Question

The distance between the cathode (filament) and the target in an X-ray tube is 1.5 m. If the cutoff wavelength is 30 pm, find the electric field between the cathode and the target.

Answer 1

Given:
Distance between the filament and the target in the X-ray tube, d = 1.5 m
Cut off wavelength, $\lambda$ = 30 pm
Energy $\left(E\right)$ is given by
$E=\frac{hc}{\lambda }$
Here,
h = Planck's constant
c = Speed of light
$\lambda$ = Wavelength of light

Thus, we have

$$\begin{gathered} E=\frac{1242\mathrm{eV}-\mathrm{nm}}{30\times {10}^{-3}} \\ \Rightarrow E=\frac{1242\times {10}^{-9}}{30\times {10}^{-12}} \\ \Rightarrow E=41.4\times {10}^3\mathrm{eV} \\ \mathrm{Now}, \\ \mathrm{Electric}\mathrm{field}=\frac{\mathrm{V}}{d}=\frac{41.4\times {10}^3}{1.5} \\ =27.6\times {10}^3\mathrm{V}/\mathrm{m} \\ =27.6\mathrm{kV}/\mathrm{m} \end{gathered}$$