An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out with kinetic energy of 10 MeV (10 × 106 eV), and heads in the +x direction straight toward a gold nucleus (containing 79 protons and 118 neutrons). The particles are initially far apart, and the gold nucleus is initially at rest. Assuming that all speeds are small compared to the speed of light, answer the following questions about the collision. (a) What is the final momentum of the alpha particle, long after it interacts with the gold nucleus? (b) What is the final momentum of the gold nucleus, long after it interacts with the alpha particle? (c) What is the final kinetic energy of the alpha particle? (d) What is the final kinetic energy of the gold nucleus? (e) Assuming that the movement of the gold nucleus is negligible, calculate how close the alpha particle will get to the gold nucleus in this head-on collision.

The given data can be listed below,

• The kinetic energy of the alpha particle is,$K_{alpha}=10\mathrm{MeV}=10\times {10}^6\mathrm{eV}$.
• The mass of the alpha particle is$m_1=4\mathrm{u}$ .
• Mass of the gold nucleus $m_2=197\mathrm{u}$.
• Initial velocity of the gold nucleus is,$u_2=0$.

The kinetic energy of the objects is also conserved along with their momentum; then, it is said to be an elastic collision.

The collision takes place between two or more objects. When objects collide, each object feels a small amount of force for a short period.

The initial velocity of the alpha particle is given by,

$$\begin{gathered} K_{alpha}=\frac{1}{2}{m}_1{u}_1^2 \\ {u}_1=\sqrt{\frac{2K_{alpha}}{m_1}} \end{gathered}$$

Here,$K_{alpha}$is the kinetic energy of the alpha particle and$m_1$is the mass of the alpha particle.

Substitute values in the above,

$$\begin{gathered} u_1=\sqrt{\frac{2\times \left({10}^{7}eV\right)\left({\displaystyle \frac{1.6\times {10}^{-19}\mathrm{J}}{1eV}}\right)}{4u\left({\displaystyle \frac{1.665\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}}\right)}} \\ =2.19\times {10}^7\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The final velocity of the alpha particle for the elastic collision of two particles is given by,

$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2{m_2}_{}}{m_1+m_2}\right)u^{}$

Here, $m_1$is the mass of the alpha particle, $m_2$is the mass of the gold particle, $u_1$is the initial velocity of the alpha particle and $u_2$is the initial velocity of the gold nucleus.

Substitute values in the above,

$$\begin{gathered} v_1=\left[\left(\frac{4\mathrm{u}-197\mathrm{u}}{4\mathrm{u}+197\mathrm{u}}\right)\right]\left(2.19\times {10}^7\mathrm{m}/\mathrm{s}\right)+\left[\left(\frac{2\times 197\mathrm{u}}{4\mathrm{u}+197\mathrm{u}}\right)\right]\times 0 \\ =\left(-0.96\right)\left(2.19\times {10}^7\mathrm{m}/\mathrm{s}\right) \\ =-2.1\times {10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the final momentum of the alpha particle is given by,

$p_1=m_1{v}_1$

Here, the$m_1$ is the mass of the alpha particle and$v_1$ is the final velocity of the particle.

Substitute all the values in the above,

$$\begin{gathered} p_1=\left(4\mathrm{u}\right)\left(-2.19\times {10}^7\mathrm{m}/\mathrm{s}\right)\left(\frac{1.6605\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}\right) \\ =-1.40\times {10}^{-19}\mathrm{kg}.\mathrm{m}/\mathrm{s} \\ \mathrm{Thus},\mathrm{the}\mathrm{momentum}\mathrm{of}\mathrm{the}\mathrm{alpha}\mathrm{particle}\mathrm{is}-1.40\times {10}^{-19}\mathrm{kg}.\mathrm{m}/\mathrm{s}. \end{gathered}$$

The final momentum of the gold nucleus is given by,

$p_2=m_2{v}_2$

Here, $m_2$is the mass of the gold particle and $v_2$is the final velocity of the gold nucleus.

The initial velocity of the gold nucleus is zero. The final velocity of the gold nucleus is given by,

$v_2=\left(\frac{2m_1}{m_1+m_2}\right)u_1+\left(\frac{m_2-m_1}{m_1+m_2}\right)u_2$

Here,$m_1$is the mass of the alpha particle,$m_2$is the mass of the gold particle,$u_1$is the initial velocity of the alpha particle and$u_2$is the initial velocity of the gold nucleus.

Substitute values in the above,

$$\begin{gathered} v_2=\left[\left(\frac{2\left(4\mathrm{u}\right)}{4\mathrm{u}+197\mathrm{u}}\right)\right]\left(2.19\times {10}^7\mathrm{m}/\mathrm{s}\right)+\left[\left(\frac{197\mathrm{u}-4\mathrm{u}}{4\mathrm{u}+197\mathrm{u}}\right)\times 0\mathrm{m}/\mathrm{s}\right] \\ =0.087\times {10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$

Substitute all the values in the final momentum equation.

$$\begin{gathered} p_2=\left(197\mathrm{u}\right)\left(0.087\times {10}^7\mathrm{m}/\mathrm{s}\right)\left(\frac{1.6605\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}\right) \\ =2.85\times {10}^{-19}\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the final momentum of the gold nucleus is $2.85\times {10}^{-19}\mathrm{kg}.\mathrm{m}/\mathrm{s}$.

The final kinetic energy of the alpha particle is given by,

$K_{finalalpha}=\frac{p_1^2}{2m_1}$

Here, $p_1$is the final momentum of the alpha particle and $m_1$is the mass of the alpha particle.

Substitute all the values in the above,

$$\begin{gathered} K_{finalalpha}=\frac{{\left(-1.40\times {10}^{-19}\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)}^2}{2\times \left(4\mathrm{u}\right)\left({\displaystyle \frac{1.6605\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}}\right)\left(1.602\times {10}^{-19}\mathrm{eV}\right)} \\ =9.21\mathrm{Mev} \end{gathered}$$

Thus, the final kinetic energy of the alpha particle is $9.21\mathrm{Mev}$.

The final kinetic energy of the gold nucleus is given by,

$K_{finalgold}=\frac{p_2^2}{2m_2}$

Here, $m_2$is the mass of the gold nucleus and$p_2$ is the momentum of the gold nucleus.

Substitute values in the above,

$$\begin{gathered} K_{finalgold}=\frac{{\left(2.85\times {10}^{-19}\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)}^2}{2\times \left(197\mathrm{u}\right)\left({\displaystyle \frac{1.6605\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}}\right)\left(1.602\times {10}^{-19}\mathrm{eV}\right)} \\ =0.78\mathrm{MeV} \end{gathered}$$

Thus, the final kinetic energy of the gold nucleus is $0.78\mathrm{MeV}$.

The closest approach between the particle is given by,

$r=\frac{q_1{q}_2}{4\pi {\epsilon }_{0}U}$

Here,$q_1$ is the charge in alpha particle whose value is 2e, $q_2$is the charge on gold nucleus whose value is given by 79e,and U is the potential energy which is converted from kinetic energy just before deflection of alpha particle.

Substitute all the values in the above equation.

$$\begin{gathered} r=\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left[\frac{\left(2\times 1.602\times {10}^{-19}\mathrm{C}\right)\left(79\times 1.60\times {10}^{-19}\mathrm{C}\right)}{\left(9.21\mathrm{MeV}\right)\left({\displaystyle \frac{1.60\times {10}^{-19}\mathrm{J}}{1\mathrm{eV}}}\right)}\right] \\ =2.47\times {10}^{-14}\mathrm{m} \end{gathered}$$

Thus, the closest approach of the alpha particle is $2.47\times {10}^{-14}\mathrm{m}$.