Question

The wavelength $\lambda$ of a photon and the de-Broglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass m, speed of light c and planck constant.

• A. $\frac{\lambda mc}{h}$
• B. $\frac{hmc}{\lambda }$
• C. $\frac{2hmc}{\lambda }$
• D. $\frac{2\lambda mc}{h}$

Answer 1

The correct option is D $\frac{2\lambda mc}{h}$

The de-Broglie wavelength
$h=\frac{h}{mv}\Rightarrow v=\frac{h}{m\lambda }$ .........(i)
Energy of photon
$E_p=\frac{hc}{\lambda }$(since $\lambda$ is same) ...........(ii)
Energy of Photon
Kinetic energy of photon $=\frac{E_p}{E_e}=\frac{hc/\lambda }{\frac{1}{2}1m{u}^2}=\frac{2hc}{\lambda m{v}^2}$
Substituting value of v from Eq. (i), we get
$\frac{E_p}{E_e}=\frac{2hc}{{\lambda }_m{\left(\frac{h}{m\lambda }\right)}^2}=\frac{2\lambda mc}{h}$.