Modern Physics 4 Question 17
17. 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
(2000, 2M)
(a) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of $\approx 0.155 \AA$
(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 $\approx 0.155 \AA$ and the characteristic $X$-ray spectrum of tungsten
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Solution:
- Minimum wavelength of continuous $X$-ray spectrum is given by $\lambda _{\min }($ in $\AA)=\frac{12375}{E(\text { in } eV)}$
Here $E=$ energy of incident electrons (in $eV$ )
$=$ energy corresponding to minimum wavelength $\lambda _{\text {min }}$ of X-rays.
$E=80 keV=80 \times 10^{3} eV$
$\therefore \quad \lambda _{\text {min }}($ in $\AA)=\frac{12375}{80 \times 10^{3}} \approx 0.155$
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 energy of incident electron is high enough to knock out the electron from $K$ or $L$-shells.
