Q. An X-ray machine is operated at 40 kV. The short wavelength limit of continuous X-rays will be $(h = 6.63 \times 10^{- 34}Js,\ c\ = \ 3 \times 10^{8}\ {ms}^{- 1},\ e = 1.6 \times 10^{- 19}\ C)$
A) $0.31\mathring{\mathrm{A}}$
B) $0.62\mathring{\mathrm{A}}$
C) $0.155\mathring{\mathrm{A}}$
D) $0.62\mathring{\mathrm{A}}$
Explanation
(a) For an accelerating voltage $V,\ $the maximum X-ray photon energy is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {hv\ }_{\max} = eV$
Also, $\gamma =$ $\frac{c}{\lambda} = \frac{velocity\ of\ light\ }{wavelength\ }$
$\therefore\ \ \ \ \ \ {\lambda\ }_{\min} = \frac{ch}{eV} = \frac{3 \times 10^{8} \times 6.63 \times 10^{- 34}}{1.6 \times 10^{- 19} \times 40 \times 10^{3}}$
= $0.31 \times 10^{- 10}$ m $\approx 0.31$ A
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.