Question

The de-Broglie wavelength of an electron is the same as that of a $50keV$ 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 $0.5MeV$):

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

Answer 1

The correct option is C $20:1$

Energy of X-ray photon $E_{photon}=50keV=5\times {10}^{4}eV$

Energy equivalent of electron mass is $0.5MeV$ i.e $m{c}^2=0.5MeV=5\times {10}^{5}eV$ ......(1)

Wavelength of the photon ${\lambda }_{photon}=\frac{hc}{E}=\frac{1242}{5\times {10}^4}$ $nm=0.0248$ $nm$

$\therefore$ de-Broglie wavelength of the electron $\lambda ={\lambda }_{photon}=0.0248$ $nm$

Kinetic energy of electron $E_{electron}=\frac{h^2}{2m\times {\lambda }^2}=\frac{h^2{c}^2}{2\left(m{c}^2\right){\lambda }^2}$ $(\because E=\frac{p^2}{2m})$

Kinetic energy of electron $E_{electron}=\frac{(1242{)}^2}{2(5\times {10}^5)\times (0.0248{)}^2}$ $eV=2.51\times {10}^{3}eV$

$\therefore$ $\frac{E_{photon}}{E_{electron}}=\frac{5\times {10}^4}{2.51\times {10}^3}=20$