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 d = 1.5 m

$\lambda =30pm=30\times {10}^{-3}$ nm

E = $\frac{hc}{\lambda }=\frac{1242}{30\times {10}^{-3}}$

= 4.14 $\times {10}^3$ eV

Electric field = $\frac{V}{d}=\frac{41.4\times {10}^3}{1.5}$

= 27.6 $\times {10}^3$ V/m

= 27.6 KV/m