Chapter 4, Problem 51P

(a)

To determine

The fraction of the alpha particles that is scattered between $1.0°$ and $2.0°$.

Answer to Problem 51P

The fraction of the alpha particles that is scattered between $1.0°$ and $2.0°$ is $0.1176$.

Explanation of Solution

Write the expression for number of alpha particles scattered.

    $f=\pi nt{\left(\frac{Z_1{Z}_2{e}^2}{8\pi {\epsilon }_{0}K}\right)}^2{\cot }^2\left(\frac{\theta }{2}\right)$        (I)

Here, $f$ is the number of alpha particles scattered, $t$ is the thickness of foil with $n$ atoms per unit volume, $Z_{1}e$ is charge on the particle, $Z_{2}e$ is the charge on scatterer, $\theta$ is the scattering angle and $K$ is the kinetic energy.

Write the expression to find the frequency between  $1.0°$ and $2.0°$.

    $f=f\left(1°\right)-f\left(2°\right)$        (II)

Conclusion:

Substitute $2$ for $Z_1$, $\text{0}\text{.32 μm}$ for $t$, $5.90\times {10}^{28}{\text{m}}^{-3}$ for $n$ , $79$ for $Z_2$, $1°$ for $\theta$ , $8\text{MeV}$ for $K$ , $9\times {10}^9$ for $\frac{1}{4\pi {\epsilon }_0}$ and $1.6\times {10}^{-19}$ for $e$  in equation (I).

  $\begin{array}{c}f\left(1°\right)=\left(\begin{array}{l}\pi \left(5.90\times {10}^{28}{\text{m}}^{-3}\right)\left(\text{0}\text{.32 μm}\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)\\ {\left(\frac{\left(2\right)\left(79\right)\left(9\times {10}^9\right)\left(1.6\times {10}^{-19}\text{eV}\cdot \text{m}\right)}{2\left(8\times {10}^6\text{eV}\right)}\right)}^2{\cot }^2\left(\frac{1}{2}\right)\end{array}\right)\\ =\pi \left(1.888\times {10}^{22}\text{m}\right){\left(\frac{\left(2\right)\left(79\right)\left(1.44\times {10}^{-9}\text{eV}\cdot \text{m}\right)}{2\left(8\times {10}^6\text{eV}\right)}\right)}^2{\cot }^2\left(0.5\right)\\ =0.157\end{array}$

Substitute $2$ for $Z_1$, $\text{0}\text{.32 μm}$ for $t$, $5.90\times {10}^{28}{\text{m}}^{-3}$ for $n$ , $79$ for $Z_2$, $2°$ for $\theta$ , $8\text{MeV}$ for $K$ , $9\times {10}^9$ for $\frac{1}{4\pi {\epsilon }_0}$ and $1.6\times {10}^{-19}$ for $e$  in equation (I).

  $\begin{array}{c}f\left(2°\right)=\left(\begin{array}{l}\pi \left(5.90\times {10}^{28}{\text{m}}^{-3}\right)\left(\text{0}\text{.32 μm}\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)\\ {\left(\frac{\left(2\right)\left(79\right)\left(9\times {10}^9\right)\left(1.6\times {10}^{-19}\text{eV}\cdot \text{m}\right)}{2\left(8\times {10}^6\text{eV}\right)}\right)}^2{\cot }^2\left(\frac{2}{2}\right)\end{array}\right)\\ =\pi \left(1.888\times {10}^{22}\text{m}\right){\left(\frac{\left(2\right)\left(79\right)\left(1.44\times {10}^{-9}\text{eV}\cdot \text{m}\right)}{2\left(8\times {10}^6\text{eV}\right)}\right)}^2{\cot }^2\left(1\right)\\ =0.0394\end{array}$

Substitute $0.157$ for $f\left(1°\right)$ and $0.0394$ for $f\left(2°\right)$ in equation (II).

    $\begin{array}{c}f=0.157-0.0394\\ =0.1176\end{array}$

Thus, the fraction of the alpha particles that is scattered between $1.0°$ and $2.0°$ is $0.1176$.

(b)

To determine

The ratio of alpha particles scattered through angles greater than $1°$  to the number scattered through angles greater than $10°$ and angles greater than $90°$

Answer to Problem 51P

The ratio of alpha particles scattered through angles greater than $1°$  to the number scattered through angles greater than $10°$ is $100.5$ and angles greater than $90°$ is $1.31\times {10}^4$.

Explanation of Solution

At two different angles, everything will remain same except the angle, so the ratio of the number of $\alpha$  particles scattered will only depends on the scattering angle.

    $\frac{f_1}{f_2}=\frac{{\cot }^2\left(\frac{{\theta }_1}{2}\right)}{{\cot }^2\left(\frac{{\theta }_2}{2}\right)}$        (III)

Conclusion:

Substitute $1°$ for ${\theta }_1$ and $10°$ for ${\theta }_2$ in equation (III).

    $\begin{array}{c}\frac{f\left(1°\right)}{f\left(10°\right)}=\frac{{\cot }^2\left(\frac{1}{2}\right)}{{\cot }^2\left(\frac{10}{2}\right)}\\ =100.5\end{array}$

Substitute $1°$ for ${\theta }_1$ and $90°$ for ${\theta }_2$ in equation (III).

    $\begin{array}{c}\frac{f\left(1°\right)}{f\left(90°\right)}=\frac{{\cot }^2\left(\frac{1}{2}\right)}{{\cot }^2\left(\frac{90}{2}\right)}\\ =1.31\times {10}^4\end{array}$

Thus, the ratio of alpha particles scattered through angles greater than $1°$  to the number scattered through angles greater than $10°$ is $100.5$ and angles greater than $90°$ is $1.31\times {10}^4$.