Question

In Rutherford's experiments on $\alpha -$particle scattering, the number of particles scattered per minute at ${90}^{\circ }$ be $28$ . Then the number of particles scattered per minute by the same foil but at ${60}^{\circ }$ are-

• A. $120$
• B. $60$
• C. $112$
• D. $56$

Answer 1

The correct option is C $112$

The number of particles scattered at angle $\theta$ is,

$N\alpha \frac{1}{{\sin }^4\left(\frac{\theta }{2}\right)}$

When ${\theta }_1={90}^{\circ };N_1=28$
${\theta }_2={60}^{\circ };N_2=?$

$\frac{N_2}{N_1}={\left[\frac{\sin \left(\frac{{\theta }_1}{2}\right)}{\sin \left(\frac{{\theta }_2}{2}\right)}\right]}^4={\left[\frac{\sin 45}{\sin 30}\right]}^4$

$={\left[\frac{\left(\frac{1}{\sqrt{2}}\right)}{\left(\frac{1}{2}\right)}\right]}^4=(\sqrt{2}{)}^4=2^2$

$\Rightarrow N_2=4N_1=4\times 28=112$

Hence, $\left(C\right)$ is the correct answer.