Question

If scattering particles are $56$ for ${90}^{\circ }$ angle then this will be at ${60}^{\circ }$ angle.

• A. $224$
• B. $256$
• C. $98$
• D. $108$

Answer 1

The correct option is A $224$

According to scattering formula
$N\propto \frac{1}{{\sin }^4\left(\frac{\theta }{2}\right)}$
where, $N=\text{no. of scattering particles}$ and $\theta =\text{scattering angle.}$
Let $N_1$ be the no. of scattering particles for scattering angle
${90}^{\circ }$ and let $N_2$ be the no. of scattering particles for scattering angle ${60}^{\circ }$
So, $N_1\propto \frac{1}{{\sin }^4\left(\frac{{\theta }_1}{2}\right)}.....\left(1\right)$
also, $N_2\propto \frac{1}{{\sin }^4\left(\frac{{\theta }_2}{2}\right)}.......\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ we get,
$\Rightarrow \frac{N_2}{N_1}={\left[\frac{\sin \left(\frac{{\theta }_1}{2}\right)}{\sin \left(\frac{{\theta }_2}{2}\right)}\right]}^4$
$\Rightarrow \frac{N_2}{N_1}={\left[\frac{\sin \left(\frac{{90}^{\circ }}{2}\right)}{\sin \left(\frac{{60}^{\circ }}{2}\right)}\right]}^4$
$\Rightarrow \frac{N_2}{N_1}={\left(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\right)}^4$
$\Rightarrow \frac{N_2}{N_1}={\left(\frac{2}{\sqrt{2}}\right)}^4$
$\Rightarrow \frac{N_2}{N_1}=4$
$N_2=4N_1$
$\Rightarrow N_2=4\times 56=224$
Hence, $\left(A\right)$ is the correct answer.