Question

In Rutherford's alpha particle scattering experiment with thin gold foil, 8100 scintillations per minute are observed at an angle of 600. The number of scintillations per minute at an angle 1200 will be

• A. 100
• B. 2025
• C. 32400
• D. 4050

Answer 1

The correct option is A 100

Number of scintillation per minute at an angle ${60}^0,n_1=8100m$

Number of scintillation per minute at an angle ${120}^0,n_2=?$

the scattering in the Rutherford's experiment is proportional to ${\cot }^4\frac{\varphi }{2}$

$\Rightarrow \frac{n_2}{n_1}=\frac{\frac{\cot {\varphi }_2}{2}}{\frac{\cot {\varphi }_1}{2}}$

therefore$,$

$\frac{n_2}{n_1}=\frac{{\cot }^4\left(\frac{{120}^0}{2}\right)}{{\cot }^4\left(\frac{{60}^0}{2}\right)}$

$=\frac{{\cot }^4{60}^0}{{\cot }^4{30}^0}$

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

$=\frac{1}{81}$

$n_2=\frac{1}{81}\times n_1$

$=\frac{1}{81}\times 8100=100$

Hence,

option $\left(A\right)$ is correct answer.