Question

If the operating potential in an X-ray tube is increased by 1%, by what percentage does the cutoff wavelength decrease?

Answer 1

Let $\lambda$ be the cut off wavelength and V be the operating potential in the X-ray tube.
Then,
$\lambda =\frac{hc}{V}$
Here,
h = Planck's constant
c = Speed of light

If the operating voltage is increased by 1%, then the new operating voltage (V') will be given by
$$\begin{gathered} V\text{'}=\mathrm{V}+\frac{1}{100}\times \mathrm{V} \\ V\text{'}=1.01\mathrm{V} \end{gathered}$$

Cut off wavelength $\left(\lambda \text{'}\right)$ on increasing the operating voltage is given by
$$\begin{gathered} \lambda \text{'}=\frac{hc}{1.01V}=\frac{\lambda }{1.01} \\ \therefore \lambda \text{'}-\lambda =\frac{0.01}{1.01}\lambda \end{gathered}$$

Percentage change in the wavelength is given by
$$\begin{gathered} \frac{0.01\lambda }{1.01\times \lambda }\times 100=\frac{1}{1.01} \\ =0.9901 \end{gathered}$$
= 1 (approx.)