Question

The de-Broglie wavelength of an electron is the same as that of a $50keVX-ray$ photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5MeV$)

• A. 1 : 50
• B. 1 : 20
• C. 20 : 1
• D. 50 : 1

Answer 1

Answer: (C)

de-Broglie wavelength $\lambda =\frac{h}{\sqrt{2mK}}$
The kinetic energy of the electron
$K_{\text{electron }}=\frac{1}{2m}\cdot \frac{h^2}{{\lambda }^2}...(i)$
where $h=$ Planck constant $\lambda =$ wavelength
The photon energy $E_{\text{photon }}=\frac{hc}{\lambda }...(ii)$
From Eqs. (ii) and (i), we get
$\frac{E_{\text{photon }}}{K_{\text{electron }}}=\frac{hc/\lambda }{h^2/2m\cdot {\lambda }^2}=\frac{hc\cdot {\lambda }^2\times 2m}{h^2\cdot \lambda }$
where $m=0.5MeV=5\times 10^{5}eV$
$\frac{h}{\lambda c}=50\times 10^{3}eV=\frac{2m\lambda c}{h}$
$=\frac{2m}{h/\lambda c}=\frac{2\times 5\times 10^5}{50\times 10^3}\left(\because m=\frac{h}{c\lambda }\right)$
$=20:1$