Chapter 29, Problem 82P

(a)

To determine

The closest approach an $\alpha$ particle with a kinetic energy $1.0\text{ MeV}$ to the gold nucleus.

Answer to Problem 82P

The closest approach an $\alpha$ particle with a kinetic energy $1.0\text{ MeV}$ to the gold nucleus is $230\text{ fm}$.

Explanation of Solution

An $\alpha$ particle is approaching the gold nucleus with a kinetic energy $K_{\alpha }=1.0\text{ MeV}$ and the charge of $\alpha$ particle is $q_{\alpha }=2e$ and gold nucleus is $q_{\text{Au}}=79e$. The charge $e=1.602\times {10}^{-19}\text{ C}$ and the kinetic energy of $\alpha$ particle, $K_{\alpha }=U_{\text{E}}$.

Write the formula for electric potential energy

    $U_E=\frac{k{q}_{\alpha }{q}_{\text{Au}}}{r}$                                                                                     (I)

Here, $U_E$ is the electric potential energy, $k$ is Coulomb constant, $q_{\alpha }\&q_{\text{Au}}$ is the charge and $r$ is the radius.

Rewrtie equation (I) to solve for $r$

$r=\frac{k{q}_{\alpha }{q}_{\text{Au}}}{K_{\alpha }}$

Substitute $1.0\text{ MeV}$ for $K_{\alpha }$, $8.988\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$, $2e$ for $q_{\alpha }$ and $79e$ for $q_{\text{Au}}$ in above equation to find the value of $r$

$\begin{array}{l}r=\frac{8.988\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\times 2e\times 79e}{1.0\text{ MeV}}\\ r=\frac{8.988\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\times 158\times 1.602\times {10}^{-19}\text{ C}}{1.0\times {10}^6\text{ eV}\times 1.602\times {10}^{-19}\text{ C}}\\ r=230\text{ fm}\end{array}$

Thus, the closest approach an $\alpha$ particle with a kinetic energy $1.0\text{ MeV}$ to the gold nucleus is $230\text{ fm}$.

(b)

To determine

Will an $\alpha$ particle with a kinetic energy $1.0\text{ MeV}$ get close enough to touch the gold nucleus.

Answer to Problem 82P

An $\alpha$ particle with a kinetic energy $1.0\text{ MeV}$ will not get close enough to touch the gold nucleus.

Explanation of Solution

The closest approach an $\alpha$ particle with a kinetic energy $1.0\text{ MeV}$ to the gold nucleus is $230\text{ fm}$. The mass number of gold is $A_{\text{Au}}=197$ and the $\alpha$ particle is $A_{\alpha }=4$.

Write the equation for radius

    $r=r_0{A}^{1/3}$                                                                              (I)

Here, $r$ is the radius, $r_0$ is the initial radius $r_0=1.2\text{ fm}$ and $A$ is the mass number.

The radius of $\alpha$ particle is $r=1.2\text{ fm}\times 4^{1/3}=1.9\text{ fm}$

The radius of gold nucleus is $r=1.2\text{ fm}\times {197}^{1/3}=7.0\text{ fm}$

Therefore, the distance between the centers of $\alpha$ particle and gold nucleus when they just touching is $8.90\text{ fm}$. And the closest approach an $\alpha$ particle with a kinetic energy $1.0\text{ MeV}$ to the gold nucleus is $230\text{ fm}$. Thus, an $\alpha$ particle with a kinetic energy $1.0\text{ MeV}$ will not get close enough to touch the gold nucleus

(c)

To determine

The minimum value of the total kinetic energy of  $\alpha$ particle and gold nucleus necessary to bring them into contact.

Answer to Problem 82P

The minimum value of the total kinetic energy of  $\alpha$ particle and gold nucleus necessary to bring them into contact is $26\text{ MeV}$.

Explanation of Solution

The distance between the centers of proton and nitrogen nucleus when they just touching is $8.90\text{ fm}$ and the charge of $\alpha$ particle is $q_{\alpha }=2e$ and gold nucleus is $q_{\text{Au}}=79e$. The charge $e=1.602\times {10}^{-19}\text{ C}$

Substitute $8.90\text{ fm}$ for $r$, $8.988\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$, $2e$ for $q_{\alpha }$ and $79e$ for $q_{\text{Au}}$ in equation (I) to find the value of $U_E$

$\begin{array}{l}{U}_E=\frac{8.988\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\times 2e\times 79e}{8.90\text{ fm}}\\ U_E=\frac{8.988\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\times 158\times 1.602\times {10}^{-19}\text{ C}}{8.90\times {10}^{-15}\text{ m}}\times \frac{1\text{ eV}}{1.602\times {10}^{-19}\text{ J}}\\ U_E=26\text{ MeV}\end{array}$

Therefore, the minimum value of the total kinetic energy of  $\alpha$ particle and gold nucleus necessary to bring them into contact is $26\text{ MeV}$.