If the potential difference applied to an X-ray tube is doubled while keeping the separation between the filament and the target as same, what will happen to the cutoff wavelength?
Will remain same
Will be doubled
Will be halved
Will be four times of the original wavelength
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC NDA-1 – 2017
According to the Duane-Hunt law: $eV = hc/\lambda_{min}$
This equation shows that the cutoff wavelength $\lambda_{min}$ is inversely proportional to the applied potential difference $V$: $\lambda_{min} \propto 1/V$.
If the potential difference is doubled, i.e., $V’ = 2V$, the new cutoff wavelength $\lambda’_{min}$ will be:
$\lambda’_{min} = \frac{hc}{e(2V)} = \frac{1}{2} \left(\frac{hc}{eV}\right) = \frac{1}{2} \lambda_{min}$
So, the cutoff wavelength will be halved. The separation between the filament and the target does not directly affect the cutoff wavelength, which is determined by the maximum energy of the electrons reaching the target, dictated by the potential difference.