Question

An X-ray tube operates at 40 kV. Suppose the electron converts 70 % of its energy into a photon at each collision. Find the lowest three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.

Answer 1

Given V = 40 KV = 40 $\times {10}^3$ V

Energy = 40 $\times {10}^3$ eV

Energy Utilised = $\frac{70}{100}\times 40\times {10}^3$

= 28 $\times {10}^3$ eV

$\lambda =\frac{hc}{E}=\frac{1242-eVnm}{28\times {10}^{3}eV}$

= 44.35 $\times {10}^{3}nm$

= 44.35 Pm

For other wavelength

E = 70 % (Left over energy)

= $\frac{70}{100}\times (40-28){10}^3$

= $\frac{70}{100}\times 12\times {10}^3$

= 84 $\times {10}^2$

$\lambda =\frac{hc}{E}=\frac{1242}{8.4\times {10}^3}$

= 147.86 $\times {10}^{-3}$ nm

= 147.86 pm = 148 pm

For third wavelength,

E = $\frac{70}{100}(12-8.4)\times {10}^3$

= 7 $\times 3.6\times {10}^2=25.2\times {10}^2$

$\lambda =\frac{hc}{E}=\frac{1242}{25.2\times {10}^2}$

= 49.2857 $\times {10}^{-2}$

= 493 pm