Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is
A. 14.4 \(\frac{Z}{V}\) Å
B. 14.4 \(\frac{\mathrm{V}}{\mathrm{Z}}\) Å
C. 1.44 \(\frac{Z}{V}\) Å
D. 1.44 \(\frac{\mathrm{V}}{\mathrm{Z}}\) Å
Answer:C
Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is
A. 14.4 \(\frac{Z}{V}\) Å B. 14.4 \(\frac{\mathrm{V}}{\mathrm{Z}}\) Å C. 1.44 \(\frac{Z}{V}\) Å D. 1.44 \(\frac{\mathrm{V}}{\mathrm{Z}}\) Å