Q. Electron with energy 80 keV are incident on the tungsten target of a $X$-rays tube. $K$ shell electrons of tungsten have $- 72.5$ keV energy. $X$-rays emitted by the tube contain only
A) A continuous $X$-rays spectrum (Bremsstrahlung) with a minimum wavelength of $\sim 0.155\mathring{\mathrm{A}}$
B) A continuous $X$-ray spectrum (Bremsstrahlung) with all wavelengths
C) The characteristic $X$-rays spectrum of tungsten
D) A continuous $X$-rays spectrum (Bremsstrahlung) with a minimum wavelength of $\sim 0.155\ \mathring{\mathrm{A}}$ and the characteristic $X$-rays spectrum of tungsten
Explanation
(d) Since the energy of incident electron, $E = 80$ keV. The minimum wavelength of $X$-rays produced is
$\lambda = \frac{hc}{E} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{80 \times 1000 \times 1.6 \times 10^{- 19}}$
$= 1.55 \times 10^{- 10}\ m = 0.155\ \mathring{\mathrm{A}}$
Since the energy of K-shell electron is $- 72.5$ keV, so the incident electron of energy 80 keV will not only produce continuous spectrum of minimum wavelength 0.155$\mathring{\mathrm{A}}$ but shell also knock electron of $K$ shell out of atom, resulting emission of characteristics $X$-rays
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.