Chapter 4, Problem 8P

(a)

To determine

The number of $\alpha$ particles will be counted at $40°,60°,80°,\text{and 100}°$.

Answer to Problem 8P

The number of $\alpha$ particles will be counted at $40°$ is $\underset{\_}{6.64\text{cpm}}$, the number of $\alpha$ particles will be counted at $60°$ is $\underset{\_}{1.45\text{cpm}}$, the number of $\alpha$ particles will be counted at $80°$ is $\underset{\_}{0.533\text{cpm}}$, and the number of $\alpha$ particles will be counted at $100°$ is $\underset{\_}{0.264\text{cpm}}$.

Explanation of Solution

Write the relation between the number of $\alpha$ particles and the scattering angle using Equation 4.16.

    $\Delta n\infty {\left(\frac{\sin \varphi }{2}\right)}^{-4}$

Here, $\Delta n$ is the number of $\alpha$ particles detected, and $\varphi$ is the scattering angle.

Given that $100\alpha$ particles per minute are detected at $20°$.

Write the expression for $\frac{\Delta n_2}{\Delta n_1}$.

    $\frac{\Delta n_2}{\Delta n_1}=\frac{{\left(\frac{\sin {\varphi }_1}{2}\right)}^4}{{\left(\frac{\sin {\varphi }_2}{2}\right)}^4}$

Here, $\Delta n_2$ is the number of particle per minute detected at $40°$, and $\Delta n_1$ is the number of particle per minute detected at $20°$.

Rearrange the above equation.

    $\Delta n_2=\Delta n_1\frac{{\left(\frac{\sin {\varphi }_1}{2}\right)}^4}{{\left(\frac{\sin {\varphi }_2}{2}\right)}^4}$        (I)

Write the expression for $\frac{\Delta n_3}{\Delta n_1}$.

    $\frac{\Delta n_3}{\Delta n_1}=\frac{{\left(\frac{\sin {\varphi }_1}{2}\right)}^4}{{\left(\frac{\sin {\varphi }_3}{2}\right)}^4}$

Here, $\Delta n_3$ is the number of particle per minute detected at $60°$, and $\Delta n_1$ is the number of particle per minute detected at $20°$.

Rearrange the above equation.

    $\Delta n_3=\Delta n_1\frac{{\left(\frac{\sin {\varphi }_1}{2}\right)}^4}{{\left(\frac{\sin {\varphi }_3}{2}\right)}^4}$        (II)

Write the expression for $\frac{\Delta n_4}{\Delta n_1}$.

    $\frac{\Delta n_4}{\Delta n_1}=\frac{{\left(\frac{\sin {\varphi }_1}{2}\right)}^4}{{\left(\frac{\sin {\varphi }_4}{2}\right)}^4}$

Here, $\Delta n_4$ is the number of particle per minute are detected at $80°$, and $\Delta n_1$ is the number of particle per minute detected at $20°$.

Rearrange the above equation.

    $\Delta n_4=\Delta n_1\frac{{\left(\frac{\sin {\varphi }_1}{2}\right)}^4}{{\left(\frac{\sin {\varphi }_4}{2}\right)}^4}$        (III)

Write the expression for $\frac{\Delta n_5}{\Delta n_1}$.

    $\frac{\Delta n_5}{\Delta n_1}=\frac{{\left(\frac{\sin {\varphi }_1}{2}\right)}^4}{{\left(\frac{\sin {\varphi }_5}{2}\right)}^4}$

Here, $\Delta n_5$ is the number of particle per minute detected at $100°$, and $\Delta n_1$ is the number of particle per minute detected at $20°$.

Rearrange the above equation.

    $\Delta n_5=\Delta n_1\frac{{\left(\frac{\sin {\varphi }_1}{2}\right)}^4}{{\left(\frac{\sin {\varphi }_5}{2}\right)}^4}$        (IV)

Conclusion:

Substitute $100\text{cpm}$ for $\Delta n_1$, and $20°$ for ${\varphi }_1$, and $40°$ for ${\varphi }_2$ in equation (I), to find $\Delta n_2$.

    $\begin{array}{c}\Delta n_2=\left(100\text{cpm}\right)\frac{{\left(\frac{\sin 20°}{2}\right)}^4}{{\left(\frac{\sin 40°}{2}\right)}^4}\\ =6.64\text{cpm}\end{array}$

Substitute $100\text{cpm}$ for $\Delta n_1$, and $20°$ for ${\varphi }_1$, and $60°$ for ${\varphi }_2$ in equation (II), to find $\Delta n_3$.

    $\begin{array}{c}\Delta n_3=\left(100\text{cpm}\right)\frac{{\left(\frac{\sin 20°}{2}\right)}^4}{{\left(\frac{\sin 60°}{2}\right)}^4}\\ =1.45\text{cpm}\end{array}$

Substitute $100\text{cpm}$ for $\Delta n_1$, and $20°$ for ${\varphi }_1$, and $80°$ for ${\varphi }_3$ in equation (III), to find $\Delta n_4$.

    $\begin{array}{c}\Delta n_4=\left(100\text{cpm}\right)\frac{{\left(\frac{\sin 20°}{2}\right)}^4}{{\left(\frac{\sin 80°}{2}\right)}^4}\\ =0.533\text{cpm}\end{array}$

Substitute $100\text{cpm}$ for $\Delta n_1$, and $20°$ for ${\varphi }_1$, and $100°$ for ${\varphi }_4$ in equation (III), to find $\Delta n_5$.

    $\begin{array}{c}\Delta n_5=\left(100\text{cpm}\right)\frac{{\left(\frac{\sin 20°}{2}\right)}^4}{{\left(\frac{\sin 100°}{2}\right)}^4}\\ =0.264\text{cpm}\end{array}$

Therefore, the number of $\alpha$ particles will be counted at $40°$ is $\underset{\_}{6.64\text{cpm}}$, the number of $\alpha$ particles will be counted at $60°$ is $\underset{\_}{1.45\text{cpm}}$, the number of $\alpha$ particles will be counted at $80°$ is $\underset{\_}{0.533\text{cpm}}$, and the number of $\alpha$ particles will be counted at $100°$ is $\underset{\_}{0.264\text{cpm}}$.

(b)

To determine

The number of $\alpha$ particles observed at $20°$, if the kinetic energy of the incident particle are doubled.

Answer to Problem 8P

If the kinetic energy of the incident particle are doubled, then the number of $\alpha$ particles observed at $20°$ is $\underset{\_}{25\text{cpm}}$.

Explanation of Solution

It is given that $100\alpha$ particles per minute are detected at $20°$.

From Equation 4.16, it is clear that by doubling the kinetic energy of the incident particle $\left(\left(\frac{1}{2}\right)m_{\alpha }{v}_{\alpha }^2\right)$, the number of particles $\Delta n$ reduced by a factor of $4$.

Using the above condition, write the equation for $\Delta n$ at $20°$.

    $\begin{array}{c}\Delta n=\left(\frac{1}{4}\right)100\text{cpm}\\ \text{=25}\text{cpm}\end{array}$

Conclusion:

Therefore, If the kinetic energy of the incident particle are doubled, then the number of $\alpha$ particles observed at $20°$ is $\underset{\_}{25\text{cpm}}$.

(c)

To determine

The number of $\alpha$ particles observed at $20°$, if the $\alpha$ particles are incident particle on a copper foil of same thickness.

Answer to Problem 8P

The number of $\alpha$ particles observed at $20°$, if the $\alpha$ particles are incident particle on a copper foil of same thickness is $\underset{\_}{19.3\text{cpm}}$.

Explanation of Solution

Write the expression for $\frac{\Delta n_{\text{Cu}}}{\Delta n_{\text{Au}}}$.

    $\frac{\Delta n_{\text{Cu}}}{\Delta n_{\text{Au}}}=\frac{Z_{\text{Cu}}^2{N}_{\text{Cu}}}{Z_{\text{Au}}^2{N}_{\text{Au}}}$

Here, $\Delta n_{\text{Cu}}$ is the number of $\alpha$ particles scattered at copper foil, $\Delta n_{\text{Au}}$ is the number of $\alpha$ particles scattered at gold foil, $Z_{\text{Cu}}$ is the atomic number of copper, $Z_{\text{Au}}$ is the atomic number of gold, $N_{\text{Cu}}$ is the number of copper nuclei per unit area and  $N_{\text{Au}}$ is the number of gold nuclei per unit area.

Rearrange the above equation.

    $\Delta n_{\text{Cu}}=\Delta n_{\text{Au}}\left(\frac{Z_{\text{Cu}}^2{N}_{\text{Cu}}}{Z_{\text{Au}}^2{N}_{\text{Au}}}\right)$        (V)

The number of copper nuclei per unit area is same as the number of copper nuclei per unit volume multiplied with foil thickness.

Write the number of copper nuclei per unit area.

    $\begin{array}{c}{N}_{\text{Cu}}=\left[\left(8.9\text{g}/{\text{cm}}^{\text{3}}\right)\left(\frac{6.02\times {10}^{23}\text{ nuclei}}{63.54\text{ g}}\right)\right]t\\ =8.43\times {10}^{22}t\end{array}$        (VI)

Write the number of gold nuclei per unit area.

    $\begin{array}{c}{N}_{\text{Au}}=\left[\left(\text{19}\text{.3 }\text{g}/{\text{cm}}^{\text{3}}\right)\left(\frac{6.02\times {10}^{23}\text{ nuclei}}{\text{197}\text{.0}\text{g}}\right)\right]t\\ =5.90\times {10}^{22}t\end{array}$        (VII)

Conclusion:

Substitute $100\text{cpm}$ for $\Delta n_{\text{Au}}$, $29$ for $Z_{\text{Cu}}$, $79$ for $Z_{\text{Au}}$, $8.43\times {10}^{22}t$ for $N_{\text{Cu}}$, $5.90\times {10}^{22}t$ for $N_{\text{Au}}$ in equation (V), to find $\Delta n_{\text{Cu}}$.

    $\begin{array}{c}\Delta n_{\text{Cu}}=\left(100\text{cpm}\right)\left(\frac{{\left(29\right)}^2\left(8.43\times {10}^{22}t\right)}{{\left(79\right)}^2\left(5.90\times {10}^{22}t\right)}\right)\\ =\left(100\text{cpm}\right)\left(\frac{{\left(29\right)}^2\left(8.43\times {10}^{22}\right)}{{\left(79\right)}^2\left(5.90\times {10}^{22}\right)}\right)\\ =\left(100\text{cpm}\right)\left(\frac{{\left(29\right)}^2\left(8.43\right)}{{\left(79\right)}^2\left(5.90\right)}\right)\\ =19.3\text{cpm}\end{array}$

Therefore, the number of $\alpha$ particles observed at $20°$, if the $\alpha$ particles are incident particle on a copper foil of same thickness is $\underset{\_}{19.3\text{cpm}}$.