yourself. Is it really that simple?
Christian Mysticism
Christian Mysticism is as old as Christianity itself, for Jesus led a mystic life (Walker, 2003). Since that time there have been many Christian mystics, but two particular groups stand out: the desert fathers and the women mystics (Chervin, 1992; Clement, 1993; Waddell, 1998). As the Christian faith became legal, following the conversion of the Roman emperor Constantine the Great late in the third century, Christianity became caught up in the politics of the empire. Soon enough, a group of spiritual men sought to escape worldly politics and secular distractions by becoming hermits. They traveled out into the Egyptian desert and began monastic lives. In order to be with God, they did not merely seek solitude, but they also sought to eliminate their sense of ego. This was attempted through what we might call contemplative prayer, or simply meditation. However, the sense of ego does not go away easily:
The ego doesn’t want us looking for God because when we find God, the illusion of being an ego will be destroyed. It will mean the end of our self-centered existence and all of its negative emotions…One cannot see God and continue to live as a separate person. (pg. 49; Walker, 2003).
Thus, a battle arises between the sense of ego and one’s efforts to immerse oneself in the Deity. This spiritual combat, essentially the battle between good and evil, exists because of our freedom to choose our path in life (Clement, 1993). It was in recognition of this challenge that the desert fathers sought the solitude of the desert. There they were able to pursue the ecstasy of unknowing, that which is beyond the boundaries of any human ability to comprehend or rationalize the experience:
At one time Zachary went to his abbot Silvanus, and found him in an ecstasy, and his hands were stretched out to heaven. And when he saw him thus, he closed the door and went away: and coming back about the sixth hour, and the ninth, he found him even so: but toward the tenth hour he knocked, and coming in found him lying quiet…the young man held his fee saying, “I shall not let thee go, until thou tell me what thou hast seen.” The old man answered him: “I was caught up into heaven, and I saw the glory of God. And I stood there until now, and now am I sent away.” (pg. 130; Sayings of the Fathers, in Waddell, 1998)
Although there have been many mystics who were men, including the desert fathers, Meister Eckhart (a favorite of Erich Fromm), and the anonymous author of The Cloud of Unknowing (see Kirvan, 1996a), there have also been a number of well-known women mystics, such as Blessed Julian of Norwich (see Chilson, 1995), Saint Teresa of Avila (see Kirvan, 1996b), Saint Therese of Lisieux (Kirvan, 1996c), and most famous of them all, Saint Joan of Arc (see Chervin, 1992). Given the patriarchal history of the Roman Catholic Church, and the beginning of mysticism with the desert fathers , it is enlightening to see that so many women were blessed by God’s grace and presence in a profound mystical fashion, and that this was recognized by the Catholic Church (as many of these women became saints). Thus, God does not discriminate based on gender. Indeed, in one mystical experience, St. Hildegard of Bingen (who lived from 1098-1179 A.D.) was instructed to use the majesty of her mystical gift to instruct men in the true meaning of faith:
…as I was gazing with great fear and trembling attention at a heavenly vision, I saw a great splendor in which resounded a voice from Heaven, saying to me, “O fragile human…Cry out and speak of the origin of pure salvation until those people are instructed, who, though they see the inmost contents of the Scriptures, do not wish to tell them or preach them, because they are lukewarm and sluggish in serving God’s justice…Burst forth into a fountain of abundance and overflow with mystical knowledge, until they who now think you contemptible because of Eve’s transgression are stirred up by the flood of your irrigation.” (pp. 17-18; St. Hildegard of Bingen, cited in Chervin, 1992)
Christian Meditation
The practice of meditation as a form of contemplative prayer has continued to the present day, thanks in part to two influential monks who lived during the twentieth century: Thomas Merton, OCSO (1915-1968) and John Main, OSB (1926-1982). Fr. Merton was a Trappist monk who wrote extensively on the monastic life, contemplation and silence, and connections between Western and Eastern spiritual philosophies (e.g., Merton, 1948, 1951, 1977; Montaldo, 2001; Nouwen, 1972). Always supportive of these various approaches to life, Fr. Merton adopted one of Jung’s objections to Freud’s discounting of religion: the observation that many people in psychoanalysis (whether Catholic, Protestant, or Jewish) had at the core of their dysfunction a religious crisis. Whereas Freud blamed this problem on the failure of religion, Fr. Merton blamed the problem on a failure of faith:
The real religious problem exists in the soul of those of us who in their hearts believe in God, and who recognize their obligation to love Him and serve Him - yet do not! (pg. 4; Merton, 1951)
While a monastic life may have allowed Fr. Merton to serve God with all his heart and soul, not everyone can be a monk or a nun. The Benedictine monk John Main sought to offer an easy path to meditation practice for the average person. Having become interested in mantra meditation following his early career in the Far East and his study of the writings of the desert fathers, Fr. Main began leading meditation groups at a monastery in London. He continued these meditation groups in Montreal after establishing the Benedictine priory there, and he also began sharing his interest in meditation through the publication of books such as Word Into Silence (Main, 1980), Moment of Christ (Main, 1984), and The Way of Unknowing (Main, 1989). Following Fr. Main’s death, his devoted student Laurence Freeman, OSB continued teaching Christian Meditation. In 1991, Fr. Freeman helped to establish the World Community for Christian Meditation, with its headquarters in London, England, and he continues to serve as its director. Fr. Freeman has written his own books on Christ as the inner source teaching us about life (Freeman, 1986, 2000), as well as some very practical books on meditation practice and establishing a meditation group (Freeman, 1994, 2002, 2004). Both Fr. Main and Fr. Freeman recommended a simple mantra meditation, using the Aramaic word “ Maranatha ,” one of the oldest Christian prayers, which means simply “Come Lord.” This simple, yet spiritually deep, form of meditation comes quite easily to those who are willing to pursue this silent path to contentment and being one with God.
discussion question \(\PageIndex{2}\)
There is a long and ongoing history of Christian meditation. Were you aware of this, or did you think that meditation and Christianity were unrelated? Does it seem appropriate to compare meditation to contemplative prayer?


7.4: Emotions in Relationships
Emotional Intelligence, Awareness, and Contagion
Emotional Intelligence
Many of us are familiar with the term IQ ("intelligence quota") that research has historically used to study the standards of human intelligence. In parallel to our IQ, researchers also study our emotional intelligence , or EI , which measures our ability to process emotions and emotional information (Alegre, et al., 2019). EI is a social intelligence that helps us manage our emotions within our social interactions. It helps us dictate aspects of our expression of emotions, as well as works to help regulate our emotions. Having a high emotional intelligence means that you are not only able to showcase self-awareness of your own emotions to control and balance them, but also able to monitor and accurately understand the emotions within others (Connor et. al, 2019). Psychologist Daniel Goleman conducted extensive research into categorizing what are now known as the five key elements of emotional intelligence : self-awareness, self-regulation, self-motivation, empathy, and social skill (Issah, 2018). Table \(\PageIndex{1}\) describes each key element of emotional intelligence as well as highlights specific communicative elements that emotional intelligence is enacted through.
Component |
Description |
|---|---|
|
Self-awareness |
The conscious understanding of our character, feeling, emotions, desires, personality, and individuality. Includes elements of humility, self-confidence, and our impact on others. |
|
Self-regulation |
The ability to respond to emotional stimuli in ways that promote facilitative control over feelings, thoughts, and physiological responses. Includes elements of emotional flexibility, comfort with ambiguity, and openness to change. |
|
Self-motivation |
The drive to set attainable goals using personal desire as opposed to external influence. Includes elements of organization, commitment, and optimism. |
|
Empathy |
Listening for understanding of the other person’s feelings and/or emotion with the goal of validating. Empathic listening is higher-level listening and therefore requires more energy. Includes elements of intercultural communication competence, and nonverbal awareness. |
|
Social skill |
The ability to build and maintain relationships. Effectively interacting with others, managing social norms, and rules within communication. Includes elements of verbal and nonverbal communication norms, intercultural communication competence, and active listening. |
Source: This work is a derivative of Change Leadership: The Role of Emotional Intelligence by Mohammad Issah, under a CC BY. Figure 7.4.8 by Elizabeth
Emotional Awareness
Although relationships affect our emotions, relationships themselves don’t experience the emotions, we do—as the interpersonal members of the relationship. Within the study of EI, researchers have expanded their understanding into more precise applications. One area that has been expanded upon from original research in EI is the concept of emotional awareness. Emotional awareness is the conscious understanding and recognition of one's own and others' emotions (Agnoli et al., 2019; Alegre, et al., 2019). This is an important aspect of emotional intelligence. In order to act, or communicate effectively within a given range of emotional interactions, we must first be acutely aware of what emotions are present to begin with.
Emotional awareness is often experienced at its height during adolescence, when we are experiencing puberty and our understanding of ourselves in relation to other people. During this time period in our age and development, as many of us probably remember all too well, it can also be difficult to regulate emotions (Agnoli et al., 2019). Because of the changes that are occurring in our physiological and emotional selves, this time period of learning and growing is usually met with lots of bumps and bruises (both physically and metaphorically). This rocky road into adulthood is beneficial for our emotional intelligence and emotional awareness, and it is crucial in our development of self-concept and self-esteem, which help to mold our emotional intelligence and awareness as well (Agnoli et al., 2019).
We are constantly being affected by the people around us and managing the emotional responses in those interactions. When looking at the communicative interaction of our emotions with other people, there are two primary ways we classify emotions: emotion traits and emotion states . Trait emotional awareness exists when we experience a specific emotion during person-to-person interactions regardless of whom we are interacting with (Alegre, et al., 2019). Emotion traits are connected to our personality and demonstrate our habitual and prolonged set of baseline emotions. Emotion traits are inherently linked to the theoretical framework of the Big Five personality traits we discussed earlier in this chapter (Alegre, et al., 2019). For example, specifically looking at conscientiousness and neuroticism, and agreeableness, an emotional trait we may label ourselves as could be as an “anxious person,” where we may not be effective in regulating the feelings of worry or fear. Or we may define ourselves as a “people-pleaser” if we are consistently focused on other people’s needs and attending to them as opposed to ourselves.
Emotion states exist in smaller “windows'' of time; experiencing them is much less dependent on our personality. Emotion states are emotions that we experience in a specific moment of time and are generally impacted by the conversations, interactions, and individuals we are in the presence of. Emotion states come and go much more easily than moods or personality, and they can change from one moment to the next. Some emotion states we experience are connected “directly to other people, like anger, envy, jealousy, hate and shame” (The Libarynth. n.d.). For example, we may find that we feel joy and comfort in the presence of certain people in our lives, while we may feel a sense of annoyance or frustration when interacting with others. Our emotional awareness that allows us to tune in to these different sets of emotional states and understand the connection and the reasoning as to why we experience these emotional reactions in the presence of certain people. This greater understanding allows us to better manage our emotions and create facilitative or positive emotional boundaries, which we will discuss later in this chapter.
Emotional Contagion
Emotions are like yawns: they are contagious. When we see someone yawn, it creates the urge to yawn ourselves—and emotions work in the same way. This “emotional yawn” is called emotional contagion . Emotional contagion occurs when we are exposed to other individuals' emotions during social interactions. Our instincts tend to impact these social interactions and create an alignment with the emotional states we are perceiving in others (Herrando & Constantinides, 2021).
Emotional contagion can be both a conscious and unconscious reaction. When we are in social settings with people whom we feel close to and would identify as significant individuals in our lives, we might be able to pick up and be impacted by their emotion states more quickly and without reason, as if it is an “automatic” connection we have with the person, whereas other interactions may require us to appraise, or assess the emotional state of the other person to establish our own reaction (The Libarynth. n.d.). Both reactions are part of our emotional awareness in balancing our communication to being effective and appropriate.
Emotions can be transferred from one individual to another through a wide array of communicative expressions: physical features, such as the urge to smile at someone when they smile at you; vocal expressions, as when someone starts to become tearful while they are talking, we might find ourselves holding back tears as well; posture or kinesics, when we see someone slump down into a chair we might feel their exhaustion; as well as through our nervous system, where our brain and nerve cells transmit information, we are able to not just perceive the emotions of others but literally feel them within ourselves.
Empathy versus Sympathy
Emotional contagion relates to the concepts of empathy and sympathy. Sympathy is the ability to understand your own experiences and emotional reactions to the environment around you and use that as a way to compare to other individuals who may be having similar emotional experiences. Sympathy is the projection of our emotions on other people, where we might feel bad for the person but we don’t see ourselves as agents in the emotional state. Empathy , in contrast, is sharing the emotional experiences with the other person. By showcasing emotional awareness to experience the emotional state of the other person. Empathy is directly connected to the concept of emotional contagion" in order to feel empathy, one must be able to rely on our instincts in order to align to the communicative expressions that the other person is conveying. Empathy is expressed in phrases such as “I understand how hard this is.”
Empathy is a learned skill that not everyone has or will enact the same way as others. Some individuals may not experience empathy at all, may have heightened levels of empathy, or may be at the same level as others. However we experience empathy, it does not mean we cannot communicate in an effective manner—but it may mean these aspects of emotional expression do not come as easily or in the same manner as others. For example, autism spectrum disorder (ASD) is characterized as having more difficulty enacting communication and social functioning as those who are neurotypical. Individuals who have ASD may express emotions and empathy in various ways that may look different, such as displaying an overabundance of emotional expression and empathy, or displaying very little. This impacts the amount of emotional contagion that is being experienced on the individual level.
Cultural Understanding
Another example of ways in which empathy and emotional contagion may be showcased in different behaviors is through cultural understanding . All cultures have different display rules regarding emotional expression, as we’ve learned, and that will translate to affecting others around them in various ways. Research has shown that within many cultures, negative emotional reactions and the subsequent impact of emotional contagion creates different display rules as well, therefore there are different measures of emotional contagion in order to understand these concepts within context (Kuang, et al., 2019). One example of different ways various cultures will showcase empathy include expressions of love. While some cultures showcase love outwardly through public displays of affection such as hand holding, kissing, or hugging, other cultures do not welcome these overt demonstrations of love in public ways.


3.5: Subtract Integers (Part 1)
- Model subtraction of integers
- Simplify expressions with integers
- Evaluate variable expressions with integers
- Translate words phrases to algebraic expressions
- Subtract integers in applications
Before you get started, take this readiness quiz.
- Simplify: \(12 − (8−1)\). If you missed this problem, review Example 2.1.8 .
- Translate the difference of \( 20\) and \(−15\) into an algebraic expression. If you missed this problem, review Example 1.3.11 .
- Add: \(−18 + 7\). If you missed this problem, review Example 3.2.6 .
Model Subtraction of Integers
Remember the story in the last section about the toddler and the cookies? Children learn how to subtract numbers through their everyday experiences. Real-life experiences serve as models for subtracting positive numbers, and in some cases, such as temperature, for adding negative as well as positive numbers. But it is difficult to relate subtracting negative numbers to common life experiences. Most people do not have an intuitive understanding of subtraction when negative numbers are involved. Math teachers use several different models to explain subtracting negative numbers.
We will continue to use counters to model subtraction. Remember, the blue counters represent positive numbers and the red counters represent negative numbers.
Perhaps when you were younger, you read \(5 − 3\) as five take away three . When we use counters, we can think of subtraction the same way.
We will model four subtraction facts using the numbers \(5\) and \(3\).
\[5 − 3 \qquad −5 − (−3) \qquad −5 − 3 \qquad 5 − (−3) \nonumber \]
Model: \(5 − 3\).
Solution
| Interpret the expression. | 5 − 3 means 5 take away 3. |
| Model the first number. Start with 5 positives. | |
| Take away the second number. So take away 3 positives. | |
| Find the counters that are left. |
The difference between \(5\) and \(3\) is \(2\).
Model the expression: \(6 − 4\)
- Answer
-
\(2\)
Model the expression: \(7 − 4\)
- Answer
-
\(3\)
Model: \(−5 − (−3)\).
Solution
| Interpret the expression. | −5 − (−3) means −5 take away −3. |
| Model the first number. Start with 5 negatives. | |
| Take away the second number. So take away 3 negatives. | |
| Find the number of counters that are left. |
The difference between \(−5\) and \(−3\) is \(−2\).
Model the expression: \(−6 − (−4)\)
- Answer
-
\(-2\)
Model the expression: \(−7 − (−4)\)
- Answer
-
\(-3\)
Notice that Example \(\PageIndex{1}\) and Example \(\PageIndex{2}\) are very much alike.
- First, we subtracted \(3\) positives from \(5\) positives to get \(2\) positives.
- Then we subtracted \(3\) negatives from \(5\) negatives to get \(2\) negatives.
Each example used counters of only one color, and the “take away” model of subtraction was easy to apply.
Figure \(\PageIndex{1}\)
Now let’s see what happens when we subtract one positive and one negative number. We will need to use both positive and negative counters and sometimes some neutral pairs, too. Adding a neutral pair does not change the value.
Model: \(−5 − 3\).
Solution
| Interpret the expression. | −5 − 3 means −5 take away 3. |
| Model the first number. Start with 5 negatives. | |
| Take away the second number. So we need to take away 3 positives. | |
| But there are no positives to take away. Add neutral pairs until you have 3 positives. | |
| Now take away 3 positives. | |
| Count the number of counters that are left. |
The difference of \(−5\) and \(3\) is \(−8\).
Model the expression: \(−6 − 4\)
- Answer
-
\(-10\)
Model the expression: \(−7 − 4\)
- Answer
-
\(-11\)
Model: \(5 − (−3)\).
Solution
| Interpret the expression. | 5 − (−3) means 5 take away −3. |
| Model the first number. Start with 5 positives. | |
| Take away the second number, so take away 3 negatives. | |
| But there are no negatives to take away. Add neutral pairs until you have 3 negatives. | |
| Then take away 3 negatives. | |
| Count the number of counters that are left. |
The difference of \(5\) and \(−3\) is \(8\).
Model the expression: \(6 − (−4)\)
- Answer
-
\(10\)
Model the expression: \(7 − (−4)\)
- Answer
-
\(11\)
Model each subtraction.
- \(8 − 2\)
- \(−5 − 4\)
- \(6 − (−6)\)
- \(−8 − (−3)\)
Solution
- \(8 − 2\): This means \(8\) take away \(2\).
| Start with 8 positives. | |
| Take away 2 positives. | |
| How many are left? | 8 − 2 = 6 |
- \(−5 − 4\): This means \(−5\) take away \(4\).
| Start with 5 negatives. | |
| You need to take away 4 positives. Add 4 neutral pairs to get 4 positives. | |
| Take away 4 positives. | |
| How many are left? | −5 − 4 = −9 |
- \(6 − (−6)\): This means \(6\) take away \(−6\).
| Start with 6 positives. | |
| Add 6 neutrals to get 6 negatives to take away. | |
| Remove 6 negatives. | |
| How many are left? | 6 − (−6) = 12 |
- \(−8 − (−3)\): This means \(−8\) take away \(−3\).
| Start with 8 negatives. | |
| Take away 3 negatives. | |
| How many are left? | −8 − (−3) = −5 |
Model each subtraction.
- \(7 - (-8)\)
- \(-7 - (-2)\)
- \(4 - 1\)
- \(-6 - 8\)
- Answer a
-
- Answer b
-
- Answer c
-
- Answer d
-
Model each subtraction.
- \(4 - (-6)\)
- \(-8 - (-1)\)
- \(7 - 3\)
- \(-4 - 2\)
- Answer a
-
- Answer b
-
- Answer c
-
- Answer d
-
Model each subtraction expression:
- \(2 − 8\)
- \(−3 − (−8)\)
Solution
| We start with 2 positives. | |
| We need to take away 8 positives, but we have only 2. | |
| Add neutral pairs until there are 8 positives to take away. | |
| Then take away eight positives. | |
| Find the number of counters that are left. There are 6 negatives. |
\(2 − 8 = −6\)
| We start with 3 negatives. | |
| We need to take away 8 negatives, but we have only 3. | |
| Add neutral pairs until there are 8 negatives to take away | |
| Then take away the 8 negatives. | |
| Find the number of counters that are left. There are 5 positives. |
\(−3 − (−8) = 5\)
Model each subtraction expression.
- \(7 − 9\)
- \(−5 − (−9)\)
- Answer a
-
\(-2\)
- Answer b
-
\(4\)
Model each subtraction expression.
- \(4 − 7\)
- \(−7 − (−10)\)
- Answer a
-
\(-3\)
- Answer b
-
\(3\)
Simplify Expressions with Integers
Do you see a pattern? Are you ready to subtract integers without counters? Let’s do two more subtractions. We’ll think about how we would model these with counters, but we won’t actually use the counters.
- Subtract \(−23 − 7\). Think: We start with \(23\) negative counters. We have to subtract \(7\) positives, but there are no positives to take away. So we add \(7\) neutral pairs to get the \(7\) positives. Now we take away the \(7\) positives. So what’s left? We have the original \(23\) negatives plus \(7\) more negatives from the neutral pair. The result is \(30\) negatives. \(−23 − 7 = −30\) Notice, that to subtract \(7\), we added \(7\) negatives.
- Subtract \(30 − (−12)\). Think: We start with \(30\) positives. We have to subtract \(12\) negatives, but there are no negatives to take away. So we add \(12\) neutral pairs to the \(30\) positives. Now we take away the \(12\) negatives. What’s left? We have the original \(30\) positives plus \(12\) more positives from the neutral pairs. The result is \(42\) positives. \(30 − (−12) = 42\) Notice that to subtract \(−12\), we added \(12\).
While we may not always use the counters, especially when we work with large numbers, practicing with them first gave us a concrete way to apply the concept, so that we can visualize and remember how to do the subtraction without the counters.
Have you noticed that subtraction of signed numbers can be done by adding the opposite? You will often see the idea, the Subtraction Property, written as follows:
\[a − b = a + (−b)\]
Look at these two examples.
Figure \(\PageIndex{2}\)
We see that \(6 − 4\) gives the same answer as \(6 + (−4)\).
Of course, when we have a subtraction problem that has only positive numbers, like the first example, we just do the subtraction. We already knew how to subtract \(6 − 4\) long ago. But knowing that \(6 − 4\) gives the same answer as \(6 + (−4)\) helps when we are subtracting negative numbers.
Simplify:
- \(13 − 8\) and \(13 + (−8)\)
- \(−17 − 9\) and \(−17 + (−9)\)
Solution
| Subtract to simplify. | 13 − 8 = 5 |
| Add to simplify. | 13 + (−8) = 5 |
| Subtracting 8 from 13 is the same as adding −8 to 13. |
| Subtract to simplify. | −17 − 9 = −26 |
| Add to simplify. | −17 + (−9) = −26 |
| Subtracting 9 from −17 is the same as adding −9 to −17. |
Simplify each expression:
- \(21 − 13\) and \(21 + (−13)\)
- \(−11 − 7\) and \(−11 + (−7)\)
- Answer a
-
\(8\), \(8\)
- Answer b
-
\(-18\), \(-18\)
Simplify each expression:
- \(15 − 7\) and \(15 + (−7)\)
- \(−14 − 8\) and \(−14 + (−8)\)
- Answer a
-
\(8\), \(8\)
- Answer b
-
\(-22\), \(-22\)
Now look what happens when we subtract a negative.
Figure \(\PageIndex{3}\)
We see that \(8 − (−5)\) gives the same result as \(8 + 5\). Subtracting a negative number is like adding a positive.
Simplify:
- \(9 − (−15)\) and \(9 + 15\)
- \(−7 − (−4)\) and \(−7 + 4\)
Solution
- \(9 − (−15)\) and \(9 + 15\)
| Subtract to simplify. | 9 − (−15) = 24 |
| Add to simplify. | 9 + 15 = 24 |
Subtracting \(−15\) from \(9\) is the same as adding \(15\) to \(9\).
- \(−7 − (−4)\) and \(−7 + 4\)
| Subtract to simplify. | −7 − (−4) = −3 |
| Add to simplify. | −7 + 4 = −3 |
Subtracting \(−4\) from \(−7\) is the same as adding \(4\) to \(−7\).
Simplify each expression:
- \(6 − (−13)\) and \(6 + 13\)
- \(−5 − (−1)\) and \(−5 + 1\)
- Answer a
-
\(19\), \(19\)
- Answer b
-
\(-4\), \(-4\)
Simplify each expression:
- \(4 − (−19)\) and \(4 + 19\)
- \(−4 − (−7)\) and \(−4 + 7\)
- Answer a
-
\(23\), \(23\)
- Answer b
-
\(3\), \(3\)
Look again at the results of Example \(\PageIndex{1}\) - Example \(\PageIndex{4}\).
| 5 – 3 | –5 – (–3) |
|---|---|
| 2 | –2 |
| 2 positives | 2 negatives |
| When there would be enough counters of the color to take away, subtract. | |
| –5 – 3 | 5 – (–3) |
| –8 | 8 |
| 5 negatives, want to subtract 3 positives | 5 positives, want to subtract 3 negatives |
| need neutral pairs | need neutral pairs |
| When there would not be enough of the counters to take away, add neutral pairs. |
Contributors and Attributions
- Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (formerly of Santa Ana College). This content produced by OpenStax and is licensed under a Creative Commons Attribution License 4.0 license.


5.5: Bohr’s Model of the Hydrogen Atom
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Learning Objectives
By the end of this section, you will be able to:
- Explain the difference between the absorption spectrum and the emission spectrum of radiation emitted by atoms
- Describe the Rutherford gold foil experiment and the discovery of the atomic nucleus
- Explain the atomic structure of hydrogen
- Describe the postulates of the early quantum theory for the hydrogen atom
- Summarize how Bohr’s quantum model of the hydrogen atom explains the radiation spectrum of atomic hydrogen
Historically, Bohr’s model of the hydrogen atom is the very first model of atomic structure that correctly explained the radiation spectra of atomic hydrogen. The model has a special place in the history of physics because it introduced an early quantum theory, which brought about new developments in scientific thought and later culminated in the development of quantum mechanics. To understand the specifics of Bohr’s model, we must first review the nineteenth-century discoveries that prompted its formulation.
When we use a prism to analyze white light coming from the sun, several dark lines in the solar spectrum are observed (Figure \(\PageIndex{1}\)). Solar absorption lines are called Fraunhofer lines after Joseph von Fraunhofer , who accurately measured their wavelengths. During 1854–1861, Gustav Kirchhoff and Robert Bunsen discovered that for the various chemical elements, the line emission spectrum of an element exactly matches its line absorption spectrum . The difference between the absorption spectrum and the emission spectrum is explained in Figure \(\PageIndex{2}\).
An absorption spectrum is observed when light passes through a gas. This spectrum appears as black lines that occur only at certain wavelengths on the background of the continuous spectrum of white light (Figure \(\PageIndex{2}\)). The missing wavelengths tell us which wavelengths of the radiation are absorbed by the gas. The emission spectrum is observed when light is emitted by a gas. This spectrum is seen as colorful lines on the black background (Figures \(\PageIndex{3}\) and \(\PageIndex{4}\)). Positions of the emission lines tell us which wavelengths of the radiation are emitted by the gas. Each chemical element has its own characteristic emission spectrum. For each element, the positions of its emission lines are exactly the same as the positions of its absorption lines. This means that atoms of a specific element absorb radiation only at specific wavelengths and radiation that does not have these wavelengths is not absorbed by the element at all. This also means that the radiation emitted by atoms of each element has exactly the same wavelengths as the radiation they absorb.
Emission spectra of the elements have complex structures; they become even more complex for elements with higher atomic numbers. The simplest spectrum, shown in Figure \(\PageIndex{4}\), belongs to the hydrogen atom. Only four lines are visible to the human eye. As you read from right to left in Figure \(\PageIndex{4}\), these lines are: red (656 nm), called the \(H-\alpha\) line; aqua (486 nm), blue (434 nm), and violet (410 nm). The lines with wavelengths shorter than 400 nm appear in the ultraviolet part of the spectrum (Figure \(\PageIndex{4}\), far left) and are invisible to the human eye. There are infinitely many invisible spectral lines in the series for hydrogen.
An empirical formula to describe the positions (wavelengths) λ of the hydrogen emission lines in this series was discovered in 1885 by Johann Balmer . It is known as the Balmer formula:
\[ \dfrac{1}{\lambda} = R_H \left( \dfrac{1}{2^2} - \dfrac{1}{n^2}\right). \label{balmer} \]
The constant \(R_H = 1.09737 \times 10^7 m^{-1}\) is called the Rydberg constant for hydrogen . In Equation \ref{balmer}, the positive integer n takes on values \(n = 3, 4,5,6\) for the four visible lines in this series. The series of emission lines given by the Balmer formula is called the Balmer series for hydrogen. Other emission lines of hydrogen that were discovered in the twentieth century are described by the Rydberg formula , which summarizes all of the experimental data:
\[ \dfrac{1}{\lambda} = R_H \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)\label{rydberg} \]
where \(n_i = n_f > n_i\) (in integer steps).
When \(n_f = 1\), the series of spectral lines is called the Lyman series . When \(n_f = 2\), the series is called the Balmer series, and in this case, the Rydberg formula coincides with the Balmer formula (Equation \ref{balmer}). When \(n_f = 3\), the series is called the Paschen series . When \(n_f = 4\), the series is called the Brackett series . When \(n_f = 5\), the series is called the Pfund series . When \(n_f = 6\), we have the Humphreys series . As you may guess, there are infinitely many such spectral bands in the spectrum of hydrogen because \(n_f\) can be any positive integer number.
The Rydberg formula for hydrogen gives the exact positions of the spectral lines as they are observed in a laboratory; however, at the beginning of the twentieth century, nobody could explain why it worked so well. The Rydberg formula remained unexplained until the first successful model of the hydrogen atom was proposed in 1913.
Example \(\PageIndex{1}\): Limits of the Balmer Series
Calculate the longest and the shortest wavelengths in the Balmer series.
Strategy
We can use either the Balmer formula (Equation \ref{balmer}) or the Rydberg formula (Equation \ref{rydberg}). The longest wavelength is obtained when \(1/n_i\)i is largest, which is when \(n_i = n_f + 1 = 3\), because \(n_f = 2\) for the Balmer series. The smallest wavelength is obtained when \(1/n_i\) is smallest, which is \(1/n_i \rightarrow 0\) when \(n_i \rightarrow \infty\).
Solution
The long-wave limit:
\[ \dfrac{1}{\lambda} = R_H \left( \dfrac{1}{2^2} - \dfrac{1}{3^2}\right) = (1.09737 \times 10^7) \dfrac{1}{m} \left( \dfrac{1}{4} - \dfrac{1}{9}\right) \Rightarrow \lambda = 656.3 \, nm \nonumber \]
The short-wave limit:
\[ \dfrac{1}{\lambda} = R_H \left( \dfrac{1}{2^2} - 0\right) = (1.09737 \times 10^7) \dfrac{1}{m} \left( \dfrac{1}{4} \right) \Rightarrow \lambda = 364.6 \, nm. \nonumber \]
Significance
Note that there are infinitely many spectral lines lying between these two limits.
Exercise \(\PageIndex{1}\)
What are the limits of the Lyman series? Can you see these spectral lines?
- Answer
-
121.5 nm and 91.1 nm; no, these spectral bands are in the ultraviolet
The key to unlocking the mystery of atomic spectra is in understanding atomic structure. Scientists have long known that matter is made of atoms. According to nineteenth-century science, atoms are the smallest indivisible quantities of matter. This scientific belief was shattered by a series of groundbreaking experiments that proved the existence of subatomic particles, such as electrons, protons, and neutrons.
The electron was discovered and identified as the smallest quantity of electric charge by J.J. Thomson in 1897 in his cathode ray experiments, also known as β-ray experiments: A β -ray is a beam of electrons. In 1904, Thomson proposed the first model of atomic structure, known as the “plum pudding” model, in which an atom consisted of an unknown positively charged matter with negative electrons embedded in it like plums in a pudding. Around 1900, E. Rutherford , and independently, Paul Ulrich Villard, classified all radiation known at that time as \(\alpha\)- rays , \(β\)- rays , and \(\gamma\)- rays (a \(\gamma\)-ray is a beam of highly energetic photons). In 1907, Rutherford and Thomas Royds used spectroscopy methods to show that positively charged particles of \(\alpha\)-radiation (called \(\alpha\)-particles) are in fact doubly ionized atoms of helium. In 1909, Rutherford, Ernest Marsden, and Hans Geiger used \(\alpha\)-particles in their famous scattering experiment that disproved Thomson’s model (see Linear Momentum and Collisions ).
In the Rutherford gold foil experiment (also known as the Geiger–Marsden experiment), α-particles were incident on a thin gold foil and were scattered by gold atoms inside the foil (see Types of Collisions ). The outgoing particles were detected by a 360° scintillation screen surrounding the gold target (for a detailed description of the experimental setup, see Linear Momentum and Collisions ). When a scattered particle struck the screen, a tiny flash of light (scintillation) was observed at that location. By counting the scintillations seen at various angles with respect to the direction of the incident beam, the scientists could determine what fraction of the incident particles were scattered and what fraction were not deflected at all. If the plum pudding model were correct, there would be no back-scattered α-particles. However, the results of the Rutherford experiment showed that, although a sizable fraction of α-particles emerged from the foil not scattered at all as though the foil were not in their way, a significant fraction of α-particles were back-scattered toward the source. This kind of result was possible only when most of the mass and the entire positive charge of the gold atom were concentrated in a tiny space inside the atom.
In 1911, Rutherford proposed a nuclear model of the atom . In Rutherford’s model, an atom contained a positively charged nucleus of negligible size, almost like a point, but included almost the entire mass of the atom. The atom also contained negative electrons that were located within the atom but relatively far away from the nucleus. Ten years later, Rutherford coined the name proton for the nucleus of hydrogen and the name neutron for a hypothetical electrically neutral particle that would mediate the binding of positive protons in the nucleus (the neutron was discovered in 1932 by James Chadwick ). Rutherford is credited with the discovery of the atomic nucleus; however, the Rutherford model of atomic structure does not explain the Rydberg formula for the hydrogen emission lines.
Bohr’s model of the hydrogen atom , proposed by Niels Bohr in 1913, was the first quantum model that correctly explained the hydrogen emission spectrum. Bohr’s model combines the classical mechanics of planetary motion with the quantum concept of photons. Once Rutherford had established the existence of the atomic nucleus, Bohr’s intuition that the negative electron in the hydrogen atom must revolve around the positive nucleus became a logical consequence of the inverse-square-distance law of electrostatic attraction. Recall that Coulomb’s law describing the attraction between two opposite charges has a similar form to Newton’s universal law of gravitation in the sense that the gravitational force and the electrostatic force are both decreasing as \(1/r^2\), where r is the separation distance between the bodies. In the same way as Earth revolves around the sun, the negative electron in the hydrogen atom can revolve around the positive nucleus. However, an accelerating charge radiates its energy. Classically, if the electron moved around the nucleus in a planetary fashion, it would be undergoing centripetal acceleration, and thus would be radiating energy that would cause it to spiral down into the nucleus. Such a planetary hydrogen atom would not be stable, which is contrary to what we know about ordinary hydrogen atoms that do not disintegrate. Moreover, the classical motion of the electron is not able to explain the discrete emission spectrum of hydrogen.
To circumvent these two difficulties, Bohr proposed the following three postulates of Bohr’s model :
- The negative electron moves around the positive nucleus (proton) in a circular orbit. All electron orbits are centered at the nucleus. Not all classically possible orbits are available to an electron bound to the nucleus.
- The allowed electron orbits satisfy the first quantization condition : In the n th orbit, the angular momentum \(L_n\) of the electron can take only discrete values: \[L_n = n\hbar, \, where \, n = 1,2,3, . . . \nonumber \] This postulate says that the electron’s angular momentum is quantized. Denoted by \(r_n\) and \(v_n\), respectively, the radius of the n th orbit and the electron’s speed in it, the first quantization condition can be expressed explicitly as \[m_ev_nr_n = n\hbar. \label{6.34} \]
- An electron is allowed to make transitions from one orbit where its energy is \(E_n\) to another orbit where its energy is \(E_m\). When an atom absorbs a photon, the electron makes a transition to a higher-energy orbit. When an atom emits a photon, the electron transits to a lower-energy orbit. Electron transitions with the simultaneous photon absorption or photon emission take place instantaneously . The allowed electron transitions satisfy the second quantization condition : \[hf = |E_n - E_m| \nonumber \] where \(hf\) is the energy of either an emitted or an absorbed photon with frequency \(f\). The second quantization condition states that an electron’s change in energy in the hydrogen atom is quantized.
These three postulates of the early quantum theory of the hydrogen atom allow us to derive not only the Rydberg formula, but also the value of the Rydberg constant and other important properties of the hydrogen atom such as its energy levels, its ionization energy, and the sizes of electron orbits. Note that in Bohr’s model, along with two nonclassical quantization postulates, we also have the classical description of the electron as a particle that is subjected to the Coulomb force, and its motion must obey Newton’s laws of motion. The hydrogen atom, as an isolated system, must obey the laws of conservation of energy and momentum in the way we know from classical physics. Having this theoretical framework in mind, we are ready to proceed with our analysis.
Electron Orbits
To obtain the size \(r_n\) of the electron’s n th orbit and the electron’s speed \(v_n\) in it, we turn to Newtonian mechanics. As a charged particle, the electron experiences an electrostatic pull toward the positively charged nucleus in the center of its circular orbit. This electrostatic pull is the centripetal force that causes the electron to move in a circle around the nucleus. Therefore, the magnitude of centripetal force is identified with the magnitude of the electrostatic force:
\[ \dfrac{m_ev_n^2}{r_n} = \dfrac{1}{4\pi \epsilon_0} \dfrac{e^2}{r_n^2}. \label{6.36} \]
Here, \(e\) denotes the value of the elementary charge. The negative electron and positive proton have the same value of charge,
\[|q| = e. \nonumber \]
When Equation \ref{6.36} is combined with the first quantization condition given by Equation \ref{6.34}, we can solve for the speed, \(v_n\), and for the radius, \(r_n\):
\[v_n = \dfrac{1}{4\pi \epsilon_0} \dfrac{e^2}{\hbar} \dfrac{1}{n} \label{6.37} \]
\[r_n = 4\pi \epsilon_0 \dfrac{\hbar^2}{m_ee^2}n^2. \label{6.38} \]
Note that these results tell us that the electron’s speed as well as the radius of its orbit depend only on the index n that enumerates the orbit because all other quantities in the preceding equations are fundamental constants. We see from Equation \ref{6.38} that the size of the orbit grows as the square of n . This means that the second orbit is four times as large as the first orbit, and the third orbit is nine times as large as the first orbit, and so on. We also see from Equation \ref{6.37} that the electron’s speed in the orbit decreases as the orbit size increases. The electron’s speed is largest in the first Bohr orbit, for \(n = 1\), which is the orbit closest to the nucleus. The radius of the first Bohr orbit is called the Bohr radius of hydrogen , denoted as \(a_0\). Its value is obtained by setting \(n = 1\) in Equation \ref{6.38}:
\[a_0 = 4\pi \epsilon_0 \dfrac{\hbar^2}{m_ee^2} = 5.29 \times 10^{-11} m = 0.529\space Å. \nonumber \]
We can substitute \(a_0\) in Equation \ref{6.38} to express the radius of the n th orbit in terms of \(a_0\):
\[r_n = a_0n^2. \label{6.40} \]
This result means that the electron orbits in hydrogen atom are quantized because the orbital radius takes on only specific values of \(a_0\), \(4a_0\), \(9a_0\), \(16a_0\)... given by Equation \ref{6.40}, and no other values are allowed.
Electron Energies
The total energy \(E_n\) of an electron in the n th orbit is the sum of its kinetic energy \(K_n\) and its electrostatic potential energy \(U_n\). Utilizing Equation \ref{6.37}, we find that
\[K_n = \dfrac{1}{2}m_ev_n^2 = \dfrac{1}{32\pi^2 \epsilon^2} \dfrac{m_ee^4}{\hbar^2} \dfrac{1}{n^2}. \label{6.41} \]
Recall that the electrostatic potential energy of interaction between two charges \(q_1\) and \(q_2\) that are separated by a distance \(r_{12}\) is \((1/4\pi \epsilon_0)q_1q_2/r_{12}\). Here, \(q_1 = +e\) is the charge of the nucleus in the hydrogen atom (the charge of the proton), \(q_2 = -e\) is the charge of the electron and \(r_{12} = r_n\) is the radius of the n- th orbit. Now we use Equation \ref{6.38} to find the potential energy of the electron:
\[U_n = - \dfrac{1}{4\pi \epsilon_0} \dfrac{e^2}{r_n} = - \dfrac{1}{16\pi^2 \epsilon_0^2} \dfrac{m_ee^4}{\hbar^2} \dfrac{1}{n^2}. \label{6.42} \]
The total energy of the electron is the sum of Equation \ref{6.41} and Equation \ref{6.42}:
\[E_n = K_n + U_n = - \dfrac{1}{32\pi^2 \epsilon_0^2} \dfrac{m_ee^4}{\hbar^2} \dfrac{1}{n^2}. \label{6.43} \]
Note that the energy depends only on the index n because the remaining symbols in Equation \ref{6.43} are physical constants. The value of the constant factor in Equation \ref{6.43} is
\[E_0 = \dfrac{1}{32\pi^2 \epsilon_0^2} \dfrac{m_ee^4}{\hbar^2} = \dfrac{1}{8\epsilon_0^2} \dfrac{m_ee^4}{h^2} = 2.17 \times 10^{-18} J = 13.6 \, eV. \label{6.44} \]
It is convenient to express the electron’s energy in the n th orbit in terms of this energy, as
\[E_n = -E_0 \dfrac{1}{n^2}. \label{6.45} \]
Now we can see that the electron energies in the hydrogen atom are quantized because they can have only discrete values of \(-E_0, \, -E_0/4, \, -E_0/9, \, -E_0/16, . . . \) given by Equation \ref{6.45}, and no other energy values are allowed. This set of allowed electron energies is called the energy spectrum of hydrogen (Figure \(\PageIndex{5}\)). The index n that enumerates energy levels in Bohr’s model is called the energy quantum number . We identify the energy of the electron inside the hydrogen atom with the energy of the hydrogen atom. Note that the smallest value of energy is obtained for \(n = 1\), so the hydrogen atom cannot have energy smaller than that. This smallest value of the electron energy in the hydrogen atom is called the ground state energy of the hydrogen atom and its value is
\[E_1 = −E_0 = −13.6 \, eV. \label{6.46} \]
The hydrogen atom may have other energies that are higher than the ground state. These higher energy states are known as excited energy states of a hydrogen atom.
There is only one ground state, but there are infinitely many excited states because there are infinitely many values of n in Equation \ref{6.45}. We say that the electron is in the “first exited state” when its energy is \(E_n\) (when \(n = 2\)), the second excited state when its energy is \(E_3\) (when \(n = 3\)) and, in general, in the n th exited state when its energy is \(E_n + 1\). There is no highest-of-all excited state; however, there is a limit to the sequence of excited states. If we keep increasing \(n\) in Equation \ref{6.45}, we find that the limit is \(- lim_{n \rightarrow \infty} \, E_0/n^2 = 0\). In this limit, the electron is no longer bound to the nucleus but becomes a free electron. An electron remains bound in the hydrogen atom as long as its energy is negative. An electron that orbits the nucleus in the first Bohr orbit, closest to the nucleus, is in the ground state, where its energy has the smallest value. In the ground state, the electron is most strongly bound to the nucleus and its energy is given by Equation \ref{6.46}. If we want to remove this electron from the atom, we must supply it with enough energy, \(E_{\infty}\), to at least balance out its ground state energy \(E_1\):
\[E_{\infty} + E_1 = 0 \Rightarrow E_{\infty} = - E_1 = - (- E_0) = E_0 = 13.6 \, eV. \label{6.47} \]
The energy that is needed to remove the electron from the atom is called the ionization energy . The ionization energy \(E_{\infty}\) that is needed to remove the electron from the first Bohr orbit is called the ionization limit of the hydrogen atom . The ionization limit in Equation \ref{6.47} that we obtain in Bohr’s model agrees with experimental value.
Spectral Emission Lines of Hydrogen
To obtain the wavelengths of the emitted radiation when an electron makes a transition from the n- th orbit to the m -th orbit, we use the second of Bohr’s quantization conditions and Equation \ref{6.45} for energies. The emission of energy from the atom can occur only when an electron makes a transition from an excited state to a lower-energy state. In the course of such a transition, the emitted photon carries away the difference of energies between the states involved in the transition. The transition cannot go in the other direction because the energy of a photon cannot be negative, which means that for emission we must have \(E_n > E_m\) and \(n > m\). Therefore, the third of Bohr’s postulates gives
\[\begin{align} hf &= |E_n - E_m| \nonumber \\[4pt] &= E_n - E_m \nonumber \\[4pt] &= -E_0 \dfrac{1}{n^2} + E_m \dfrac{1}{m^2} \nonumber \\[4pt] &= E_0 \left( \dfrac{1}{m^2} - \dfrac{1}{n^2} \right). \label{6.48} \end{align} \]
Now we express the photon’s energy in terms of its wavelength, \(hf = hc/\lambda\), and divide both sides of Equation \ref{6.48} by \(hc\). The result is
\[ \dfrac{1}{\lambda} = \dfrac{E_0}{hc} \left( \dfrac{1}{m^2} - \dfrac{1}{n^2} \right). \label{6.49} \]
The value of the constant in this equation is
\[\dfrac{E_0}{hc} = \dfrac{13.6 \, eV}{(4.136 \times 10^{-15} eV \cdot s)(2.997 \times 10^8 m/s)} = 1.097 \times 10^7 \dfrac{1}{m}. \label{6.50} \]
This value is exactly the Rydberg constant \(R_H\) in the Rydberg heuristic formula Equation \ref{rydberg}. In fact, Equation \ref{6.49} is identical to the Rydberg formula, because for a given m , we have \(n = m + 1, \, m + 2, . . .\). In this way, the Bohr quantum model of the hydrogen atom allows us to derive the experimental Rydberg constant from first principles and to express it in terms of fundamental constants. Transitions between the allowed electron orbits are illustrated in Figure \(\PageIndex{5}\).
We can repeat the same steps that led to Equation \ref{6.49} to obtain the wavelength of the absorbed radiation; this again gives Equation \ref{6.49} but this time for the positions of absorption lines in the absorption spectrum of hydrogen. The only difference is that for absorption, the quantum number \(m\) is the index of the orbit occupied by the electron before the transition (lower-energy orbit) and the quantum number \(n\) is the index of the orbit to which the electron makes the transition (higher-energy orbit). The difference between the electron energies in these two orbits is the energy of the absorbed photon.
Example \(\PageIndex{2}\): Size and Ionization Energy of the Hydrogen Atom in an Excited State
If a hydrogen atom in the ground state absorbs a 93.7-nm photon, corresponding to a transition line in the Lyman series, how does this affect the atom’s energy and size? How much energy is needed to ionize the atom when it is in this excited state? Give your answers in absolute units, and relative to the ground state.
Strategy
Before the absorption, the atom is in its ground state. This means that the electron transition takes place from the orbit \(m = 1\) to some higher n th orbit. First, we must determine nn for the absorbed wavelength \(\lambda = 93.7 \, nm\). Then, we can use Equation \ref{6.45} to find the energy \(E_n\) of the excited state and its ionization energy \(E_{\infty,n}\), and use Equation \ref{6.40} to find the radius \(r_n\) of the atom in the excited state. To estimate n , we use Equation \ref{6.49}.
Solution
Substitute \(m = 1\) and λ = 93.7 nm in Equation \ref{6.49} and solve for n. You should not expect to obtain a perfect integer answer because of rounding errors, but your answer will be close to an integer, and you can estimate n by taking the integral part of your answer:
\[ \begin{align*} \dfrac{1}{\lambda} &= R_H \left( \dfrac{1}{1^1} - \dfrac{1}{n^2}\right) \\[4pt] \Rightarrow n &= \dfrac{1}{\sqrt{1 - \dfrac{1}{\lambda R_H}}} \\[4pt] &= \dfrac{1}{\sqrt{1 - \dfrac{1}{(93.7 \times 10^{−9}m)(1.097 \times 10^7 m^{−1})}}} \\[4pt] &= 6.07 \\[4pt] \Rightarrow n &= 6. \end{align*} \]
The radius of the \(n = 6\) orbit is
\[r_n = a_0n^2= a_06^2 = 36a_0 = 36(0.529 \times 10^{−10} \, m) = 19.04 \times 10^{−10}\space m ≅ 19.0 \, Å. \nonumber \]
Thus, after absorbing the 93.7-nm photon, the size of the hydrogen atom in the excited \(n = 6\) state is 36 times larger than before the absorption, when the atom was in the ground state. The energy of the fifth excited state (\(n = 6\)) is:
\[E_n = - \dfrac{E_0}{n^2} = - \dfrac{E_0}{6^2} = - \dfrac{E_0}{36} = - \dfrac{13.6 \, eV}{36} ≅ − 0.378 \, eV. \nonumber \]
After absorbing the 93.7-nm photon, the energy of the hydrogen atom is larger than it was before the absorption. Ionization of the atom when it is in the fifth excited state (\(n = 6\)) requites 36 times less energy than is needed when the atom is in the ground state:
\[E_{\infty,6} = -E_6 = -(-0.378 \, eV) = 0.378 \, eV. \nonumber \]
Significance
We can analyze any spectral line in the spectrum of hydrogen in the same way. Thus, the experimental measurements of spectral lines provide us with information about the atomic structure of the hydrogen atom.
Exercise \(\PageIndex{2}\)
When an electron in a hydrogen atom is in the first excited state, what prediction does the Bohr model give about its orbital speed and kinetic energy? What is the magnitude of its orbital angular momentum?
- Answer
-
\(v_2 = 1.1 \times 10^6 m/s ≅0.0036 \, c;\)
\(L_2 = 2\hbar K_2 = 3.4 \, eV\)
Bohr’s model of the hydrogen atom also correctly predicts the spectra of some hydrogen-like ions. Hydrogen-like ions are atoms of elements with an atomic number Z larger than one (\(Z = 1\) for hydrogen) but with all electrons removed except one. For example, an electrically neutral helium atom has an atomic number \(Z = 2\). This means it has two electrons orbiting the nucleus with a charge of \(q = +Ze\). When one of the orbiting electrons is removed from the helium atom (we say, when the helium atom is singly ionized), what remains is a hydrogen-like atomic structure where the remaining electron orbits the nucleus with a charge of \(q = +Ze\). This type of situation is described by the Bohr model. Assuming that the charge of the nucleus is not \(+e\) but \(+Ze\), we can repeat all steps, beginning with Equation \ref{6.36}, to obtain the results for a hydrogen-like ion:
\[r_n = \dfrac{a_0}{Z}n^2 \label{6.51} \]
where \(a_0\) is the Bohr orbit of hydrogen, and
\[E_n = -Z^2E_0 \dfrac{1}{n^2} \label{6.52} \]
where \(E_0\) is the ionization limit of a hydrogen atom. These equations are good approximations as long as the atomic number Z is not too large.
The Bohr model is important because it was the first model to postulate the quantization of electron orbits in atoms. Thus, it represents an early quantum theory that gave a start to developing modern quantum theory. It introduced the concept of a quantum number to describe atomic states. The limitation of the early quantum theory is that it cannot describe atoms in which the number of electrons orbiting the nucleus is larger than one. The Bohr model of hydrogen is a semi-classical model because it combines the classical concept of electron orbits with the new concept of quantization. The remarkable success of this model prompted many physicists to seek an explanation for why such a model should work at all, and to seek an understanding of the physics behind the postulates of early quantum theory. This search brought about the onset of an entirely new concept of “matter waves.”


2.14: Treating Depression
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A variety of antidepressant medications and psychotherapies are used to treat depression. The most effective treatment for most people is a combination of medication and psychotherapy. Many of us are aware that medications are available to treat depressive disorders—we see the ads on television and in magazines. Up to 70 percent of people with depression can be treated effectively with medication.
Medications used to treat depressive disorders usually act on the neurotransmission pathway. For example, some medications affect the activity of certain neurotransmitters, such as serotonin or norepinephrine. Different depressive disorders require different medication therapies. For example, individuals who have bipolar disorder are often treated with a mood-stabilizing drug, such as lithium, during their manic phase and a combination of mood-stabilizer and antidepressant medications during their depressive phase.
Medications usually lead to relief from the symptoms of depression within six to eight weeks. If one drug doesn’t relieve symptoms, doctors can prescribe a different antidepressant drug. As with drugs to treat other mental illnesses, patients are monitored closely by their doctor for symptoms of depression and for side effects. Patients who continue to take their medication for at least six months after recovery from major depression are 70 percent less likely to experience a relapse. Psychotherapy helps patients learn more effective ways to deal with the problems in their lives. These therapies usually involve 6 to 20 weekly meetings. These treatment plans should be revised if there is no improvement of symptoms within three or four months.


3: Basic Editing
The LibreTexts libraries are powered by the Mindtouch system. An exhaustive set of references for the tools on this powerful system can be found online at https://success.mindtouch.com/Author/Editor . The pages below summarize a subset of these tools that are useful in day-to-day use of the LibreTexts project.
An Instructor's account allows editing capability on essentially all pages in the library
- The wiki infrastructure of the LibreTexts makes it difficult to break anything. So feel free to experiment, play around, and test things out.
- All edits are saved as revisions. You can always revert to any older revision via the “Revision history” button under “Options”
- All edits get sent to the LibreTexts development team for eventual review (not approval).
- You can sign up with your personal account to “Watch” page(s) if any edits occur (using the little star next to forward/back right below the page title).


2.4: Reading- Atomic Bonds
Chemical Reactions
Minerals form as a result of chemical reactions. Chemical reactions are driven mainly by the arrangement and rearrangement of electrons in atoms. In a mineral, the atoms are held together by chemical bonds, which derive from the electrons.
Electrons can be thought of as occupying energy levels, or shells, in an atom. The lowest-energy shell is closest to the nucleus. Each shell can accommodate only a limited number of electrons. The first shell can hold two electrons, the second and third shells can hold eight electrons, and the next two shells can hold eighteen electrons. Unless energy is added to an atom to excite it from its low-energy state, the electrons in the atom will occupy the lowest-energy shells available to them.
The Bohr model (see figure 1) was developed by Niels Bohrs in 1913. In this model, electrons exist within principal shells. An electron normally exists in the lowest energy shell available, which is the one closest to the nucleus. Energy from a photon of light can bump it up to a higher energy shell, but this situation is unstable, and the electron quickly decays back to the ground state.
Chemical Bonds
If atoms interact with other atoms, they can gain or lose electrons to the other atoms, or share electrons with other atoms. In an individual atom, the most stable arrangement is a full outer shell of electrons. Therefore, chemical reactions will occur, and chemical bonds will form that hold atoms together to each other, when atoms encounter other atoms and change their electron configurations toward more stable, lower-energy arrangements, which generally involves achieving full outer electron shells in the atoms.
This stable configuration—a full outer shell of electrons—is exemplified by the inert gases. In the periodic table the inert gases are the elements of group 18 or VIIIA, the last column on the right. Inert gases do not have to undergo any chemical reactions or form any chemical bonds with other atoms in order to have a full outer shell of electrons. The inert gases already have full outer shells of electrons. That is why they are chemically inert. Their electrons are as stable as can be arranged. For this reason, inert gases are extremely unlikely to undergo any chemical reactions and it is almost impossible for them to bond with any other atoms. Because they do not bond with any other atoms to form a liquid, a solid, a molecule, or a mineral, the inert gases consist of atoms that stay separate from each other, in the gas state.
Individual atoms of all the other chemical elements, when they are neutral atoms, do not have full outer shells of electrons like the inert gases do. Therefore, they do not have the most stable arrangement of electrons that they possibly can. That is why most chemical elements have a strong tendency to either gain or lose electrons, or to enter into other arrangements of their valence electrons, the electrons in their outer shell. Chemical reactions and chemical bonds are generally a result of electrons being rearranged within and among atoms to give the atoms full outer electron shells.
For an atom to lose or gain one electron takes less energy than to lose or gain two, which in turn takes less energy than to lose or gain a third electron. For an individual atom to gain or lose four electrons will only occur in extremely high-energy environments such as in a star. In common chemical reactions on earth, and in the formation of chemical bonds, no element will completely gain or lose four electrons. This limits the charges of atomic cations to +1, +2 or +3 and the charges of atomic anions to –1, –2, or –3.
Reading this far, you have learned about one group of elements in the periodic table, group 18, the inert gases. Another group of chemical elements in the periodic table is the alkali elements. The alkali elements compose group 1 or IA, the left hand column, including the elements sodium (Na) and potassium (K).
Hydrogen is not usually considered as an alkali element because, even though it is shown in group 1 in the periodic table. Hydrogen is so light and small, with just a single proton in its nucleus, that it has some unique behaviors and is considered in a class by itself.
The alkali elements have a single electron in their outer electron shell. If an alkali element loses a single electron, it becomes an ion with a +1 charge and a full outer shell. If an opportunity arises, alkali elements will easily turned into +1 cations.
Ionic bonds
Now look at group 17 or VIIA in the periodic table, which includes the chemical elements fluorine (F), chlorine (Cl) and so on. These are the halogen elements. If a halogen element gains a single electron, it becomes an ion with a–1 charge and a full outer electron shell. If an opportunity arises, halogen elements have a strong tendency to take in an extra electron and become -1 anions because by doing so they achieve a full outer shell of electrons, which is the most stable arrangement of electrons possible.
If sodium and chlorine atoms get together in the right conditions, such as in an evaporating solution of salt water, each sodium atoms will give up an electron to a chlorine atom. This turns the sodium atoms into sodium ions, Na + , and the chlorine atoms into chloride ions, Cl – . Opposite electrical charges attract, so the sodium ions and chloride ions will tend to stick together with each other, joined by what are called ionic bonds.
Not only will the sodium and chloride ions have a very strong tendency to join together with each other via ionic bonds, in most situations they will naturally arrange into a configuration where there is no wasted space and no wasted energy. This leads them to form the crystal lattice of the mineral halite. Halite is a mineral with the chemical formula NaCl, sodium chloride, in which the bonds between the atoms are all ionic bonds.
Look at the diagram of halite showing the sodium and chloride ions arranged into the crystal lattice. All the ionic bonds are at the same angle and the same distance, so they are all of equal strength. This is the lowest-energy arrangement of the ions, the most stable arrangement. If any of the ions were spaced located at different angles or at different distances, there would be extra energy available. This extra energy would drive the ions toward equal angles and distances from each other, until the extra energy is used up and the ions are arranged into their lowest energy state. That is why minerals form, as a natural way for atoms to arrange themselves into the lowest energy state currently available to them.
Covalent bonds
Some elements, such as carbon (C) and silicon (Si) have a half-full valence shell. (The valence shell is another name for the outer shell, where the most reactive electrons are.) If an element such as carbon were to gain 4 electrons or lose 4 electrons, it would have a full valence shell. However, it is very difficult for an atom to gain or lose four electrons—the energy barrier becomes too strong. Therefore, carbon and silicon, along with a few other elements, tend to form a different type of bond in which they share their outer electrons with other atoms, which in turn share their outer electrons with the carbon (or silicon) atom. The atoms all end up with a full outer shell of electrons, even though some or all of those electrons are being shared with neighboring atoms. This electron sharing keeps the atoms bonded together. This type of chemical bond is called a covalent bond.
It is not uncommon for covalent bonds to be relatively strong. An extreme example can be in diamond. Diamond is a mineral consisting of nothing but carbon atoms, so its chemical formula is simply C. Each carbon atom in the diamond crystal lattice is covalently bonded to—sharing its valence electrons with—four neighboring carbon atoms. A diamond crystal is held together by nothing but extremely strong covalent bonds in all directions, which makes diamond a very hard mineral, the hardest known.
Metallic bonds
Gold forms a naturally occurring mineral of more or less pure gold, Au, held together by another type of bond, the metallic bond. Metallic elements such as gold and copper, when they bond with other metallic elements, are sharing some of their electrons not just with adjacent atoms, but throughout the whole substance. That is why metallic substances such as copper, gold, and aluminum make such good electrical conductors, because it is so easy to get the “loose” electrons to respond through the whole extent of the metal.
Hydrogen bonds
Another type of chemical bond that occurs in some minerals is the hydrogen bond. Hydrogen bonds are caused by the positive and negative ends of polar molecules attracting each other strongly enough to hold each other in fixed positions. For example, water molecules can join together through hydrogen bonds to form the mineral known as ice. In a water molecule, H 2 O, each of the hydrogen atoms forms a covalent bond with the oxygen atom.
Contributors and Attributions
- Basics - Minerals. Authored by : Ralph L. Dawes and Cheryl D. Dawes. Provided by : Wenatchee Valley College. Located at : http://commons.wvc.edu/rdawes/G101OCL/Basics/minerals.html . Project : Geology 101 - Introduction to Physical Geology. License : CC BY: Attribution
- Atoms, Isotopes, Ions, and Molecules: The Building Blocks. Provided by : OpenStax. Located at<EMAIL_ADDRESS>. Project : Biology. License : CC BY: Attribution


9.10: Chapters Eight and Nine- Review Questions
Chapter s 8 – 9 : Review Questions
- Define each of the following terms: Incrementalism, Sunk Costs, and Cannibalization.
- In words, explain what is meant by Free Cash Flow.
-
Why is
FCF
important? Give two reasons.
-
- How do we use this model – for individual projects, for the entire corporation, or both? Explain.
-
- What options does the company have regarding how it may choose to utilize its Free Cash Flow?
-
-
Create a
Free Cash
Flow
template
and spread the forecasted numbers based on the following assumptions:
- Last year’s sales were $15.5 million and are expected to grow for the next two years at 15% per year, followed by three years of 8% growth.
- Cost of Goods Sold last year were $12.6 million and are expected to grow at a 7% rate per year indefinitely.
- Depreciation is $550,000 per year at a straight-line rate; in the fifth year, the building will have been fully depreciated. This company has no depreciable equipment.
- There is no amortization.
- Selling and General Administrative expenses last year were $200,000 and will grow modestly at an annual 2% rate.
- This company is in the 30% tax bracket, including Federal and State. There are no local taxes.
- The company expects to spend $2 million each year on “ CapEx , ” all of which will be necessary.
- Last year’s Current Assets, excluding Cash, were $2.5 million, and is expected to grow at a 5% rate per year indefinitely.
- Last year’s Current Liabilities were $2 million and are expected to grow at a 3% rate for at least five years.
- How does the analyst handle depreciation in the FCF Model? Why does s/he handle it that way? Note that depreciation occurs twice in the formula.
- Can you list all four capital items , which are included in the Balance Sheet?
- Why don’t we include capital costs in the FCF Model?
- An increase in Current Assets provides for/uses funds. Which is it? Why?
- On what basis do we distinguish between “internal” and “external” funds?
- List some internal and external funds.
-
Calculate the
External Funds Needed
formula for the LCM Company (below), based on the following assumptions.
- Last year’s sales were $5,000 million.
- Next year’s objective is to increase sales by 30%.
- Variable costs will be 70% of sales . (Variable costs change with sales volume.)
- Fixed Costs are expected to run 30% of P, P, & E (Fixed costs do not change and are unrelated to sales volume.)
- Interest Expense is 5% of Notes Payable and 7% of Long-term debt.
- Taxes are 40%.
- The company expects to maintain its Payout Ratio at 20% of income.
What is the company’s EFN if it is to meet its growth objective?
- What do this year’s three Solvency ratios look like?
- Why w ill the company’s financial ratios change next year ?
Selected Answers To Chapters 8 and 9
FCF Table
- You are ill-advised to do this by XL . Do it by hand. Place it in a Word table.
- One needs to figure EBIT by adding in the Income Statement data to the template in the chapter.
- “Last Year’s” numbers are not illustrated in this Spread Sheet .
- “Year 1’s” numbers follow “Last Year’s.” For example, Last Year’s Sales were $15.5 Million. “This Year’s” sales increased by 15%. Therefore: (15.5) (1. 15 ) = $17.825.
- Be careful about the Current Assets and Current Liabilities numbers. We first calculate increases or decreases, not the gross numbers. Which data add to FCF ?
($ Millions)
EFN Formula
-
- D/E = (900 + 475) / 1,415 = 0.97
- D / TA = (900 + 475) / (900 + 475 + 1,415) = 0.4828
- TIE = EBIT / I = 1.050 / 0.068 = 15.44x
- EFN = [(A 0 /S 0 ) Δ S ] – [ (AP 0 /S 0 ) Δ S ] – [ (M 0 ) (S 1 ) ( RR 0 )]
- We will assume “Static Analysis.”
($ Millions)
- Sales and VC will change next year, but FC will remain the same. Therefore, we should see a change in the net profit margin, and changes in the dividend paid and retention rates – assuming no change in payout (percent of earnings) policy.