$\begin {array} {1 1} (a)\;E^0 & \quad (b)\;E^{\large\frac{1}{2}} \\ (c)\;E^{-1} & \quad (d)\;E^{-2} \end {array}$

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}}$

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