6.1: Ethical Practice Introduction
Learning Objectives
- Recognize ethical dilemmas and take appropriate action
- Practice in a manner consistent with a code of ethics for nurses
- Compare and contrast theories of ethical decision-making
- Examine resources to resolve ethical dilemmas
- Examine competent practice within the legal/ethical/regulatory framework of health care
- Defend the course of action or give a rationale for the action
- Apply the ANA Code of Ethics to diverse situations in health care
- Analyze the impact of cultural diversity in ethical decision-making
- Identify advocacy as part of the nursing role when responding to ethical dilemmas
The nursing profession is guided by a code of ethics. As you practice nursing, how will you determine “right” from “wrong” actions? What is the difference between morality, values, and ethical principles? What additional considerations impact your ethical decision-making? What are ethical dilemmas and how should nurses participate in resolving them? This chapter answers these questions by reviewing concepts related to ethical nursing practice and describing how nurses can resolve ethical dilemmas. By the end of this chapter, you will be able to describe how to make ethical decisions using the Code of Ethics established by the American Nurses Association.


2.5: Measuring Height on Irregular Trees
Species considerations: It is quite easy to measure tree height on conifers because conifers have a very distinct top. Each year’s whorl of growth produces a clear tip with short lateral branches, even on hemlocks ( Tsuga spp.)(Figure 2.8).
Hardwoods on the other hand, have rounded crowns that are often a function of the amount of sun they are able to capture (Figure 2.9). Under shaded conditions, they may be very one-sided. On hardwoods, it is extremely important to get a clear view of the whole crown, so that side branches are not mistaken as the top.
Broken tops
Trees in which the top has blown out can be tricky. The short or nonexistent tip is often hidden by long lateral branches near the top of the tree. If the top cannot be seen clearly, it is easy to mistake the tips of the lateral branches for the top. A rounded or flat top in a conifer suggests a missing top, and this type of tree should always be examined closely (Figure 2.10). As we saw in Figure 2.6, measuring a lateral branch instead of the tip can overestimate the tree’s height. The closer one is to the tree, the greater the error. This is another reason why it is important to walk a distance far enough from the tree to get a clear view of the top.
When measuring total height on trees with broken tops, the tree top must be “reconstructed,” in order to maintain the tree’s correct taper, or “original” shape. Incorrect taper will affect wood volume estimates. A normal tree that is 124’ tall (A below) has a very different shape than a tree whose top has broken out at 124’ (B below). The standard method for reconstructing a tree’s top is to look at the surrounding trees and estimate the broken tree’s missing height from their growth. Let’s say a tree similar in diameter and taper to the broken-top tree below (C below), runs 20 feet from a diameter of five inches to its tip (Figure 2.11). Using this as a guide, one could add 20 feet to the broken-top tree for a reconstructed total height of 144 feet.
Leaning trees
For a leaning tree, we have to adjust our image of the tree-to-ground triangle. In this case, the leaning tree is the hypotenuse of the triangle instead of the rise (Figure 2.12). The height of the tree can be estimated using the Pythagorean Theorem and the following steps:
- Measure out a horizontal distance from the tip of the tree until it is clearly in view.
- Calculate the perpendicular distance from the tip of the tree to the ground (rise), using %slope readings as before.
- Measure the horizontal distance from the perpendicular drop to the base of the tree (the run).
- Once these two sides of the triangle have been determined, estimate the total tree height using the Pythagorean Theorem to solve for the hypotenuse.
See the example in Figure 2.12 below:
1. The technician walks out a horizontal distance from fall line AB (in this case 100 ft.).
2. A %slope reading is taken to the tip of the tree (A, +102%), and then to the point on the ground where the AB fall line intersects the ground (B, (-14%). Using the two %slope readings, the rise of the triangle is determined; in this case, 116 feet.
3. The horizontal distance between Point B and the stump of the tree (C) is measured with a tape to determine the run; in this case 42 feet.
4. Finally, using the Pythagorean Theorem, the hypotenuse or height of the tree can be determined.
or
so
and
c = 123 feet.
Forked Trees
On forked trees, the tallest, or dominant fork is measured (Figure 2.13). In some cases, the second fork occurs low enough in the tree to be counted as a second tree, but for most trees, the tallest fork is the only merchantable fork.


9.2: Newton’s Universal Theory of Gravity
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Newton supposedly gained insight into the gravitational force by observing an apple falling from a tree and concluding that if it is the same force that makes apples fall at sea level and at the top of a mountain, perhaps that force can be exerted all the way up to the moon. It is rather remarkable that Newton was able to make the connection between falling apples and the motion of the moon around the Earth to find a single theory to describe both situations.
We should be clear that the theory of gravity is a different theory than Newton’s “Laws of Motion” (Newton’s Three Laws). The Laws of Motion introduce the concepts of force and inertial mass, and prescribe how to use those concepts in order to model motion using kinematics. Newton’s Universal Theory of Gravity is a theory that describes the force of gravity that two bodies with (gravitational) mass exert on each other.
Newton’s Universal Theory of Gravity states that if two bodies, with masses \(M_1\) and \(M_2\) , located at positions \(\vec r_1\) and \(\vec r_2\) , respectively, are separated by a distance, \(r\) , then \(M_2\) will exert an attractive force on \(M_1\) , \(\vec F_{12}\) , given by:
\[\begin{aligned} \vec F_{12}=-G\frac{M_1M_2}{r^2}\hat r_{21}\end{aligned}\]
where \(\hat r_{21}\) is the unit vector from \(M_2\) to \(M_1\) :
\[\begin{aligned} \vec r_{21} &= \vec r_2 - \vec r_1\\[4pt] \hat r_{21} &= \frac{1}{r} \vec r_{21}\end{aligned}\]
as shown in Figure \(\PageIndex{1}\) . \(\vec F_{12}\) should be read as “the force on body 1 from body 2”. \(G=6.67\times 10^{-11}\text{Nm}^{2}/\text{kg}^{2}\) is Newton’s Universal Constant of Gravity. It should be noted that Newton’s theory for the force of gravity written in this form only applies to either point masses (separated by a distance \(r\) ) or spherical bodies whose centers are separated by a distance \(r\) that is larger than the radius of either sphere.
Originally, Newton argued that the force of gravity would be proportional to the masses of the bodies, and inversely proportional to the square of the distance between them:
\[\begin{aligned} F_{12}\propto \frac{M_1M_2}{r^2}\end{aligned}\]
and \(G\) is simply the constant of proportionality.
Because of Newton’s Third Law, body 1 exerts a force on body 2 that is equal in magnitude but opposite in direction:
\[\begin{aligned} \vec F_{12} = -\vec F_{21}\end{aligned}\]
Example \(\PageIndex{1}\)
Calculate the magnitude of the force of gravity between yourself and a person standing \(50\text{cm}\) from you and compare that to your weight at the surface of the Earth (the force of gravity exerted by the Earth on you).
Solution
If we assume that the two people have a mass of \(50\text{kg}\) , the force of gravity exerted by one on the other, if they are separated by \(50\text{cm}\) , is given by:
\[\begin{aligned} F=G\frac{M_1M_2}{r^2}=(6.67\times 10^{-11}\text{Nm}^{2}/\text{kg}^{2})\frac{(50\text{kg})(50\text{kg})}{(0.5\text{m})^2}=6.67\times 10^{-7}\text{N}\end{aligned}\]
This is a very small force, compared to their weight, \(F_g\) :
\[\begin{aligned} F_g=M_1g=(50\text{kg})(9.8\text{N/kg})=490\text{N}\end{aligned}\]
which is approximately 700 million times bigger.
Discussion
The force of gravity is a very weak force when compared to other forces in Nature, such as the electric force between two charges. Newton’s Universal Constant of Gravity is very small, so the force of gravity between two objects is very small unless either of the masses involved are very large, or the distance between them is very small. In general, when modeling the motion of objects on the Earth, it is generally safe to ignore the forces of gravity between objects and only include their weight (the force of gravity from the Earth).
Exercise \(\PageIndex{4}\)
The radius of the Earth is \(6371\) km. What is the order of magnitude of the Earth’s mass?
- \(10^{24}\text{kg}\)
- \(10^{18}\text{kg}\)
- \(10^{19}\text{kg}\)
- \(10^{21}\text{kg}\)
Not enough information.
- Answer
Example \(\PageIndex{2}\)
Determine the constant in Kepler’s Third Law for planets orbiting the Sun, namely the value of the ratio:
\[\begin{aligned} \frac{s^3}{T^2}\end{aligned}\]
where \(s\) is the semi-major axis and \(T\) is the orbital period.
Solution
Since Kepler’s Third Law holds for any body orbiting the Sun, we can easily determine the ratio by considering a planet of mass \(m\) in a circular orbit of radius \(R\) centerd about the Sun. The semi-major axis of the orbit is equal to the radius of the orbit for a circular orbit ( \(s=R\) ).
If the planet is in a circular orbit about the Sun, its speed, \(v\) , will be constant, by Kepler’s Second Law, and it will thus be executing uniform circular motion. The only force exerted on the planet is the force of gravity exerted by the Sun. Thus the force of gravity must be equal to the mass of the planet times its radial (centripetal) acceleration, \(a_R\) , which is given by:
\[\begin{aligned} a_R=\frac{v^2}{R}\end{aligned}\]
Newton’s Second Law for the planet can be written as:
\[\begin{aligned} \sum F = F_g &= ma_R\\[4pt] G\frac{Mm}{R^2}&=m\frac{v^2}{R}\\[4pt] G\frac{M}{R}&=v^2\end{aligned}\]
where \(M\) is the mass of the Sun. The speed of the planet is given by the circumference of the orbit divided by the orbital period \(T\) , since it is constant:
\[\begin{aligned} v=\frac{2\pi R}{T}\end{aligned}\]
Re-arranging the equation from Newton’s Second Law:
\[\begin{aligned} G\frac{M}{R}&=v^2\\[4pt] G\frac{M}{R}&=\frac{4\pi^2 R^2}{T^2}\\[4pt] \therefore \frac{R^3}{T^2}&=G\frac{M}{4\pi^2}\end{aligned}\]
Thus, we find that the ratio of the cube of the orbital radius to the period squared is a constant that depends only on the mass of the Sun, which will then be the same for all planets (as it does not depend on, say, the mass of the planet that we considered).
Discussion
The relation above can, for example, be used to determine the mass of the Sun, since we can use geometry to determine the semi-major axis for the orbit of a planet, as Kepler did with the data from Tycho Brahe.
Example \(\PageIndex{3}\)
The acceleration due to Earth’s gravity depends on the force that the Earth exerts on an object. Using the Earth’s mass and radius, determine the acceleration of falling objects due to Earth’s gravity at the surface of the Earth. Also, determine the altitude where the acceleration due to Earth’s gravity is half of that at the Earth’s surface.
Solution
We can find the acceleration due to Earth’s gravity by determining the acceleration of a mass \(m\) upon which gravity is the only acting force. In other words, we model an object that is in free-fall a distance \(R\) away from the center of the Earth. Newton’s Second Law can be used in one dimension (corresponding to the direction that connects the falling mass to the center of the Earth):
\[\begin{aligned} \sum F &= G\frac{Mm}{R^2}=ma\\[4pt] \therefore a&=G\frac{M}{R^2}\end{aligned}\]
where \(M=5.97\times 10^{24}\text{kg}\) is the mass of the Earth. At the surface of the Earth, \(R=R_\oplus=6.371\times 10^{6}\text{m}\) :
\[\begin{aligned} a&=G\frac{M}{R_\oplus^2}=(6.67\times 10^{-11}\text{Nm}^{2}/\text{kg}^{2})\frac{(5.97\times 10^{24}\text{kg})}{(6.371\times 10^{6}\text{m})^2}\\[4pt] &=9.81\text{m/s}^{2}\end{aligned}\]
which, of course, is the value of \(g\) that we have been using so far for the acceleration due to gravity near the surface of the Earth. To find the altitude at which this is reduced by half, we first find the value of \(R\) that corresponds to this acceleration:
\[\begin{aligned} \frac{1}{2}G\frac{M}{R_\oplus^2}&=G\frac{M}{R^2}\\[4pt] \therefore R &=\sqrt{2}R_\oplus = 9.0\times 10^{6}\text{m}\end{aligned}\]
which corresponds to an altitude of \(h=R-R_\oplus=2640\text{km}\) , well above the Earth’s atmosphere.
Discussion
The acceleration of falling objects decreases as one moves further from the center of the Earth. It is thus an approximation to assume that \(g\) is a constant, although in most cases this is a very good approximation. In practice, the value of \(g\) will depend both on the distance from the center of the Earth and the composition (density) of the material in the Earth’s crust below where \(g\) is being measured. Precise measurements of \(g\) have bee used to determine the composition of the Earth’s crust and to search for mineral and oil deposits.
Weight and apparent weight
You have probably seen images of astronauts floating around the International Space Station (ISS) or other orbiting vessels, and heard of the term “weightlessness” to describe their motion. The ISS is in orbit at an altitude of approximately \(400\text{km}\) , where the force of Earth’s gravity is far from negligible (in Example 9.2.3 we showed that one needs to go to an altitude of \(2640\text{km}\) for the force to be reduced by half of that at the surface of the Earth). The contradiction between being weightless and the fact that weight is not zero is resolved by understanding that the popular term “weightless” is imprecise from a physics perspective.
The correct term to use from a physics perspective is to say that the apparent weight of the astronauts is zero when they are floating around. Weight is the magnitude of the force of gravity exerted by the Earth. Apparent weight is, for example, the force that one measures when standing on a spring scale, which is equal to the normal force exerted by the spring scale on the person. Apparent weight could also be determined by the tension in a string from which a person is suspended. The apparent weight is the sum of the forces exerted on a person minus the force of gravity. If gravity is the only force exerted on a person (or object), that person’s apparent weight is zero, which is what is popularly called being weightless.
One way to feel weightless is when you are in free-fall (e.g. the first few seconds of a parachute jump from an airplane). One can think of being in orbit as continuously falling towards the center of the Earth, but with an initial velocity in a direction such that you never actually collide with the Earth. The feeling of weightlessness will occur any time that the only force exerted on you is the force of gravity. If you are in a spacecraft in any orbit and the only force on the spacecraft is from gravity (i.e. no rockets or wings are exerting any forces), then everything in the spacecraft will have the same acceleration, since gravity is the only force acting on anything in the spacecraft, and it will appear that everything is just floating. To an outside observer, it would be clear that the spacecraft and its contents are all accelerating.
Effects of Earth’s rotation
Earth’s rotation affects the apparent weight of objects near the Earth’s surface. Consider a person standing on a spring scale at the North pole (top free-body diagram in Figure \(\PageIndex{2}\) ). The only two forces exerted on the person are their weight, \(\vec F_g\) , and the normal force, \(\vec N\) , exerted by the spring scale. Since the person is not accelerating, the normal force and the weight have the same magnitude and opposite directions. The scale will thus read the actual weight of the person 1 .
Consider, instead, a person standing on a spring scale at the equator (Figure \(\PageIndex{2}\)). That person is accelerating because they are undergoing uniform circular motion as they rotate along with the Earth. Again, the only forces acting on the person are their weight and the normal force exerted by the scale. The sum of the forces must now be equal to the person’s mass, \(m\) , times the radial acceleration, \(a_r\) , that is necessary for them to follow the surface of the Earth as the Earth rotates about its axis. Newton’s Second Law allows us to find the magnitude of the normal force acting on the person:
\[\begin{aligned} \sum F &= F_g-N=ma_r=m\frac{v^2}{R}\\[4pt] \therefore N &= F_g - m\frac{v^2}{R}\\[4pt] &=G\frac{Mm}{R^2} - m\frac{v^2}{R}\\[4pt] &=m\left(G\frac{M}{R^2} - \frac{v^2}{R} \right)\\[4pt] &=m\left(g - \frac{v^2}{R} \right)\end{aligned}\]
where \(M\) is the mass of the Earth, \(R\) is the radius of the Earth, and \(v\) is the speed at the surface of the Earth due to the Earth’s rotation. In the last line, we used the result from Example 9.2.3 where we determined the value of \(g\) in terms of the mass and radius of the Earth.
We see that the normal force is reduced compared to what it would be if the Earth were not rotating ( \(v=0\) ) or if one is standing at one of the poles. Your apparent weight, which you can measure by standing on a spring scale, is thus smaller at the equator than it is at the poles. The quantity in parenthesis can be thought of as a modified or “effective” value of \(g\) at the equator.
Exercise \(\PageIndex{5}\)
What is the effective value of \(g\) at the equator?
- \(9.80 \text{m/s}^2\)
- \(9.78 \text{m/s}^2\)
- \(9.70 \text{m/s}^2\)
- \(9.51 \text{m/s}^2\)
- Answer
If you are circling the Earth a distance \(R\) from the center of the Earth at a constant speed \(v\) , it is possible for your apparent weight to be zero. Imagine standing on a scale in an aircraft that is circling the Earth and measuring your apparent weight with the spring scale. As the speed of the aircraft increases, your apparent weight, \(N\) , decreases according to the formula that we just found:
\[\begin{aligned} N=m\left(G\frac{M}{R^2} - \frac{v^2}{R} \right)\end{aligned}\]
At a certain speed, \(v\) , your apparent weight is zero and you feel weightless:
\[\begin{aligned} G\frac{M}{R^2} &= \frac{v^2}{R}\\[4pt] \therefore v&= \sqrt{G\frac{M}{R} }\end{aligned}\]
This speed corresponds to a centripetal acceleration that is exactly equal to that due to gravity. This makes sense, since gravity is the only force that is acting on you (the normal force is now zero), which is exactly what we call being in orbit.
Earth’s rotation has some interesting consequences for stationary objects. At any position on Earth that is not at the equator or the poles, the sum of the forces on any stationary object (meaning stationary relative to the Earth’s surface) cannot be zero. This is because the object must rotate along with the Earth, so the net force on the object must point toward the center of the circle about which that location on Earth is rotating.
Take, for example, a plumb line, which consists of a mass hanging from a string. The two forces acting on the mass are gravity and tension. The force of gravity must point towards the center of the Earth. We would expect that the force of tension, and therefore the string, would point directly away from the center of the Earth. However, we find that if the plumb line is located at some angle \(\theta\) from the equator (but not at the equator or poles), as in Figure \(\PageIndex{3}\), then the string will point slightly away from the center of the Earth. In order for the mass to remain stationary relative to the ground, it must rotate along with the Earth (radius \(R\) ) around a circle of radius \(R\cos\theta\) . Thus, the tension from the string cannot point away from the center of the Earth, because the net force must point towards the center of the circle of radius \(R\cos\theta\) .
Exercise \(\PageIndex{6}\)
You cut the string of the plumb line. Where does the mass land relative to its initial latitude (the angle \(\theta\) in Figure \(\PageIndex{3}\))?
- At the same latitude.
- Closer to the nearest pole.
- Closer to the equator.
- Answer
The gravitational field
The gravitational force exerted on a mass \(m\) by a mass \(M\) can be written as:
\[\begin{aligned} \vec F(\vec r) = -G\frac{Mm}{r^2}\hat r\end{aligned}\]
if we define a coordinate system with the origin located at the center of mass \(M\) so that \(\vec r\) is the position of \(m\) relative to \(M\) . We can define the “gravitational field”, \(\vec g(\vec r)\) , at position, \(\vec r\) , due to the presence of mass \(M\) as the gravitational force per unit mass exerted by \(M\) :
\[\vec g(\vec r)=\frac{\vec F(\vec r)}{m}=-G\frac{Mm}{r^{2}}\hat r\]
The word “field” is just a mathematical term for a function that depends on position. Since \(\vec g(\vec r)\) is a vector, we would refer to it as a “vector field”.
Defining the gravitational field makes it easy to calculate the force of gravity from \(M\) on any mass \(m\) :
\[\begin{aligned} \vec F_g = m\vec g(\vec r)\end{aligned}\]
At the surface of the Earth, the magnitude of the gravitational field is given by:
\[\begin{aligned} g(R_\oplus)=\frac{GM}{R_\oplus^2}=9.81\text{N/kg}\end{aligned}\]
where \(R_\oplus\) is the radius of the Earth. Of course, this also corresponds to the acceleration of objects in free-fall near the surface of the Earth, which we can find from Newton’s Second Law:
\[\begin{aligned} \sum \vec F &= \vec F_g = m\vec a\\[4pt] m\vec g(R_\oplus)&= m\vec a\\[4pt] \therefore \vec a &= \vec g(R_\oplus)\end{aligned}\]
but we see here why it more precise to refer to \(g\) as the “magnitude of the gravitational field at the surface of the Earth” rather than “the acceleration due to Earth’s gravity”. It is also worth noting that the two are only equal if the gravitational mass (on the left of the equation in the second line) is the same as the inertial mass (on the right of the equation). The gravitational mass is the mass that appears in the gravitational force defined by Newton, whereas the inertial mass is the mass that appears with the acceleration in Newton’s Second Law.
Exercise \(\PageIndex{7}\)
Two small objects with different masses, \(m_1\) and \(m_2\) , are located a distance \(r\) from a nearby star. What can you say about the magnitude of the gravitational field and the magnitude of the gravitational force on \(m_1\) and \(m_2\) ?
- The field is different and the forces are different.
- The field is different but the forces are the same.
- The field is the same but the forces are different.
- The field is the same and the forces are the same.
- Answer
Suppose that there are two large mass bodies, \(M_1\) and \(M_2\) , and a smaller mass body, \(m\) . We can calculate the net gravitational force on \(m\) by summing the gravitational force vectors from \(M_1\) and \(M_2\) separately. If the gravitational fields from \(M_1\) and \(M_2\) are given by \(\vec g_1(\vec r)\) and \(\vec g_2(\vec r)\) , respectively, then the total gravitational force on \(m\) is given by:
\[\begin{aligned} \vec F &= m\vec g_1(\vec r) + m\vec g_2(\vec r)=m(\vec g_1(\vec r)+\vec g_2(\vec r))\\[4pt] &=m \vec g(\vec r)\end{aligned}\]
where we have introduced the total gravitational field:
\[\begin{aligned} \vec g(\vec r) = \vec g_1(\vec r)+\vec g_2(\vec r)\end{aligned}\]
In other words, if there are multiple bodies that result in a non-negligible force of gravity, we can calculate their gravitational fields independently and sum them together to define a net gravitational field, \(\vec g(\vec r)\) , that models the net force of gravity from all of the bodies. The net gravitational force on a new body of mass \(m'\) is then simply given by \(m'\vec g(\vec r)\) , and we do not need to add any more vectors together. For example, when calculating the motion of satellites that can be influenced by the force of gravity from both the Earth and the Moon, we simply need to calculate the net gravitational field from the Earth and Moon, and the motion of any satellite can then be modeled using that net gravitational field.
Exercise \(\PageIndex{8}\)
T here are two planets with equal mass located a distance \(d\) apart. Position \(A\) is located midway between the two planets. Position \(B\) is located a distance \(d/2\) from one of the planets, in the position shown in Figure \(\PageIndex{4}\). How does the field at \(A\) compare to the field at \(B\) ?
- The magnitude of the field is the same at \(A\) and \(B\) .
- The magnitude of the field is greater at \(A\) than at \(B\) .
- The magnitude of the field is greater at \(B\) than at \(A\) .
- Answer
Gauss’ Law
Newton’s Universal Theory of Gravity postulates that the force of gravity between two bodies decreases as the squared of the distance between those two bodies. Using the terminology of a field, we would say that the strength of the gravitational field from an object decreases as the inverse of the square of the distance to that object. This is an example of a what we generally call an “inverse-square law”. The electric force between two charges is also given by an inverse-square law, and we now understand that these forces behave as if they were “transmitted” by waves or particles.
If a force is given by an inverse-square law, then Gauss’ Law gives a way to determine the strength of the field that is associated with that force. In the case of gravity, Gauss’ Law states that:
\[\begin{aligned} \oint \vec g(\vec r) \cdot d\vec A = 4\pi G M^{enc}\end{aligned}\]
where the integral on the left is an integral over a “closed surface” that completely surrounds the mass for which we want to determine the gravitational field. To evaluate the integral, imagine taking a closed surface, \(S\) , that surrounds the mass. The vector \(d\vec A\) is defined as the vector that goes with a small element of that surface and points outwards from the closed surface. The magnitude of the vector is equal to the area of that small surface ( \(dA\) ), as illustrated in Figure \(\PageIndex{5}\). You can then take the scalar product of \(d\vec A\) with the gravitational field, \(\vec g(\vec r)\) , at that point on the surface. If you sum all of those scalar products, you get the value of the integral on the left. Gauss’ Law states that the value of that integral is equal \(4 \pi G\) times the total mass that is enclosed by the surface.
Olivia's Thoughts
If you want to know if a surface is closed, ask yourself if you could put water inside the surface and not be worried about it spilling out. For example, if you put water in a sphere or a cube , the water would not spill out even if you shook it around, so they are closed surfaces. A flat square is an open surface because there is no “inside” to even put the water in. A bowl is an open surface because, though you can put water in it, the water could spill out.
We will go into more detail about Gauss’ Law when we cover electromagnetism, but it is worth seeing how to use it in a simple scenario. Figure \(\PageIndex{5}\) shows a spherical body of mass \(M\) and radius \(R\) for which we would like to determine the value of the gravitational field at a distance \(r\) from the center of the body.
To do so, we draw a “Gaussian surface”, \(S\) , that is a sphere with a radius \(r\) , and centered about the body. At any point on the surface, the area element vector \(d\vec A\) points away from the center of the spherical surface. The gravitational field vector, \(\vec g(\vec r)\) will always point towards the center of the spherical surface, as illustrated. Furthermore, by symmetry, the magnitude of \(\vec g(\vec r)\) is constant along the whole Gaussian surface. Thus, the scalar product \(\vec g(\vec r) \cdot d\vec A=-g(r)dA\) everywhere along the surface (it is negative because the two vectors are anti-parallel). The integral is thus given by:
\[\begin{aligned} \oint \vec g(\vec r) \cdot d\vec A = -g(r)\oint dA \end{aligned}\]
where we factored \(g(r)\) out of the integral, since the magnitude of \(\vec g(\vec r)\) is constant for all of the area elements \(dA\) on the sphere. Remember that an integral is a sum. The integral \(\oint dA\) thus means “sum all of the area elements \(dA\) over the entire surface \(S\) ”. Thus, the integral is the total area of the spherical surface \(S\) 2 :
\[\begin{aligned} \oint \vec g(\vec r) \
d\vec A = -g(r)\oint dA =-g(r)(4\pi r^2)\end{aligned}\]
Inserting this into Gauss’ Law, we find:
\[\begin{aligned} \oint \vec g(\vec r) \cdot d\vec A &= 4\pi G M^{enc}\\[4pt] -g(r)(4\pi r^2) &= 4\pi G M^{enc}\\[4pt] \therefore g(r) &= - \frac{GM}{r^2}\end{aligned}\]
where \(M^{enc}=M\) is the total mass enclosed by the Gaussian surface (in this case, the entire mass \(M\) is enclosed). This is of course the result that we expected and obtained earlier from Newton’s formulation. Note that Gauss’ Law is only easy to use if the system is highly symmetric (e.g. spherically symmetric), and that it does not give the direction of the field vector, which must be obtained from symmetry arguments.
Olivia's Thoughts
Here’s an analogy to describe Gauss’s Law for gravity: A famous celebrity is doing an event, and they attract a certain number of fans who want to get as close to the celebrity as possible. You put up a barricade around the celebrity. The gravitational field is represented by how crowded it is somewhere along the barricade. If a second celebrity is at the event, they will attract their own fans, so there will be more people around the barricade. The number of celebrities is kind of like the enclosed mass \(M^{enc}\) .
A photographer is coming to the event, and you told him to stand at some location that is a distance \(r\) from the celebrities. The photographer wants to know how crowded it will be when he is standing behind the barricade at that location. Gauss’s law gives us a way to figure this out. If you know which celebrities are at the event ( \(M^{enc}\) ), you can determine how many people will be there (this is like finding \(4\pi GM^{enc}\) ). Then, if you can build a barricade such that the fans are evenly distributed around it, and you know how long that barricade is ( \(\oint dA\) ), you can easily calculate how crowded it will be at some point along the barricade (you can just divide the number of people by the length of the barricade). The barricade represents our Gaussian surface and, like a Gaussian surface, it can be whatever shape we want as long as it encloses the celebrities and passes through the point we are interested in. If we want to make sure the people are spread out evenly, the shape of the barricade is going to depend on the specific case. Let’s take the example of our single spherical body. This is analagous to having one celebrity at the event.
Figure \(\PageIndex{6}\) shows two possible barricades we could build. Although we can technically build the barricade on the left, it doesn’t help us because the areas closer to the celebrity will be more crowded. Instead, we want to build the barricade on the right, which is a circle of radius \(r\) , because the fans are evenly spread out. This is why we use a spherical Gaussian surface when we’re considering the field due to a spherical body - at any point a distance \(r\) from the body, the field will be the same. (Note: Remember that, unlike the barricade, the Gaussian surface isn’t a physical thing, so it won’t affect the gravitational field. It is just a mathematical tool that allows us to take advantage of what the field already looks like.)
We can also use Gauss’ Law to determine the gravitational field inside of the body of mass \(M\) and radius \(R\) . This is illustrated in Figure \(\PageIndex{7}\), which shows a spherical Gaussian surface of radius \(r\) that is inside of the body of mass \(M\) .
The gravitational field inside of the body of mass \(M\) is also symmetric and constant in magnitude across the whole surface, so that the integral is the same as before:
\[\begin{aligned} \oint \vec g(\vec r) \cdot d\vec A=-g(r)(4\pi r^2)\end{aligned}\]
However, in order to use Gauss’ Law, we need to determine the mass of the body that is enclosed within the spherical surface, which will be less than \(M\) . If we assume that the mass density, \(\rho\) , of the object is constant (the body is made of a uniform material), then the density is simply the mass of the object over its volume:
\[\begin{aligned} \rho = \frac{M}{\frac{4}{3}\pi R^3}\end{aligned}\]
The amount of mass enclosed by the spherical surface of radius \(r\) is the density multiplied by the volume of a sphere of radius \(r\) :
\[\begin{aligned} M^{enc} = \rho \frac{4}{3}\pi r^3 = M\frac{r^3}{R^3}\end{aligned}\]
Applying Gauss’ Law, we can now find the magnitude of the gravitational field inside of the spherical body at a distance \(r\) from the center:
\[\begin{aligned} \oint \vec g(\vec r) \cdot d\vec A &= 4\pi G M^{enc}\\[4pt] -g(r)(4\pi r^2) &= 4\pi G M\frac{r^3}{R^3}\\[4pt] \therefore g(r) &= - \frac{G M}{R^3}r\end{aligned}\]
And we find that, inside a uniform spherical body of mass \(M\) , the gravitational field increases linearly with radius as one moves out from the center. At the center of the body, the gravitational field is zero.
Exercise \(\PageIndex{9}\)
What can you say about the magnitude of the gravitational field inside a spherical shell of mass \(M\) ?
- It increases as you move out from the center of the spherical shell.
- It decreases as you move out from the center of the spherical shell.
- It is equal to zero.
- It is nonzero and constant in magnitude.
- Answer


5.13: Works Consulted and Further Reading
Burkert, Walter. The Orientalizing Revolution: Near Eastern Influence on Greek Culture in the Early Archaic Age . Cambridge: Harvard University Press, 1998.
Bury, J. B. and Russell Meiggs. A History of Greece to the Death of Alexander the Great . New York: St. Martin’s Press.
Cartledge, Paul. Sparta and Laconia: A Regional History 1300 – 362 BC. New York: Routledge, 2002.
Dickinson, Oliver. The Aegean Bronze Age . Cambridge: Cambridge University Press, 1994.
Gruen, Erich. Diaspora: Jews amidst Greeks and Romans . Cambridge: Harvard University Press, 2004.
Hanson, Victor Davis. A War Like No Other: How the Athenians and Spartans Fought the Peloponnesian War . New York: Random House Trade Paperbacks, 2006.
Horden, Peregrine and Nicholas Purcell, The Corrupting Sea: A Study of Mediterranean History . Oxford: Wiley-Blackwell, 2000.
Kagan, Donald. The Archidamian War . Ithaca: Cornell University Press, 1990.
———. The Fall of the Athenian Empire . Ithaca: Cornell University Press, 1991.
———. The Outbreak of the Peloponnesian War . Ithaca: Cornell University Press, 1989.
———. The Peace of Nicias and the Sicilian Expedition . Ithaca: Cornell University Press, 1991.
Lane Fox, Robin. Alexander the Great . New York: Penguin, 2004.
Lendon, J. E. Soldiers and Ghosts: A History of Battle in Classical Antiquity . New Haven: Yale University Press, 2006.
Lewis, Naphthali. Greeks in Ptolemaic Egypt . Durham: American Society of Papyrologists, 2001.
Long, A. A. Hellenistic Philosophy: Stoics, Epicureans, Sceptics . Berkeley: University of California Press, 1986.
Meiggs, Russell. The Athenian Empire. Oxford: Oxford University Press, 1979.
Morkot, Robert. The Penguin Historical Atlas of Ancient Greece . London: Penguin Books, 1997.
Murray, Oswyn. Early Greece . Cambridge: Harvard University Press, 1993.
Ober, Josiah. Mass and Elite in Democratic Athens: Rhetoric, Ideology, and the Power of the People . Princeton: Princeton University Press, 1991.
de Polignac, François. Cults, Territory, and the Origins of the Greek City-State . Chicago: University of Chicago Press, 1995.
Pomeroy, Sarah, Stanley Burstein, Walter Donlan, Jennifer Tolbert Roberts, and David Tandy. Ancient Greece: A Political, Social, and Cultural History . Oxford: Oxford University Press, 2011.
Walbank, F. W. The Hellenistic World . Cambridge: Harvard University Press, 1993.


9.2: Radioactive Decay
- Write and balance nuclear equations
- To know the different kinds of radioactive decay.
- To balance a nuclear reaction.
Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of \(\ce{^1_1H}\), neutrons. Recall that the number of protons in the nucleus is called the atomic number (\(Z\)) of the element, and the sum of the number of protons and the number of neutrons is the mass number (\(A\)). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation:
\[ \large \ce{^{A}_{Z}X} \label{Eq1a} \]
where
- \(X\) is the symbol for the element,
- \(A\) is the mass number, and
- \(Z\) is the atomic number.
Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, \(\ce{^{14}_6C}\) is called “carbon-14.”
Protons and neutrons, collectively called nucleons , are packed together tightly in a nucleus. With a radius of about 10 −15 meters, a nucleus is quite small compared to the radius of the entire atom, which is about 10 −10 meters. Nuclei are extremely dense compared to bulk matter, averaging \(1.8 \times 10^{14}\) grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm 3 . If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately \(6.4 \times 10^6\) meters, 30,000 times larger).
Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions . To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.
Nuclear Equations
A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of subatomic particles rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways:
- The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.
- The sum of the charges of the reactants equals the sum of the charges of the products.
If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that \(\ce{^{17}_8O}\) is a product of the nuclear reaction of \(\ce{^{14}_7N}\) and \(\ce{^4_2He}\) if we knew that a proton, \(\ce{^1_1H}\), was one of the two products. Example \(\PageIndex{1}\) shows how we can identify a nuclide by balancing the nuclear reaction.
The reaction of an α particle with magnesium-25 \( (\ce{^{25}_{12}Mg})\) produces a proton and a nuclide of another element. Identify the new nuclide produced.
Solution
The nuclear reaction can be written as:
\[\ce{^{25}_{12}Mg + ^4_2He \rightarrow ^1_1H + ^{A}_{Z}X} \nonumber \]
where
- \(\ce A\) is the mass number and
- \(\ce Z\) is the atomic number of the new nuclide, \(\ce X\).
Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products:
\[\mathrm{25+4=A+1} \nonumber \]
so
\[ \mathrm{A=28} \nonumber \]
Similarly, the charges must balance, so:
\[\mathrm{12+2=Z+1} \nonumber \]
so
\[\mathrm{Z=13} \nonumber \]
Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is \(\ce{^{28}_{13}Al}\).
The nuclide \(\ce{^{125}_{53}I}\) combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction?
- Answer
-
\[\ce{^{125}_{53}I + ^0_{−1}e \rightarrow ^{125}_{52}Te} \nonumber \]
The two general kinds of nuclear reactions are nuclear decay reactions and nuclear transmutation reactions . In a nuclear decay reaction , also called radioactive decay, an unstable nucleus emits radiation and is transformed into the nucleus of one or more other elements. The resulting daughter nuclei have a lower mass and are lower in energy (more stable) than the parent nucleus that decayed. In contrast, in a nuclear transmutation reaction , a nucleus reacts with a subatomic particle or another nucleus to form a product nucleus that is more massive than the starting material. As we shall see, nuclear decay reactions occur spontaneously under all conditions, but nuclear transmutation reactions occur only under very special conditions, such as the collision of a beam of highly energetic particles with a target nucleus or in the interior of stars. We begin this section by considering the different classes of radioactive nuclei, along with their characteristic nuclear decay reactions and the radiation they emit.
Nuclear decay reactions occur spontaneously under all conditions, whereas nuclear transmutation reactions are induced.
Nuclear Decay Reactions
Just as we use the number and type of atoms present to balance a chemical equation, we can use the number and type of nucleons present to write a balanced nuclear equation for a nuclear decay reaction. This procedure also allows us to predict the identity of either the parent or the daughter nucleus if the identity of only one is known. Regardless of the mode of decay, the total number of nucleons is conserved in all nuclear reactions.
To describe nuclear decay reactions, chemists have extended the \(^A _Z \textrm{X}\) notation for nuclides to include radioactive emissions. Table \(\PageIndex{1}\) lists the name and symbol for each type of emitted radiation. The most notable addition is the positron , a particle that has the same mass as an electron but a positive charge rather than a negative charge.
| Identity | Symbol | Charge | Mass (amu) |
|---|---|---|---|
| helium nucleus | \(^4_2\alpha\) | +2 | 4.001506 |
| electron | \(^0_{-1}\beta\) or \(\beta ^-\) | −1 | 0.000549 |
| photon | \(_0^0\gamma\) | — | — |
| neutron | \(^1_0\textrm n\) | 0 | 1.008665 |
| proton | \(^1_1\textrm p \) | +1 | 1.007276 |
| positron | \(^0_{+1}\beta\) or \(\beta ^+\) | +1 | 0.000549 |
Like the notation used to indicate isotopes, the upper left superscript in the symbol for a particle gives the mass number, which is the total number of protons and neutrons. For a proton or a neutron, A = 1. Because neither an electron nor a positron contains protons or neutrons, its mass number is 0. The numbers should not be taken literally, however, as meaning that these particles have zero mass; ejection of a beta particle (an electron) simply has a negligible effect on the mass of a nucleus.
Similarly, the lower left subscript gives the charge of the particle. Because protons carry a positive charge, Z = +1 for a proton. In contrast, a neutron contains no protons and is electrically neutral, so Z = 0. In the case of an electron, Z = −1, and for a positron, Z = +1. Because γ rays are high-energy photons, both A and Z are 0. In some cases, two different symbols are used for particles that are identical but produced in different ways. For example, the symbol \(^0_{-1}\textrm e\), which is usually simplified to e − , represents a free electron or an electron associated with an atom, whereas the symbol \(^0_{-1}\beta\), which is often simplified to β − , denotes an electron that originates from within the nucleus, which is a β particle. Similarly, \(^4_{2}\textrm{He}^{2+}\) refers to the nucleus of a helium atom, and \(^4_{2}\alpha\) denotes an identical particle that has been ejected from a heavier nucleus.
There are six fundamentally different kinds of nuclear decay reactions, and each releases a different kind of particle or energy. The essential features of each reaction are shown in Figure \(\PageIndex{1}\) . The most common are alpha and beta decay and gamma emission, but the others are essential to an understanding of nuclear decay reactions.
Alpha \(\alpha\) Decay
Many nuclei with mass numbers greater than 200 undergo alpha (α) decay , which results in the emission of a helium-4 nucleus as an alpha (α) particle , \(^4_{2}\alpha\). The general reaction is as follows:
\[\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A-4}_{Z-2} \textrm X'}+\underset{\textrm{alpha}\\ \textrm{particle}}{^4_2 \alpha}\label{Eq1} \]
The daughter nuclide contains two fewer protons and two fewer neutrons than the parent. Thus α-particle emission produces a daughter nucleus with a mass number A − 4 and a nuclear charge Z − 2 compared to the parent nucleus. Radium-226, for example, undergoes alpha decay to form radon-222:
\[^{226}_{88}\textrm{Ra}\rightarrow ^{222}_{86}\textrm{Rn}+^{4}_{2}\alpha\label{Eq2} \]
Because nucleons are conserved in this and all other nuclear reactions, the sum of the mass numbers of the products, 222 + 4 = 226, equals the mass number of the parent. Similarly, the sum of the atomic numbers of the products, 86 + 2 = 88, equals the atomic number of the parent. Thus the nuclear equation is balanced.
Just as the total number of atoms is conserved in a chemical reaction, the total number of nucleons is conserved in a nuclear reaction.
Beta \(\beta^-\) Decay
Nuclei that contain too many neutrons often undergo beta (β) decay , in which a neutron is converted to a proton and a high-energy electron that is ejected from the nucleus as a β particle:
\[\underset{\textrm{unstable} \\ \textrm{neutron in} \\ \textrm{nucleus}}{^1_0 \textrm n}\rightarrow \underset{\textrm{proton} \\ \textrm{retained} \\ \textrm{by nucleus}}{^{1}_{1} \textrm p}+\underset{\textrm{beta particle} \\ \textrm{emitted by} \\ \textrm{nucleus}}{^0_{-1} \beta}\label{Eq3} \]
The general reaction for beta decay is therefore
\[\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z+1} \textrm X'}+\underset{\textrm{beta particle}}{^0_{-1} \beta}\label{Eq4} \]
Although beta decay does not change the mass number of the nucleus, it does result in an increase of +1 in the atomic number because of the addition of a proton in the daughter nucleus. Thus beta decay decreases the neutron-to-proton ratio, moving the nucleus toward the band of stable nuclei. For example, carbon-14 undergoes beta decay to form nitrogen-14:
\[^{14}_{6}\textrm{C}\rightarrow ^{14}_{7}\textrm{N}+\,^{0}_{-1}\beta \nonumber \]
Once again, the number of nucleons is conserved, and the charges are balanced. The parent and the daughter nuclei have the same mass number, 14, and the sum of the atomic numbers of the products is 6, which is the same as the atomic number of the carbon-14 parent.
Positron \(\beta^+\) Emission
Because a positron has the same mass as an electron but opposite charge, positron emission is the opposite of beta decay. Thus positron emission is characteristic of neutron-poor nuclei, which decay by transforming a proton to a neutron and emitting a high-energy positron:
\[^{1}_{1}\textrm{p}^+\rightarrow ^{1}_{0}\textrm{n}+\,^{0}_{+1}\beta^+\label{Eq6} \]
The general reaction for positron emission is therefore
\[\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\underset{\textrm{positron}}{^0_{+1} \beta^+} \nonumber \]
Like beta decay, positron emission does not change the mass number of the nucleus. In this case, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Thus the neutron-to-proton ratio has increased, again moving the nucleus closer to the band of stable nuclei. For example, carbon-11 undergoes positron emission to form boron-11:
\[^{11}_{6}\textrm{C}\rightarrow ^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta^+ \nonumber \]
Nucleons are conserved, and the charges balance. The mass number, 11, does not change, and the sum of the atomic numbers of the products is 6, the same as the atomic number of the parent carbon-11 nuclide.
Electron Capture
A neutron-poor nucleus can decay by either positron emission or electron capture (EC) , in which an electron in an inner shell reacts with a proton to produce a neutron:
\[^{1}_{1}\textrm{p} +\; ^{0}_{-1}\textrm{e}\rightarrow \, ^{1}_{0}\textrm n\label{Eq9} \]
When a second electron moves from an outer shell to take the place of the lower-energy electron that was absorbed by the nucleus, an x-ray is emitted. The overall reaction for electron capture is thus
\[\underset{\textrm{parent}}{^A_Z \textrm X}+\underset{\textrm{electron}}{^0_{-1} \textrm e}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\textrm{x-ray} \nonumber \]
Electron capture does not change the mass number of the nucleus because both the proton that is lost and the neutron that is formed have a mass number of 1. As with positron emission, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Once again, the neutron-to-proton ratio has increased, moving the nucleus toward the band of stable nuclei. For example, iron-55 decays by electron capture to form manganese-55, which is often written as follows:
\[^{55}_{26}\textrm{Fe}\overset{\textrm{EC}}{\rightarrow}\, ^{55}_{25}\textrm{Mn}+\textrm{x-ray} \nonumber \]
The atomic numbers of the parent and daughter nuclides differ in Equation 20.2.11 , although the mass numbers are the same. To write a balanced nuclear equation for this reaction, we must explicitly include the captured electron in the equation:
\[^{55}_{26}\textrm{Fe}+\,^{0}_{-1}\textrm{e}\rightarrow \, ^{55}_{25}\textrm{Mn}+\textrm{x-ray} \nonumber \]
Both positron emission and electron capture are usually observed for nuclides with low neutron-to-proton ratios, but the decay rates for the two processes can be very different.
Gamma \(\gamma\) Emission
Many nuclear decay reactions produce daughter nuclei that are in a nuclear excited state, which is similar to an atom in which an electron has been excited to a higher-energy orbital to give an electronic excited state. Just as an electron in an electronic excited state emits energy in the form of a photon when it returns to the ground state, a nucleus in an excited state releases energy in the form of a photon when it returns to the ground state. These high-energy photons are γ rays. Gamma (\(\gamma\)) emission can occur virtually instantaneously, as it does in the alpha decay of uranium-238 to thorium-234, where the asterisk denotes an excited state:
\[^{238}_{92}\textrm{U}\rightarrow \, \underset{\textrm{excited} \\ \textrm{nuclear} \\ \textrm{state}}{^{234}_{90}\textrm{Th*}} + ^{4}_{2}\alpha\xrightarrow {\textrm{relaxation}\,}\,^{234}_{90}\textrm{Th} + \ce{^0_0\gamma} \nonumber \]
If we disregard the decay event that created the excited nucleus, then
\[^{234}_{88}\textrm{Th*} \rightarrow\, ^{234}_{88}\textrm{Th} + ^{0}_{0}\gamma \nonumber \]
or more generally,
\[^{A}_{Z}\textrm{X*} \rightarrow\, ^{A}_{Z}\textrm{X} + ^{0}_{0}\gamma \nonumber \]
Gamma emission can also occur after a significant delay. For example, technetium-99 m has a half-life of about 6 hours before emitting a \(γ\) ray to form technetium-99 (the m is for metastable). Because γ rays are energy, their emission does not affect either the mass number or the atomic number of the daughter nuclide. Gamma-ray emission is therefore the only kind of radiation that does not necessarily involve the conversion of one element to another, although it is almost always observed in conjunction with some other nuclear decay reaction.
Spontaneous Fission
Only very massive nuclei with high neutron-to-proton ratios can undergo spontaneous fission , in which the nucleus breaks into two pieces that have different atomic numbers and atomic masses. This process is most important for the transactinide elements, with Z ≥ 104. Spontaneous fission is invariably accompanied by the release of large amounts of energy, and it is usually accompanied by the emission of several neutrons as well. An example is the spontaneous fission of \(^{254}_{98}\textrm{Cf}\), which gives a distribution of fission products; one possible set of products is shown in the following equation:
\[^{254}_{98}\textrm{Cf}\rightarrow \,^{118}_{46}\textrm{Pd}+\,^{132}_{52}\textrm{Te}+4^{1}_{0}\textrm{n} \nonumber \]
Once again, the number of nucleons is conserved. Thus the sum of the mass numbers of the products (118 + 132 + 4 = 254) equals the mass number of the reactant. Similarly, the sum of the atomic numbers of the products [46 + 52 + (4 × 0) = 98] is the same as the atomic number of the parent nuclide.
Write a balanced nuclear equation to describe each reaction.
- the beta decay of \(^{35}_{16}\textrm{S}\)
- the decay of \(^{201}_{80}\textrm{Hg}\) by electron capture
- the decay of \(^{30}_{15}\textrm{P}\) by positron emission
Given: radioactive nuclide and mode of decay
Asked for: balanced nuclear equation
Strategy:
A Identify the reactants and the products from the information given.
B Use the values of A and Z to identify any missing components needed to balance the equation.
Solution
a.
A We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as \(^{A}_{Z}\textrm{X}\): \[^{35}_{16}\textrm{S}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-1}\beta \nonumber \]
B Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of A = 35 − 0 = 35 and an atomic number of Z = 16 − (−1) = 17. The element with Z = 17 is chlorine, so the balanced nuclear equation is as follows: \[^{35}_{16}\textrm{S}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta \nonumber \]
b.A We know the identities of both reactants: \(^{201}_{80}\textrm{Hg}\) and an inner electron, \(^{0}_{-1}\textrm{e}\). The reaction is as follows: \[^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X} \nonumber \]
B Both protons and neutrons are conserved, so the mass number of the product must be A = 201 + 0 = 201, and the atomic number of the product must be Z = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus \[^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au} \nonumber \]
c.A As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore \[^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta \nonumber \]
B The mass number of the second product is A = 30 − 0 = 30, and its atomic number is Z = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows: \[^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta \nonumber \]
Write a balanced nuclear equation to describe each reaction.
- \(^{11}_{6}\textrm{C}\) by positron emission
- the beta decay of molybdenum-99
- the emission of an α particle followed by gamma emission from \(^{185}_{74}\textrm{W}\)
- Answer a
-
\(^{11}_{6}\textrm{C}\rightarrow\,^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta\)
- Answer d
-
\(^{99}_{42}\textrm{Mo}\rightarrow\,^{99m}_{43}\textrm{Tc}+\,^{0}_{-1}\beta\)
- Answer c
-
\(^{185}_{74}\textrm{W}\rightarrow\,^{181}_{72}\textrm{Hf}+\,^{4}_{2}\alpha +\,^{0}_{0}\gamma\)
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
- \(^{45}_{22}\textrm{Ti}\)
- \(^{242}_{94}\textrm{Pu}\)
- \(^{12}_{5}\textrm{B}\)
- \(^{256}_{100}\textrm{Fm}\)
Given: nuclide
Asked for: type of nuclear decay
Strategy:
Based on the neutron-to-proton ratio and the value of Z , predict the type of nuclear decay reaction that will produce a more stable nuclide.
Solution
- This nuclide has a neutron-to-proton ratio of only 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that have low neutron-to-proton ratios decay by converting a proton to a neutron. The two possibilities are positron emission, which converts a proton to a neutron and a positron, and electron capture, which converts a proton and a core electron to a neutron. In this case, both are observed, with positron emission occurring about 86% of the time and electron capture about 14% of the time.
- Nuclei with Z > 83 are too heavy to be stable and usually undergo alpha decay, which decreases both the mass number and the atomic number. Thus \(^{242}_{94}\textrm{Pu}\) is expected to decay by alpha emission.
- This nuclide has a neutron-to-proton ratio of 1.4, which is very high for a light element. Nuclei with high neutron-to-proton ratios decay by converting a neutron to a proton and an electron. The electron is emitted as a β particle, and the proton remains in the nucleus, causing an increase in the atomic number with no change in the mass number. We therefore predict that \(^{12}_{5}\textrm{B}\) will undergo beta decay.
- This is a massive nuclide, with an atomic number of 100 and a mass number much greater than 200. Nuclides with A ≥ 200 tend to decay by alpha emission, and even heavier nuclei tend to undergo spontaneous fission. We therefore predict that \(^{256}_{100}\textrm{Fm}\) will decay by either or both of these two processes. In fact, it decays by both spontaneous fission and alpha emission, in a 97:3 ratio.
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
- \(^{32}_{14}\textrm{Si}\)
- \(^{43}_{21}\textrm{Sc}\)
- \(^{231}_{91}\textrm{Pa}\)
- Answer a
-
beta decay
- Answer d
-
positron emission or electron capture
- Answer c
-
alpha decay
Radioactive Decay Series
The nuclei of all elements with atomic numbers greater than 83 are unstable. Thus all isotopes of all elements beyond bismuth in the periodic table are radioactive. Because alpha decay decreases Z by only 2, and positron emission or electron capture decreases Z by only 1, it is impossible for any nuclide with Z > 85 to decay to a stable daughter nuclide in a single step, except via nuclear fission. Consequently, radioactive isotopes with Z > 85 usually decay to a daughter nucleus that is radiaoctive, which in turn decays to a second radioactive daughter nucleus, and so forth, until a stable nucleus finally results. This series of sequential alpha- and beta-decay reactions is called a radioactive decay series . The most common is the uranium-238 decay series, which produces lead-206 in a series of 14 sequential alpha- and beta-decay reactions ( Figure \(\PageIndex{2}\) ). Although a radioactive decay series can be written for almost any isotope with Z > 85, only two others occur naturally: the decay of uranium-235 to lead-207 (in 11 steps) and thorium-232 to lead-208 (in 10 steps). A fourth series, the decay of neptunium-237 to bismuth-209 in 11 steps, is known to have occurred on the primitive Earth. With a half-life of “only” 2.14 million years, all the neptunium-237 present when Earth was formed decayed long ago, and today all the neptunium on Earth is synthetic.
Due to these radioactive decay series, small amounts of very unstable isotopes are found in ores that contain uranium or thorium. These rare, unstable isotopes should have decayed long ago to stable nuclei with a lower atomic number, and they would no longer be found on Earth. Because they are generated continuously by the decay of uranium or thorium, however, their amounts have reached a steady state, in which their rate of formation is equal to their rate of decay. In some cases, the abundance of the daughter isotopes can be used to date a material or identify its origin.
Induced Nuclear Reactions
The discovery of radioactivity in the late 19th century showed that some nuclei spontaneously transform into nuclei with a different number of protons, thereby producing a different element. When scientists realized that these naturally occurring radioactive isotopes decayed by emitting subatomic particles, they realized that—in principle—it should be possible to carry out the reverse reaction, converting a stable nucleus to another more massive nucleus by bombarding it with subatomic particles in a nuclear transmutation reaction.
The first successful nuclear transmutation reaction was carried out in 1919 by Ernest Rutherford, who showed that α particles emitted by radium could react with nitrogen nuclei to form oxygen nuclei. As shown in the following equation, a proton is emitted in the process:
\(^{4}_{2}\alpha + \, ^{14}_{7}\textrm{N} \rightarrow \,^{17}_{8}\textrm{O}+\,^{1}_{1}\textrm{p}\label{Eq17}\)
Rutherford’s nuclear transmutation experiments led to the discovery of the neutron. He found that bombarding the nucleus of a light target element with an α particle usually converted the target nucleus to a product that had an atomic number higher by 1 and a mass number higher by 3 than the target nucleus. Such behavior is consistent with the emission of a proton after reaction with the α particle. Very light targets such as Li, Be, and B reacted differently, however, emitting a new kind of highly penetrating radiation rather than a proton. Because neither a magnetic field nor an electrical field could deflect these high-energy particles, Rutherford concluded that they were electrically neutral. Other observations suggested that the mass of the neutral particle was similar to the mass of the proton. In 1932, James Chadwick (Nobel Prize in Physics, 1935), who was a student of Rutherford’s at the time, named these neutral particles neutrons and proposed that they were fundamental building blocks of the atom. The reaction that Chadwick initially used to explain the production of neutrons was as follows:
\[^{4}_{2}\alpha + \, ^{9}_{4}\textrm{Be} \rightarrow \,^{12}_{6}\textrm{C}+\,^{1}_{0}\textrm{n}\label{Eq18} \]
Because α particles and atomic nuclei are both positively charged, electrostatic forces cause them to repel each other. Only α particles with very high kinetic energy can overcome this repulsion and collide with a nucleus ( Figure \(\PageIndex{3}\) ). Neutrons have no electrical charge, however, so they are not repelled by the nucleus. Hence bombardment with neutrons is a much easier way to prepare new isotopes of the lighter elements. In fact, carbon-14 is formed naturally in the atmosphere by bombarding nitrogen-14 with neutrons generated by cosmic rays:
\[^{1}_{0}\textrm{n} + \, ^{14}_{7}\textrm{N} \rightarrow \,^{14}_{6}\textrm{C}+\,^{1}_{1}\textrm{p}\label{Eq19} \]
In 1933, Frédéric Joliot and Iréne Joliot-Curie (daughter of Marie and Pierre Curie) prepared the first artificial radioactive isotope by bombarding aluminum-27 with α particles. For each 27 Al that reacted, one neutron was released. Identify the product nuclide and write a balanced nuclear equation for this transmutation reaction.
Given: reactants in a nuclear transmutation reaction
Asked for: product nuclide and balanced nuclear equation
Strategy:
A Based on the reactants and one product, identify the other product of the reaction. Use conservation of mass and charge to determine the values of Z and A of the product nuclide and thus its identity.
B Write the balanced nuclear equation for the reaction.
Solution
A Bombarding an element with α particles usually produces an element with an atomic number that is 2 greater than the atomic number of the target nucleus. Thus we expect that aluminum ( Z = 13) will be converted to phosphorus ( Z = 15). With one neutron released, conservation of mass requires that the mass number of the other product be 3 greater than the mass number of the target. In this case, the mass number of the target is 27, so the mass number of the product will be 30. The second product is therefore phosphorus-30, \(^{30}_{15}\textrm{P}\).
B The balanced nuclear equation for the reaction is as follows:
\[^{27}_{13}\textrm{Al} + \, ^{4}_{2}\alpha \rightarrow \,^{30}_{15}\textrm{P}+\,^{1}_{0}\textrm{n} \nonumber \]
Because all isotopes of technetium are radioactive and have short half-lives, it does not exist in nature. Technetium can, however, be prepared by nuclear transmutation reactions. For example, bombarding a molybdenum-96 target with deuterium nuclei \((^{2}_{1}\textrm{H})\) produces technetium-97. Identify the other product of the reaction and write a balanced nuclear equation for this transmutation reaction.
- Answer
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neutron, \(^{1}_{0}\textrm{n}\) ; \(^{96}_{42}\textrm{Mo} + \, ^{2}_{1}\textrm{H} \rightarrow \,^{97}_{43}\textrm{Tc}+\,^{1}_{0}\textrm{n}\) :
We noted earlier in this section that very heavy nuclides, corresponding to Z ≥ 104, tend to decay by spontaneous fission. Nuclides with slightly lower values of Z , such as the isotopes of uranium ( Z = 92) and plutonium ( Z = 94), do not undergo spontaneous fission at any significant rate. Some isotopes of these elements, however, such as \(^{235}_{92}\textrm{U}\) and \(^{239}_{94}\textrm{Pu}\) undergo induced nuclear fission when they are bombarded with relatively low-energy neutrons, as shown in the following equation for uranium-235 and in Figure \(\PageIndex{4}\) :
\[^{235}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{236}_{92}\textrm{U}\rightarrow \,^{141}_{56}\textrm{Ba}+\,^{92}_{36}\textrm{Kr}+3^{1}_{0}\textrm{n}\label{Eq20} \]
Any isotope that can undergo a nuclear fission reaction when bombarded with neutrons is called a fissile isotope .
During nuclear fission, the nucleus usually divides asymmetrically rather than into two equal parts, as shown in Figure \(\PageIndex{4}\) . Moreover, every fission event of a given nuclide does not give the same products; more than 50 different fission modes have been identified for uranium-235, for example. Consequently, nuclear fission of a fissile nuclide can never be described by a single equation. Instead, as shown in Figure \(\PageIndex{5}\) , a distribution of many pairs of fission products with different yields is obtained, but the mass ratio of each pair of fission products produced by a single fission event is always roughly 3:2.
Synthesis of Transuranium Elements
Uranium ( Z = 92) is the heaviest naturally occurring element. Consequently, all the elements with Z > 92, the transuranium elements , are artificial and have been prepared by bombarding suitable target nuclei with smaller particles. The first of the transuranium elements to be prepared was neptunium ( Z = 93), which was synthesized in 1940 by bombarding a 238 U target with neutrons. As shown in Equation 20.21 , this reaction occurs in two steps. Initially, a neutron combines with a 238 U nucleus to form 239 U, which is unstable and undergoes beta decay to produce 239 Np:
\(^{238}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{239}_{92}\textrm{U}\rightarrow \,^{239}_{93}\textrm{Np}+\,^{0}_{-1}\beta\label{Eq21}\)
Subsequent beta decay of 239 Np produces the second transuranium element, plutonium ( Z = 94):
\(^{239}_{93}\textrm{Np} \rightarrow \,^{239}_{94}\textrm{Pu}+\,^{0}_{-1}\beta\label{Eq22}\)
Bombarding the target with more massive nuclei creates elements that have atomic numbers significantly greater than that of the target nucleus ( Table \(\PageIndex{2}\) ). Such techniques have resulted in the creation of the superheavy elements 114 and 116, both of which lie in or near the “island of stability."
| \(^{239}_{94}\textrm{Pu}+\,^{4}_{2}\alpha \rightarrow \,^{242}_{96}\textrm{Cm}+\,^{1}_{0}\textrm{n}\) |
|---|
| \(^{239}_{94}\textrm{Pu}+\,^{4}_{2}\alpha \rightarrow \,^{241}_{95}\textrm{Am}+\,^{1}_{1}\textrm{p}+\,^{1}_{0}\textrm{n}\) |
| \(^{242}_{96}\textrm{Cm}+\,^{4}_{2}\alpha \rightarrow \,^{243}_{97}\textrm{Bk}+\,^{1}_{1}\textrm{p}+2^{1}_{0}\textrm{n}\) |
| \(^{253}_{99}\textrm{Es}+\,^{4}_{2}\alpha \rightarrow \,^{256}_{101}\textrm{Md}+\,^{1}_{0}\textrm{n}\) |
| \(^{238}_{92}\textrm{U}+\,^{12}_{6}\textrm{C} \rightarrow \,^{246}_{98}\textrm{Cf}+4^{1}_{0}\textrm{n}\) |
| \(^{252}_{98}\textrm{Cf}+\,^{10}_{5}\textrm{B} \rightarrow \,^{256}_{103}\textrm{Lr}+6^{1}_{0}\textrm{n}\) |
A device called a particle accelerator is used to accelerate positively charged particles to the speeds needed to overcome the electrostatic repulsions between them and the target nuclei by using electrical and magnetic fields. Operationally, the simplest particle accelerator is the linear accelerator ( Figure \(\PageIndex{6}\) ), in which a beam of particles is injected at one end of a long evacuated tube. Rapid alternation of the polarity of the electrodes along the tube causes the particles to be alternately accelerated toward a region of opposite charge and repelled by a region with the same charge, resulting in a tremendous acceleration as the particle travels down the tube. A modern linear accelerator such as the Stanford Linear Accelerator (SLAC) at Stanford University is about 2 miles long.
To achieve the same outcome in less space, a particle accelerator called a cyclotron forces the charged particles to travel in a circular path rather than a linear one. The particles are injected into the center of a ring and accelerated by rapidly alternating the polarity of two large D-shaped electrodes above and below the ring, which accelerates the particles outward along a spiral path toward the target.
The length of a linear accelerator and the size of the D-shaped electrodes in a cyclotron severely limit the kinetic energy that particles can attain in these devices. These limitations can be overcome by using a synchrotron, a hybrid of the two designs. A synchrotron contains an evacuated tube similar to that of a linear accelerator, but the tube is circular and can be more than a mile in diameter. Charged particles are accelerated around the circle by a series of magnets whose polarities rapidly alternate.
Summary and Key Takeaway
- Nuclear decay reactions occur spontaneously under all conditions and produce more stable daughter nuclei, whereas nuclear transmutation reactions are induced and form a product nucleus that is more massive than the starting material.
In nuclear decay reactions (or radioactive decay) , the parent nucleus is converted to a more stable daughter nucleus. Nuclei with too many neutrons decay by converting a neutron to a proton, whereas nuclei with too few neutrons decay by converting a proton to a neutron. Very heavy nuclei (with A ≥ 200 and Z > 83) are unstable and tend to decay by emitting an α particle . When an unstable nuclide undergoes radioactive decay, the total number of nucleons is conserved, as is the total positive charge. Six different kinds of nuclear decay reactions are known. Alpha decay results in the emission of an α particle, \(^4 _2 \alpha\), and produces a daughter nucleus with a mass number that is lower by 4 and an atomic number that is lower by 2 than the parent nucleus. Beta decay converts a neutron to a proton and emits a high-energy electron, producing a daughter nucleus with the same mass number as the parent and an atomic number that is higher by 1. Positron emission is the opposite of beta decay and converts a proton to a neutron plus a positron. Positron emission does not change the mass number of the nucleus, but the atomic number of the daughter nucleus is lower by 1 than the parent. In electron capture (EC) , an electron in an inner shell reacts with a proton to produce a neutron, with emission of an x-ray. The mass number does not change, but the atomic number of the daughter is lower by 1 than the parent. In gamma emission , a daughter nucleus in a nuclear excited state undergoes a transition to a lower-energy state by emitting a γ ray. Very heavy nuclei with high neutron-to-proton ratios can undergo spontaneous fission , in which the nucleus breaks into two pieces that can have different atomic numbers and atomic masses with the release of neutrons. Many very heavy nuclei decay via a radioactive decay series —a succession of some combination of alpha- and beta-decay reactions. In nuclear transmutation reactions , a target nucleus is bombarded with energetic subatomic particles to give a product nucleus that is more massive than the original. All transuranium elements —elements with Z > 92—are artificial and must be prepared by nuclear transmutation reactions. These reactions are carried out in particle accelerators such as linear accelerators, cyclotrons, and synchrotrons.
Key Equations
alpha decay
\[^A_Z \textrm X\rightarrow \, ^{A-4}_{Z-2} \textrm X'+\,^4_2 \alpha \nonumber \]
beta decay
\[^A_Z \textrm X\rightarrow \, ^{A}_{Z+1} \textrm X'+\,^0_{-1} \beta \nonumber \]
positron emission
\[^A_Z \textrm X\rightarrow \, ^{A}_{Z-1} \textrm X'+\,^0_{+1} \beta \nonumber \]
electron capture
\[^A_Z \textrm X+\,^{0}_{-1} \textrm e\rightarrow \, ^{A}_{Z-1} \textrm X'+\textrm{x-ray} \nonumber \]
gamma emission
\[^A_Z \textrm{X*}\rightarrow \, ^{A}_{Z} \textrm X+\,^0_{0} \gamma \nonumber \]


11: Solutions and Colloids
In this chapter, we will consider the nature of solutions, and examine factors that determine whether a solution will form and what properties it may have. In addition, we will discuss colloids—systems that resemble solutions but consist of dispersions of particles somewhat larger than ordinary molecules or ions.
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- 11.1: Prelude to Solutions and Colloids
- Solutions are crucial to the processes that sustain life and to many other processes involving chemical reactions. In this chapter, we will consider the nature of solutions, and examine factors that determine whether a solution will form and what properties it may have. In addition, we will discuss colloids—systems that resemble solutions but consist of dispersions of particles somewhat larger than ordinary molecules or ions.
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- 11.2: The Dissolution Process
- A solution forms when two or more substances combine physically to yield a mixture that is homogeneous at the molecular level. The solvent is the most concentrated component and determines the physical state of the solution. The solutes are the other components typically present at concentrations less than that of the solvent. Solutions may form endothermically or exothermically, depending upon the relative magnitudes of solute and solvent intermolecular attractive forces.
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- 11.3: Electrolytes
- Substances that dissolve in water to yield ions are called electrolytes. Electrolytes may be covalent compounds that chemically react with water to produce ions (for example, acids and bases), or they may be ionic compounds that dissociate to yield their constituent cations and anions, when dissolved. Dissolution of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules.
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- 11.4: Solubility
- The extent to which one substance will dissolve in another is determined by several factors, including the types and relative strengths of intermolecular attractive forces that may exist between the substances’ atoms, ions, or molecules. This tendency to dissolve is quantified as substance’s solubility, its maximum concentration in a solution at equilibrium under specified conditions. A saturated solution contains solute at a concentration equal to its solubility.
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- 11.5: Colligative Properties
- Properties of a solution that depend only on the concentration of solute particles are called colligative properties. They include changes in the vapor pressure, boiling point, and freezing point of the solvent in the solution. The magnitudes of these properties depend only on the total concentration of solute particles in solution, not on the type of particles. The total concentration of solute particles in a solution also determines its osmotic pressure.
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- 11.6: Colloids
- Colloids are mixtures in which one or more substances are dispersed as relatively large solid particles or liquid droplets throughout a solid, liquid, or gaseous medium. The particles of a colloid remain dispersed and do not settle due to gravity, and they are often electrically charged. Colloids are widespread in nature and are involved in many technological applications.
-
- 11.E: Solutions and Colloids (Exercises)
- These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
Thumbnail: Nile red solution. Image use with permission (CC BY-SA 3.0; Armin Kübelbeck).
Contributors
-
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>).


2.4: Images Formed by Refraction
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Learning Objectives
By the end of this section, you will be able to:
- Describe image formation by a single refracting surface
- Determine the location of an image and calculate its properties by using a ray diagram
- Determine the location of an image and calculate its properties by using the equation for a single refracting surface
When rays of light propagate from one medium to another, these rays undergo refraction, which is when light waves are bent at the interface between two media. The refracting surface can form an image in a similar fashion to a reflecting surface, except that the law of refraction (Snell’s law) is at the heart of the process instead of the law of reflection.
Refraction at a Plane Interface—Apparent Depth
If you look at a straight rod partially submerged in water, it appears to bend at the surface. The reason behind this curious effect is that the image of the rod inside the water forms a little closer to the surface than the actual position of the rod, so it does not line up with the part of the rod that is above the water. The same phenomenon explains why a fish in water appears to be closer to the surface than it actually is.
To study image formation as a result of refraction, consider the following questions:
- What happens to the rays of light when they enter or pass through a different medium?
- Do the refracted rays originating from a single point meet at some point or diverge away from each other?
To be concrete, we consider a simple system consisting of two media separated by a plane interface (Figure \(\PageIndex{2}\)). The object is in one medium and the observer is in the other. For instance, when you look at a fish from above the water surface, the fish is in medium 1 (the water) with refractive index 1.33, and your eye is in medium 2 (the air) with refractive index 1.00, and the surface of the water is the interface. The depth that you “see” is the image height \(h_i\) and is called the apparent depth . The actual depth of the fish is the object height \(h_o\).
The apparent depth h i depends on the angle at which you view the image. For a view from above (the so-called “normal” view), we can approximate the refraction angle \(θ\) to be small, and replace \(\sin θ\) in Snell’s law by \(\tan θ\). With this approximation, you can use the triangles \(ΔOPR\) and \(ΔOQR\) to show that the apparent depth is given by
\[h_i= \left(\dfrac{n_2}{n_1}\right)h_o. \nonumber \]
The derivation of this result is left as an exercise. Thus, a fish appears at 3/4 of the real depth when viewed from above.
Refraction at a Spherical Interface
Spherical shapes play an important role in optics primarily because high-quality spherical shapes are far easier to manufacture than other curved surfaces. To study refraction at a single spherical surface, we assume that the medium with the spherical surface at one end continues indefinitely (a “semi-infinite” medium).
Refraction at a Convex Surface
Consider a point source of light at point P in front of a convex surface made of glass (Figure \(\PageIndex{3}\)). Let \(R\) be the radius of curvature, n 1 be the refractive index of the medium in which object point P is located, and n 2 be the refractive index of the medium with the spherical surface. We want to know what happens as a result of refraction at this interface.
Because of the symmetry involved, it is sufficient to examine rays in only one plane. The figure shows a ray of light that starts at the object point \(P\), refracts at the interface, and goes through the image point \(P′\). We derive a formula relating the object distance \(d_o\), the image distance \(d_i\), and the radius of curvature \(R\).
Applying Snell’s law to the ray emanating from point \(P\) gives
\[n_1\sin θ_1=n_2 \sin θ_2. \nonumber \]
Within the small-angle approximation
\[\sin θ≈θ, \nonumber \]
Snell’s law then takes the form
\[n_1θ_1≈n_2θ_2. \label{eq8} \]
From the geometry of Figure \(\PageIndex{3}\), we see that
\[θ_1=α+ϕ, \nonumber \]
\[θ_2=ϕ−β. \nonumber \]
Inserting both expressions into Equation \ref{eq8} gives
\[n_1(α+ϕ)≈n_2(ϕ−β). \label{eq10} \]
Using Figure \(\PageIndex{3}\), we calculate the tangent of the angles \(α\), \(β\), and \(ϕ\):
- \(\tan α≈\dfrac{h}{d_o}\)
- \(\tan β≈\dfrac{h}{d_i}\)
- \(\tan ϕ≈\dfrac{h}{R}\)
Again using the small-angle approximation, we find that \(\tan θ≈ θ\), so the above relationships become
- \(α≈\dfrac{h}{d_o}\)
- \(~β≈\dfrac{h}{d_i}\)
- \(~ϕ≈\dfrac{h}{R}.\)
Putting these angles into Equation \ref{eq10} gives
\[n_1\left(\dfrac{h}{d_o}+\dfrac{h}{R}\right)=n_2 \left(\dfrac{h}{R}−\dfrac{h}{d_i}\right). \nonumber \]
We can write this more conveniently as
\[\dfrac{n_1}{d_o}+\dfrac{n_2}{d_i}=\dfrac{n_2−n_1}{R}. \]
If the object is placed at a special point called the first focus, or the object focus \(F_1\), then the image is formed at infinity, as shown in Figure \(\PageIndex{4a}\).
We can find the location \(f_1\) of the first focus \(F_1\) by setting \(d_i=\infty\) in Equation \ref{eq20}.
\[ \begin{align} \dfrac{n_1}{f_1}+\dfrac{n_2}{\infty} &=\dfrac{n_2−n_1}{R} \\[4pt] f_1 &=\dfrac{n_1R}{n_2−n_1} \end{align} \nonumber \]
Similarly, we can define a second focus or image focus \(F_2\) where the image is formed for an object that is far away (Figure \(\PageIndex{4b}\)). The location of the second focus \(F_2\) is obtained from Equation \ref{eq20} by setting \(d_0=\infty\):
\[ \begin{align} \dfrac{n_1}{\infty}+\dfrac{n_2}{f_2}=\dfrac{n_2−n_1}{R} \\[4pt] f_2=\dfrac{n_2R}{n_2−n_1}. \end{align} \nonumber \]
Note that the object focus is at a different distance from the vertex than the image focus because \(n_1≠n_2\).
Sign convention for single refracting surfaces
Although we derived this equation for refraction at a convex surface, the same expression holds for a concave surface, provided we use the following sign convention:
- \(R>0\) if surface is convex toward object; otherwise, \(R<0\).
- \(d_i>0\) if image is real and on opposite side from the object; otherwise, \(d_i<0\).


5.1: Overview and Purpose
Overview and Purpose
This lab is designed to introduce students to the principles of genomics, DNA analysis, and data interpretation. Through hands-on lab activities and group discussion, students will gain practical skills and a deeper understanding of how genomics impacts various industries and society at large.
Learning Outcomes
- Describe how genomes can be used in science and society
- Explore and summarize the social and ethical issues around the use of genomes
- Conduct and interpret DNA Gel Electrophoresis
- Run and analyze a DNA Electrophoresis Gel
- Read how genomes are utilized
- Discuss in groups the social and ethical issues of genomics
Vocabulary
DNA
Gene
Genome
Genomics
Human Genome Project
Genetic disease
Genetic testing
Gel electrophoresis


<IP_ADDRESS>C: Identity Formation
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Identity formation is the development of an individual’s distinct personality by which he or she is recognized or known.
Learning Objectives
- Discuss the formation of a person’s identity, as well as the ideas of self-concept and self-consciousness
Key Points
- Cultural identity is the feeling of identity with a group or culture, or of an individual as far as he or she is influenced by his or her belonging to a group or culture.
- An ethnic identity is an identification with a certain ethnicity, usually on the basis of a presumed common genealogy or ancestry.
- National identity is an ethical and philosophical concept whereby all humans are divided into groups called nations.
- A religious identity is the set of beliefs and practices generally held by an individual, involving adherence to codified beliefs and rituals and study of ancestral or cultural traditions.
- Self-concept is the sum of a being’s knowledge and understanding of his or her self.
- Cultural identity is the feeling of identity of a group or culture, or of an individual as far as he or she is influenced by his or her belonging to a group or culture.
- An ethnic identity is the identification with a certain ethnicity, usually on the basis of a presumed common genealogy or ancestry.
- National identity is an ethical and philosophical concept whereby all humans are divided into groups called nations.
- A religious identity is the set of beliefs and practices generally held by an individual, involving adherence to codified beliefs and rituals and study of ancestral or cultural traditions
Key Terms
- cultural identity : One’s feeling of identity affiliation to a group or culture.
- national identity : An ethical and philosophical concept whereby all humans are divided into groups called nations.
- religious identity : The set of beliefs and practices generally held by an individual,involving adherence to codified beliefs and rituals and study of ancestral or cultural traditions, writings, history, and mythology, as well as faith and mystic experience.
Identity formation is the development of an individual’s distinct personality, which is regarded as a persisting entity in a particular stage of life by which a person is recognized or known. This process defines individuals to others and themselves. Pieces of the individual’s actual identity include a sense of continuity, a sense of uniqueness from others, and a sense of affiliation. Identity formation clearly influences personal identity by which the individual thinks of him or herself as a discrete and separate entity. This may be through individuation whereby the undifferentiated individual tends to become unique, or undergoes stages through which differentiated facets of a person’s life tend toward becoming a more indivisible whole.
Individuals gain a social identity and group identity by their affiliations. Self-concept is the sum of a being’s knowledge and understanding of himself. Self-concept is different from self-consciousness, which is an awareness of one’s self. Components of self-concept include physical, psychological, and social attributes, which can be influenced by the individual’s attitudes, habits, beliefs, and ideas. Cultural identity is one’s feeling of identity affiliation to a group or culture.
Similarly, an ethnic identity is the identification with a certain ethnicity, usually on the basis of a presumed common genealogy or ancestry. Further, national identity is an ethical and philosophical concept whereby all humans are divided into groups called nations. Members of a nation share a common identity and usually a common origin in their sense of ancestry, parentage, or descent. Lastly, a religious identity is the set of beliefs and practices generally held by an individual, involving adherence to codified beliefs and rituals and study of ancestral or cultural traditions, writings, history, and mythology, as well as faith and mystic experience.
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13.6: Hubble’s Law
The appearance of Hubble’s Law was not without controversy, either. Two years prior to Hubble’s formulation, Georges Lemaître , a Belgian Catholic priest, published a paper in the journal Annales de la Société Scientifique de Bruxelles , which detailed the relationship between distance and velocity, supporting an expanding Universe Model. Lemaître earned his doctorate in Astronomy at Massachusetts Institute of Technology, MIT, under the direction of Dr. Harlow Shapley.
It appeared, however, that astronomers wanted to acknowledge Hubble’s work instead of Lemaître. This might have been due to the fact that Lemaître’s doctorate advisor, Harlow Shapley, strongly opposed Hubble’s earlier extragalactic objects research and findings.