Question

The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?

Answer 1

Let $\lambda$ be the initial wavelength, V be the initial potential, $\lambda \text{'}$ be the new wavelength and V' be the new operating voltage when the operating voltage is increased in the X-ray tube.
Given:
$\lambda \text{'}$ = $\lambda$$-$26 pm
V' = 1.5 V
Energy $\left(E\right)$ is given by
$E=\frac{hc}{\lambda }$
$\Rightarrow eV=\frac{hc}{\lambda }$
Here,
h = Planck's constant
c = Speed of light
$\lambda$ = Wavelength of light
V = Operating potential

$$\begin{gathered} \therefore \lambda =\frac{hc}{eV} \\ \Rightarrow \lambda V=\lambda \text{'}\mathrm{V}\text{'}\left[\because \mathrm{\lambda }\propto \frac{1}{V}\right] \\ \Rightarrow \lambda V=\left(\lambda -26\right)\times 1.5\mathrm{V} \\ \Rightarrow 0.5\lambda =26\times 1.5 \\ \Rightarrow \lambda =26\times 3 \\ \Rightarrow \lambda =78\mathrm{pm} \end{gathered}$$
Hence, the initial wavelength is 78$\times$10^$-$12 m.

Now, the operating voltage (V) is given by
$$\begin{gathered} V=\frac{hc}{e\lambda } \\ \Rightarrow V=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 78\times {10}^{-12}} \\ \Rightarrow V=0.15937\times {10}^5 \\ \Rightarrow V=15.9\mathrm{kV} \end{gathered}$$