Question

A certain element emits K_α X-ray of energy 3.69 keV. Use the data from the previous problem to identify the element.

Answer 1

Given:
Energy of K_α X-ray, E = 3.69 keV = 3690 eV

Wavelength $\left(\lambda \right)$ is given by
$$\begin{gathered} \lambda =\frac{hc}{E} \\ \lambda =\frac{1242}{3690} \\ \lambda =0.33658\mathrm{nm} \\ \Rightarrow \lambda =0.34\times {10}^{-9}\mathrm{m} \end{gathered}$$

From Moseley's equation,
$\sqrt{\frac{c}{\lambda }}=a(Z-b)$
Here, c = speed of light
$\lambda$ = wavelength of light
Z = atomic number of element

On substituting the respective values,
$$\begin{gathered} \sqrt{\frac{3\times {10}^8}{0.34\times {10}^{-9}}}=5\times {10}^7(Z-1.39) \\ \Rightarrow \sqrt{8.82\times {10}^{17}}=5\times {10}^7(Z-1.39) \\ \Rightarrow 9.39\times {10}^8=5\times {10}^7(Z-1.39) \\ \Rightarrow \frac{93.9}{5}=Z-1.39 \\ \Rightarrow Z=\frac{93.9}{5}+1.39 \\ =20.17=20 \end{gathered}$$
So, the element is calcium.