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 there wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.

Answer 1

Given:
Potential of the X-ray tube, V = 40 kV = 40 × 10³ V
Energy = 40 × 10³ eV
Energy utilised by the electron is given by
E =$\frac{70}{100}\times 40\times {10}^3$ = 28 $\times$ 10³ eV
Wavelength $\left(\lambda \right)$ is given by
$\lambda =\frac{hc}{E}$
Here,
h = Planck's constant
c = Speed of light
E = Energy of the electron
$$\begin{gathered} \therefore \lambda =\frac{hc}{E} \\ \Rightarrow \lambda =\frac{1242\mathrm{eV}-\mathrm{nm}}{28\times {10}^3\mathrm{eV}} \\ \Rightarrow \lambda =\frac{1242\times {10}^{-9}\mathrm{eV}}{28\times {10}^3\mathrm{eV}} \\ \Rightarrow \lambda =44.35\times {10}^{-12} \\ \Rightarrow \lambda =44.35\mathrm{pm} \end{gathered}$$

For the second wavelength,
E = 70% (Leftover energy)
$$\begin{gathered} =\frac{70}{100}\times (40-28){10}^3 \\ =\frac{70}{100}\times 12\times {10}^3 \\ =84\times {10}^2\mathrm{eV} \\ \mathrm{And}, \\ \lambda =\frac{hc}{E}=\frac{1242}{8.4\times {10}^3} \\ =147.86\times {10}^{-3}\mathrm{nm} \\ =147.86\mathrm{pm}=148\mathrm{pm} \end{gathered}$$

For the third wavelength,
$$\begin{gathered} E=\frac{70}{100}(12-8.4)\times {10}^3 \\ =7\times 3.6\times {10}^2=25.2\times {10}^2\mathrm{eV} \\ \mathrm{And}, \\ \lambda =\frac{hc}{E}=\frac{1242}{25.2\times {10}^2} \\ =49.2857\times {10}^{-2} \\ =493\mathrm{pm} \end{gathered}$$