91Ó°ÊÓ

The given data can be listed below,

• The kinetic energy of the alpha particle is,$K_{alpha}=10\mathrm{MeV}=10×{10}^6\mathrm{eV}$.
• 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,

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

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,

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

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,

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

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,

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

Substitute all the values in the final momentum equation.

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

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

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,

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

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,

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

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

The closest approach between the particle is given by,

$r=\frac{q_1{q}_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}U}$

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.

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

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