Question

The wavelength of K_α X-ray of tungsten is 21.3 pm. It takes 11.3 keV to knock out an electron from the L shell of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having tungsten target which allows production of K_α X-ray?

Answer 1

Given:
Wavelength of X-ray of tungsten, $\lambda$ = 21.3 pm
Energy required to take out electron from the L shell of a tungsten atom, E_L = 11.3 keV
Voltage required to take out electron from the L shell of a tungsten atom, V_L = 11.3 kV
Let E_K and E_L be the energies of K and L, respectively.
$$\begin{gathered} E_{\mathrm{K}}-E_{\mathrm{L}}=\frac{hc}{\lambda } \\ \mathrm{Here},h=\mathrm{Planck}\text{'}\mathrm{s}\mathrm{constant} \\ c=\mathrm{Speed}\mathrm{of}\mathrm{light} \\ {E}_{\mathrm{K}}-E_L=\frac{1242\mathrm{eV}-\mathrm{nm}}{21.3\times {10}^{-12}} \\ {E}_{\mathrm{K}}-E_L=\frac{1242\times {10}^{-9}\mathrm{eV}}{21.3\times {10}^{-12}} \\ {E}_{\mathrm{K}}-E_L=58.309\mathrm{keV} \\ {E}_{\mathrm{L}}=11.3\mathrm{keV} \\ \therefore E_{\mathrm{K}}=69.609\mathrm{keV} \end{gathered}$$
Thus, the accelerating voltage across an X-ray tube that allows the production of K_α X-ray is given by
V_K = 69.609 kV