The K_α X-rays of aluminium ( Z = 13 ) and zinc ( Z = 30 ) possess...

According to Moseley's law for $K_\alpha$ X-rays, the frequency $\nu$ is given by $\sqrt{\nu} = a(Z - b)$.

For $K_\alpha$ lines, the screening constant $b \approx 1$. The energy of the emitted X-ray is given by:

$E = h\nu = Rhc(Z-1)^2$\left(\dfrac{1}{1^2} - \dfrac{1}{2^2}\right)

Substituting $Rhc \approx 13.6 \text{ eV}$:

$E \approx 13.6 \times$\dfrac{3}{4} ($Z-1)^2 \text{ eV} = 10.2 (Z-1)^2 \text{ eV}$

Given the energy of the unknown element's $K_\alpha$ X-ray is $E = 3.69 \text{ keV} = 3690 \text{ eV}$:

$3690 = 10.2 (Z-1)^2$

$(Z-1)^2 =$\dfrac{3690}{10.2} \approx$361.76$

$Z - 1 \approx$\sqrt{361.76} \approx$19.02$

$Z \approx 20$

The atomic number $Z = 20$ corresponds to Calcium.

(Note: Using the given wavelengths for Al and Zn to find the exact empirical values of $a$ and $b$ also yields $Z \approx 20.2$, confirming the element is Calcium).