Question

Electrons with energy 80 keV are incident on the tungsten
target of an X-ray tube. X-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 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 = 0.155A and the characteristic X-ray spectrum of tungsten

Answer 1

Answer: (D)

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