Question

Q3. What is the wavelength of the x-ray emitted from a Tungsten target (Z=74) when an electron moves from the M-shell (n=3) to the K-shell (n=1)

Answer 1

The change in electron's energy is

$$\Delta E=(-13.6\text{ eV})Z^2\bigg(\frac{1}{n_M^2}-\frac{1}{n_K^2}\bigg).$$

The wavelength is

$$\lambda=\frac{hc}{\Delta E}=\frac{hc}{(-13.6\text{ eV})(1.602\cdot10^{-19}\text{ J/eV})Z^2\bigg(\frac{1}{n_M^2}-\frac{1}{n_K^2}\bigg)}=\\\space\\ =444\text{ μm}.$$