Question

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

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

Answer 1

Answer: (D)

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