Question

The de-broglie wavelength of an electron and the wavelength of a photon are same. The ratio between the energy of the photon and the momentum of the electron is

• A. h
• B. c
• C. 1/h
• D. 1/c

Answer 1

The correct option is B c

De-broglie wavelength :
$\lambda =\frac{h}{p}$

$\therefore {\lambda }_e=\frac{h}{m_{e}V}$

and for proton

${\lambda }_p=\frac{hc}{E}$

$given{\lambda }_e={\lambda }_p$

$\frac{h}{m_{e}V}=\frac{hc}{E}$

$\frac{E}{P_e}=c$

So, the answer is option (B).