Question

What potential difference should be applied across an X-ray tube to get X-ray of wavelength not less than 0.10 nm? What is the maximum energy of a photon of this X-ray in joule?

Answer 1

Given:
Wavelength of the X-ray, $\lambda$ = 0.10 nm
Planck's constant, h = 6.63$\times$10^$-34$ J-s
Speed of light, c = 3$\times {10}^8$ m/s

Minimum wavelength is given by
$$\begin{gathered} {\lambda }_{\min }=\frac{hc}{eV} \\ \Rightarrow V=\frac{hc}{e{\lambda }_{\min }} \\ \Rightarrow V=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times {10}^{-10}} \\ \Rightarrow V=12.43\times {10}^3\mathrm{V}=12.4\mathrm{kV} \end{gathered}$$

Maximum energy of the photon $\left(E\right)$ is given by
E = $\frac{hc}{\lambda }$
$$\begin{gathered} \Rightarrow E=\frac{6.68\times {10}^{-34}\times 3\times {10}^8}{{10}^{-10}} \\ \Rightarrow E=19.89\times {10}^{-16} \\ \Rightarrow E=1.989\times {10}^{-15}\approx 2\times {10}^{-15}\mathrm{J} \end{gathered}$$