Ans : (b)
$ \lambda = \large\frac{h}{p} \: \; \: \; \: $ photon energy $E = hv$
So, $ \lambda_1 = \large\frac{h}{p} = \large\frac{h}{ \sqrt {2mE}}$………..(i)
$ \Rightarrow E = hv_2 = \large\frac{hc}{ \lambda_2}$
$ \lambda_2 = \large\frac{hc}{E}$……………(ii)
From Eqns. (i) & (ii),
$ \large\frac{\lambda_1}{ \lambda_2} = \large\frac{h\sqrt {2mE}}{ \bigg( \large\frac{hc}{E} \bigg)} \sim \large\frac{E}{ \sqrt E} = E^{\large\frac{1}{2}}$