Question

Electrons with energy $80 \, keV$ are incident on the tungsten target of an X-ray tube. K-shell electrons of tungsten have $72.5 \, keV$ energy. X-rays emitted by the tube contain only

• A. A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of $0.155 \mathring{A}$
• B. A continuous X-ray spectrum (Bremsstrahlung) with all wavelengths
• C. The characteristic X-ray spectrum of tungsten
• D. A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of $0.155 \mathring{A}$ and the characteristic X-ray spectrum of tungsten

Answer 1

Answer: (D)

Minimum wavelength of continuous $X$ - ray spectrum is given by $\lambda \min \left(\right.$ in $\mathring{A}$ ) $=\frac{12375}{E}$
Here $E$ = energy of incident electrons (in eV) = energy corresponding to minimum wavelength $E =80 keV =80 \times 10^3 eV$ Therefore, $\lambda \min \left(\right.$ ni $\mathring{A}$ ) $=\frac{12375}{80 \times 10^3}=0.155$
Also, the energy of the incident electrons $(80 keV )$ is more than the ionization energy of the $K$ - shell electrons (i. e., $72.5$ $keV$ ), Therefore, characteristic $X$-ray spectrum will also be obtained because the energy of the incident electron is high enough to knock out the electron from $K$ or $L$ - shells.