Question

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

Given:

Z=74

n₁=1

n₂=2

The energy is

$$\Delta E=hcR_\infty Z^2\bigg(\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg)$$

The wavelength is

$$\lambda=\frac{hc}{\Delta E}=\frac{1}{R_\infty Z^2\bigg(\frac{1}{n_2^2}-\frac{1}{n_1^2}\bigg)}=444\cdot10^{-6}\text{ m}.$$

In this equation, $R_\infty$ is the  Rydberg constant, $R_\infty=10973731.57 \text{ m}^{-1}$.