Question

In an X-ray tube, electrons emitted from a filament (cathode) carrying current $i$ hit a target (anode) at a distance $d$ from the cathode. The target is kept at a potential $V$ higher than the cathode resulting in emission of continuous and characteristic X-rays. If the filament current $i$ is decreased to $\frac{i}{2}$, the potential difference $V$ is increased to $2V$, and the separation distance $d$ is reduced to $\frac{d}{2}$, then

• A. the cut-off wavelength will reduce to half, and the wavelengths of the characteristic X-rays will remain the same
• B. the cut-off wavelength as well as the wavelengths of the characteristic X-rays will remain the same
• C. the cut-off wavelength will reduce to half, and the intensity of all the X-rays will decrease
• D. the cut-off wavelength will become two times larger, and the intensity of all the X-rays will decrease

Answer 1

Answer: (A), (C)

the cut-off wavelength will reduce to half, and the intensity of all the X-rays will decrease

From the energy conservation on accelerated beam for minimum wavelength,
$\frac{hc}{\lambda }=eV$
${\lambda }_{\text{min}}=\frac{hc}{eV}$
$\Rightarrow {\lambda }_{\text{min}}\propto \frac{1}{V}$
$\Rightarrow \frac{({\lambda }_{\text{min}}{)}_i}{({\lambda }_{\text{min}}{)}_f}=\frac{V_f}{V_i}=2$
$\Rightarrow {\lambda }_f=\frac{{\lambda }_i}{2}$
Hence cutoff wavelegth is reduced to half.
On the other hand, the characteristic wavelength of X-ray does not depend on voltage.

since current is reducing to half that means the rate of emission of photoelectrons has decreased.

$\Rightarrow I\propto \frac{dN}{dt}$
$\because \frac{dN}{dt}$ decreases ,hence Intensity$\left(I\right)$ decreases