An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)
A
$$\dfrac{1}{c} \left(\dfrac{E}{2m} \right)^{1/2}$$
C
$$\dfrac{1}{c} \left(\dfrac{2E}{m} \right)^{1/2}$$
D
$$\left(\dfrac{E}{2m} \right)^{1/2}$$
Correct option is A. $$\dfrac{1}{c} \left(\dfrac{E}{2m} \right)^{1/2}$$
For electron
De - Broglie wavelength $$\lambda_c = \dfrac{h}{p}$$
where p is momentum $$p = mv$$
also by energy we have $$E = \dfrac{1}{2} mv^2$$
$$\Rightarrow E = \dfrac{1}{2} \dfrac{p^2}{m}$$
$$\Rightarrow p = \sqrt{2 mE}$$
$$\therefore \lambda_c = \dfrac{h}{\sqrt{2 mE}}$$
For photon energy $$\Rightarrow E = \dfrac{hc}{\lambda}$$
$$\Rightarrow \lambda = \dfrac{h c}{E}$$
$$\therefore \dfrac{\lambda_c}{\lambda} = \dfrac{h}{\sqrt{2 mE}} \dfrac{E}{hc}$$
$$= \dfrac{1}{C} \sqrt{\dfrac{E}{2m}}$$
option (A) is correct.