Question

The intensity of x-rays from Coolidge tube is plotted against wavelength $\lambda$ for a molybdenum target at an accelerating voltage of 35 kV. The minimum wavelength found is ${\lambda }_0$ and the wavelength of the $K_{\alpha }$ line is ${\lambda }_{K\alpha }$. As the accelerating voltage is increased from 35 kV to 70 kV -

• A. $({\lambda }_{K\alpha }-{\lambda }_0)$ increases
• B. $({\lambda }_{K\alpha }-{\lambda }_0)$ decreases
• C. ${\lambda }_{K\alpha }$ increases.
• D. ${\lambda }_{K\alpha }$ decreases.

Answer 1

The correct option is A $({\lambda }_{K\alpha }-{\lambda }_0)$ increases

The $K_{\alpha }$ peak corresponds to radiation when the atom transitions between two energy levels due to an internal jump of an electron from the Lshell to the Kshell. It is characteristic for a particular element and will not depend on any external parameters, like the accelerating potential.
But increasing the accelerating potential from 35kV to 70kV will now allow x-rays carrying energies in the range 35-70 eV, which was not allowed earlier. There will now be a higher cut-off on the energy, and hence, a lower cutoff on the wavelength (since $E\infty \frac{1}{\lambda })$. The new cut-off wavelength ${\lambda }_0$ will be lower than ${\lambda }_0$, and farther from ${\lambda }_{K\alpha }$.

Thus, among the given options, the only thing that will change when you increase the accelerationg potential is that $({\lambda }_{K\alpha }-{\lambda }_0)$ will increase.