Ans : (c)
$ \lambda=\large\frac{h}{mv} \: \: and\: \: K =\large\frac{1}{2}\: mv^2 = \large\frac{(mv)^2} {2m}$
$ \Rightarrow mv = \sqrt{2mK}$
So, $ \lambda=\large\frac{h}{\sqrt{2mK}}$
$ \Rightarrow \lambda \sim \large\frac{1}{\sqrt K}$
$ \large\frac{ \lambda_2}{\lambda_1}= \large\frac{ \sqrt{K_1}}{\sqrt{K_2}} = \large\frac{\sqrt{K_1}}{\sqrt{2K_1}}\: \: \: \: (K_2=2K_1)$
$ \Rightarrow \large\frac{\lambda_2}{ \lambda_1} = \large\frac{1}{\sqrt 2}$