Explanation

(c) We know
$\lambda = \frac{h}{mv}$
and $\ \ \ \ \ \ \ \ \ K = \frac{1}{2}mv^{2} = \frac{(mv)^{2}}{2m}$
$mv = \ \sqrt{2mK}\ \ \ $
$Thus,\ \ \ \ \ \ \lambda\ \ = \frac{h}{\sqrt{2mK}}\ \ \ \ \ \ $
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \lambda \propto \frac{1}{\sqrt{K}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
$\ \therefore\ \ \ \ \ \ \ \ \ \ \ \frac{\lambda_{2}}{\lambda_{1}} = \frac{{\sqrt{K}}_{1}}{{\sqrt{K}}_{2}} = \frac{{\sqrt{K}}_{1}}{{\sqrt{2K}}_{1}}\ \ \ \ \ \ \ \ \ \ \ \ \left( \because K_{2} = 2K_{1} \right)\ $
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \frac{\lambda_{2}}{\lambda_{1}} = \frac{1}{\sqrt{2}}\ $
$\ \therefore\ \ \ \ \ \ \ \ \ \ \ \ \lambda_{2} = \frac{\lambda_{1}}{\sqrt{2}}\ $