The Crrect Answer is

Impact parameter $\propto$>$\cot \frac{\theta }{2}$

$\Rightarrow \sqrt{\frac{d_1}{d_2}}=\frac{\sqrt{3}}{1}$

$\Rightarrow d_1=3d_2$

$\Rightarrow x=3$

The Correct Answer is Option (C)

From Ernest Rutherford formula. Which indicates that number of particles (Y) that would be deflected by an angle ‘$\theta$’ due to scattering is
$Y_{\theta }=\frac{K}{{\sin }^4\frac{\theta }{2}}$ where K = constant

Option (C) is the correct answer.

The correct answer is option (C)

Work done to stop the $\alpha$ particle is equal to $K.E.$

$\therefore$ $qV=\frac{1}{2}m{v}^2\Rightarrow q\times \frac{K\left(Ze\right)}{r}=\frac{1}{2}m{v}^2$

$\Rightarrow r=\frac{2\left(2e\right)K\left(Ze\right)}{m{V}^2}=\frac{4KZ{e}^2}{m{v}^2}$

$\Rightarrow r\propto \frac{1}{v^2}$ and $r\propto \frac{1}{m}.$

The correct answer is option (A)

KEY NOTE :

Distance of closest approach

$r_0=\frac{Ze\left(2e\right)}{4\pi {\epsilon }_{0}E}$

Energy, $E=5\times {10}^6\times 1.6\times {10}^{-19}J$

$\therefore$ $r_0=\frac{9\times {10}^9\times \left(92\times 1.6\times {10}^{-19}\right)\left(2\times 1.6\times {10}^{-19}\right)}{5\times {10}^6\times 1.6\times {10}^{-19}}$

$\Rightarrow r=5.2\times {10}^{-14}m=5.3\times {10}^{-12}cm$