Question

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

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

Answer 1

Answer: (C)

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