Question

What potential difference should be applied across an X-ray tube to get an X-ray of wavelength not less than $0.10\text{nm}$? $(h=6.63\times {10}^{-34}\text{Js,}c=3\times {10}^8\text{m/s},e=1.6\times {10}^{-19}\text{C})$

• A. $12.4\text{kV}$
• B. $12.4\text{MV}$
• C. $12.4\text{V}$
• D. $12.4\text{eV}$

Answer 1

The correct option is A $12.4\text{kV}$

As, ${\lambda }_{mn}=\frac{hc}{eV}\Rightarrow V=\frac{hc}{e{\lambda }_{min}}$

Given,
${\lambda }_{min}=0.10\text{nm}={10}^{-10}\text{m}$

$\Rightarrow V=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times {10}^{-10}}$

$=12.43\times {10}^3\text{V}=12.4\text{kV}$.

Hence, $\left(\text{A}\right)$ is the correct answer.

Why this question? This question gives basic idea regarding the production of x-ray and cut-off wavelength.