Question

In Rutherford's alpha particle scattering experiment with thin gold foil, 8100 scintillations per minute are observed at an angle of ${60}^o$. The number of scintillations minute at an angle of ${120}^o$ 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.