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An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelength associated with them is (c being velocity of light)
A
$\left(\frac{E}{2m}\right)^{\frac{1}{2}}$
B
$c(2mE)^{\frac{1}{2}}$
C
$\frac{1}{c}\left(\frac{2m}{E}\right)^{\frac{1}{2}}$
D
$\frac{1}{c}\left(\frac{E}{2m}\right)^{\frac{1}{2}}$
Explanation
Electron: $h/\sqrt{2mE}$; photon: $hc/E$.
Detailed Solution
For the electron: $\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$ ...(iii) (since $E = \frac{p^2}{2m}$)
For the photon: $E = \frac{hc}{\lambda_p} \Rightarrow \lambda_p = \frac{hc}{E}$ ...(iv)
$\frac{\lambda_e}{\lambda_p} = \frac{h/\sqrt{2mE}}{hc/E} = \frac{E}{c\sqrt{2mE}} = \frac{1}{c}\left(\frac{E}{2m}\right)^{\frac{1}{2}}$
For the photon: $E = \frac{hc}{\lambda_p} \Rightarrow \lambda_p = \frac{hc}{E}$ ...(iv)
$\frac{\lambda_e}{\lambda_p} = \frac{h/\sqrt{2mE}}{hc/E} = \frac{E}{c\sqrt{2mE}} = \frac{1}{c}\left(\frac{E}{2m}\right)^{\frac{1}{2}}$