Question

Electrons with energy $80\text{KeV}$ are incident on the tungsten target of an $\text{X}$-ray tube. $K$ shell electrons of tungsten have ionization energy $7.25\text{KeV}$. $\text{X}$-rays emitted by the tube contain only :

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

Answer 1

The correct option is A A continuous $\text{X}$- rays spectrum (Bremsstrahlung) with a minimum wavelength of $0.155\dot{\text{A}}$ and the characteristic $\text{X}$- rays spectrum of tungsten.

We know that,

$E=\frac{hc}{\lambda }\Rightarrow \lambda =\frac{hc}{E}$

As we know that continuous $\text{X}$- rays will always be present, so the minimum wavelength of continuous $\text{X}$- rays spectrum is given by

${\lambda }_{min}\left(\text{in}\dot{\text{A}}\right)=\frac{12400}{80\times {10}^3}=0.155$

As the energy of the incident electrons $\left(80\text{KeV}\right)$ is more than the ionization energy of the $K-$ shell electrons $(i.e.72.5\text{KeV})$. Therefore, characteristic $\text{X}$- rays spectrum will also be obtained because energy of incident electron is enough to knock out the electron from $K\text{or}L$ shells.

Therefore option $\left(\text{D}\right)$ is correct.