Question

In gold foil experiment if number deflected of $\alpha$-particle with respect to $120^\circ$ are 12 then calculate number of deflected $\alpha$-particle with respect to $60^\circ$-

• A. 36
• B. 108
• C. 24
• D. None of these

Answer 1

Answer: (B)

108

Answer 2

The number of $\alpha$-particles scattered at an angle $\theta$ in Rutherford's gold foil experiment is given by the Rutherford scattering formula:

$N(\theta) \propto \frac{1}{\sin^4(\frac{\theta}{2})}$

Given:

Number of $\alpha$-particles deflected at $\theta_1 = 120^\circ$ is $N_1 = 12$.
We need to calculate the number of $\alpha$-particles deflected at $\theta_2 = 60^\circ$, let's call it $N_2$.

Using the proportionality relation for the two angles:

$\frac{N_2}{N_1} = \frac{\frac{1}{\sin^4(\frac{\theta_2}{2})}}{\frac{1}{\sin^4(\frac{\theta_1}{2})}}$

$\frac{N_2}{N_1} = \frac{\sin^4(\frac{\theta_1}{2})}{\sin^4(\frac{\theta_2}{2})}$

Now, substitute the given angles:

$\theta_1 = 120^\circ \implies \frac{\theta_1}{2} = 60^\circ$

$\theta_2 = 60^\circ \implies \frac{\theta_2}{2} = 30^\circ$

Calculate the sine values:

$\sin(60^\circ) = \frac{\sqrt{3}}{2}$

$\sin(30^\circ) = \frac{1}{2}$

Substitute these values into the ratio:

$\frac{N_2}{12} = \frac{\left(\frac{\sqrt{3}}{2}\right)^4}{\left(\frac{1}{2}\right)^4}$

$\frac{N_2}{12} = \frac{\frac{(\sqrt{3})^4}{2^4}}{\frac{1^4}{2^4}}$

$\frac{N_2}{12} = \frac{(\sqrt{3})^4}{1^4}$

Since $(\sqrt{3})^4 = (3^{1/2})^4 = 3^2 = 9$:

$\frac{N_2}{12} = \frac{9}{1}$

$N_2 = 12 \times 9$

$N_2 = 108$

Thus, the number of deflected $\alpha$-particles with respect to $60^\circ$ is 108.