# The energy of a photon is equal to the KE of a proton. The energy of the photon is E. Let $\lambda_1$ be the de- Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. So, $\large\frac{\lambda_1}{\lambda_2}$ is

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

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),
$\large\frac{\lambda_1}{ \lambda_2} = \large\frac{h\sqrt {2mE}}{ \bigg( \large\frac{hc}{E} \bigg)} \sim \large\frac{E}{ \sqrt E} = E^{\large\frac{1}{2}}$

edited Mar 13, 2014