Chapter 44, Problem 3P

(a)

To determine

The distance of the closest approach of alpha particle and a gold nucleus.

Answer to Problem 3P

The distance of the closest approach of alpha particle and a gold nucleus is $455\text{ fm}$.

Explanation of Solution

On a head-on collision between an alpha particle and a gold nucleus ${}_{79}^{197}\text{Au}$. The initial energy of the alpha particle is $0.55\text{ MeV}$ and during the collision the gold nucleus remains at rest. Then at the distance of the closest approach, the kinetic energy of alpha particle must be equal to the electrostatic potential energy, $K_i=U_f$.

Write to formula for electrostatic potential energy

    $U=k_e\frac{q_1{q}_2}{d}$                                                                         (I)

Write to formula for kinetic energy

    $K=\frac{1}{2}m{v}^2$                                                                         (II)

Here, $U$ is the potential energy, $K$ is the kinetic energy, $k_e$ is the electrostatic constant $\left[k_e=8.99\times {10}^9\text{N}\cdot {\text{m}}^2/C^2\right]$, $q_1\text{ and }{q}_2$ are the charge of the nucleus, $d$ is the distance of separation, $m$ is the mass of the particle and $v$ is the velocity of the particle.

Conclusion:

Here, $K_i=U_f$

$\begin{array}{c}{K}_i=U_f=\frac{k_e{q}_1{q}_2}{d}\\ d=\frac{k_e{q}_1{q}_2}{K_i}\end{array}$

Substituting $8.99\times {10}^9\text{ N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $1.60\times {10}^{-19}\text{ C}$ for $e$ and $0.500\text{ MeV}$ for $K_i$ in the above equation to find the value of $d$. [Note: $q_1=2e$, $q_2=79e$ and $1\text{ MeV}=1.60\times {10}^{-13}\text{ J}$]

$\begin{array}{c}d=\frac{\left(8.99\times {10}^9\text{ N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(2\right)\left(79\right){\left(1.60\times {10}^{-19}\text{ C}\right)}^2}{\left(0.500\text{ MeV}\right)\left(1.60\times {10}^{-13}\text{ J/MeV}\right)}\\ =4.55\times {10}^{-13}\text{ m}\\ =455\times {10}^{-15}\text{ m}\\ =455\text{ fm}\end{array}$

Thus, the distance of the closest approach of alpha particle and a gold nucleus is $455\text{ fm}$.

(b)

To determine

The initial minimum initial speed for an alpha particle to approach as closed as $300\text{ fm}$ to the gold nucleus.

Answer to Problem 3P

The initial minimum initial speed for an alpha particle to approach as closed as $300\text{ fm}$ to the gold nucleus is $6.05\times {10}^6\text{ m/s}$.

Explanation of Solution

Consider the closed distance between the alpha particle and the gold nucleus is $300\text{ fm}$ and the kinetic energy of alpha particle must be equal to the electrostatic potential energy, $K_i=U_f$. The mass of the alpha particle is $6.645\times {10}^{-27}\text{ kg}$.

Conclusion:

Equating (I) and (II),

$\begin{array}{c}{K}_i=U_f\\ \frac{1}{2}{m}_{\alpha }{v}_i^2=\frac{k_e{q}_1{q}_2}{d}\\ v_i^2=\frac{2k_e{q}_1{q}_2}{m_{\alpha }d}\\ v_i=\sqrt{\frac{2k_e{q}_1{q}_2}{m_{\alpha }d}}\end{array}$

Substituting $8.99\times {10}^9\text{ N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $1.60\times {10}^{-19}\text{ C}$ for $e$, $6.645\times {10}^{-27}\text{ kg}$ for $m_{\alpha }$ and $300\text{ fm}$ for $d$ in the above equation to find the value of $v_i$. [Note: $q_1=2e$ and $q_2=79e$]

$\begin{array}{c}{v}_i=\sqrt{\frac{2\left(8.99\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^2\right)\left(2\right)\left(79\right){\left(1.602\times {10}^{-19}\text{ C}\right)}^2}{\left(6.645\times {10}^{-27}\text{ kg}\right)\left(300\times {10}^{-15}\text{ m}\right)}}\\ =6.05\times {10}^6\text{ m/s}\end{array}$

Thus, the initial minimum initial speed for an alpha particle to approach as closed as $300\text{ fm}$ to the gold nucleus is $6.05\times {10}^6\text{ m/s}$.