Question

In Rutherford scattering experiment, the number of $\alpha$-particles scattered at ${60}^{\circ }$ is $5\times {10}^6$. The number of $\alpha$-particles scattered at ${120}^{\circ }$ will be

• A. $15\times {10}^6$
• B. $\frac{3}{5}\times {10}^6$
• C. $\frac{5}{9}\times {10}^6$
• D. None of these

Answer 1

Answer: (C)

$\frac{5}{9}\times {10}^6$

$N\propto \frac{1}{si{n}^4\theta /2}$

$\frac{N_2}{N_1}=\frac{si{n}^4\left({\theta }_1/2\right)}{si{n}^4\left({\theta }_2/2\right)}$

$\frac{N_2}{5\times {10}^6}=\frac{si{n}^4\left({60}^{\circ }/2\right)}{si{n}^4\left({120}^{\circ }/2\right)}$

$\frac{N_2}{5\times {10}^6}=\frac{si{n}^4{30}^{\circ }}{si{n}^4{60}^{\circ }}$

$N_2=5\times {10}^6\times {\left(\frac{1}{2}\right)}^4{\left(\frac{2}{\sqrt{3}}\right)}^4=\frac{5}{9}\times {10}^6$