Question

An alpha particle with kinetic energy 10MeV collides head-on with a copper nucleus $(Z = 29)$ and is deflected back. Then the minimum distance of approach between the centers of the two is:
(A) $8.4 \times {10^{ - 15}}cm$
(B) $8.4 \times {10^{ - 15}}m$
(C) $4.2 \times {10^{ - 15}}m$
(D) $4.2 \times {10^{ - 15}}cm$

Answer 1

Hint: As alpha particles and copper nuclei are positively charged, they experience a potential due to the electric field which repels them from each other. Equate this potential energy to the kinetic energy of the alpha particle.

Complete step by step solution:
An alpha particle is a helium nucleus with two protons and two neutrons. The charge of a copper nucleus is also positive. So as the alpha particle approaches the nucleus of copper, the alpha particle experiences a repulsive force due to the electric field. The alpha particle will continually experience a repulsive force that grows stronger as it approaches closer to the copper nucleus. This decreases the velocity of the alpha particle. The alpha particle finally approaches zero velocity and then accelerates in the opposite direction with the obtained potential energy. The distance where the alpha particle approaches zero velocity will be the minimum distance of approach. The kinetic energy gets converted into potential energy. So we can equate the kinetic energy of the alpha particle with the potential energy due to the electric force.
Let $KE$ be the kinetic energy of the alpha particle. Let the obtained potential energy, $PE$ be
$PE = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{{q_1}{q_2}}}{r}$
Where $r$ is the distance between the alpha particle and copper nucleus,
${q_{1,}}{q_2}$ are the total charge of the alpha particle and the copper nucleus respectively.
Equating the potential energy with the kinetic energy of the alpha particle gives us
$KE = PE$
Substitute the given value of the kinetic energy and substitute the already written equation for potential energy,
$\therefore 10MeV = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{{q_1}{q_2}}}{r}$
By changing the units into joules(multiply by electron charge), and by substituting ${q_1} = 2e$ and ${q_2} = 29e$ gives us
${10^7} \times e = 9 \times {10^9} \times {\dfrac{{2 \times 29 \times e}}{r}^2}$
Canceling one $e$ from both sides and rearranging, we get
$r = \dfrac{{9 \times {{10}^9} \times 2 \times 29 \times 1.6 \times {{10}^{ - 19}}}}{{{{10}^7}}}$
$ \Rightarrow r = 8.352 \times {10^{ - 15}}m \approx 8.4 \times {10^{ - 15}}m$
$\therefore $ the minimum distance of approach is $8.4 \times {10^{ - 15}}m$ .

$ \Rightarrow $ Option (B) is the correct answer.

Note:
To convert electron volt $(eV)$ to joules, multiply by electron charge, i.e., $e = 1.6 \times {10^{ - 19}}C$ . Note that the distance of the closest approach is different from the impact parameter. When the complete kinetic energy of a system whose nucleus and alpha particle are at rest is converted to potential energy, then the distance between the nucleus and alpha particle is called the distance of the closest approach.