Answer
$\frac{(Z-1)^2}{(Z^{\prime}-1)^2}$
Work Step by Step
According to Moseley plot, the graph of the square root of the characteristic-emission frequencies $\sqrt f$ versus atomic number $Z$ of the target atoms is a straight line and follows the following equation:
$\sqrt f=C(Z-1)$, .......................$(1)$
where $C$ is a constant $(=4.96\times10^7\;Hz^{1/2})$
If $E$ be the photon energy due to $K_{\alpha}$ transitions, then substituting $\frac{E}{h}$ for $f$ in Eq. $1$, we obtain
$\sqrt {\frac{E}{h}}=C(Z-1)$
Therefore,
for a atom having atomic number $Z$: $\sqrt {\frac{E}{h}}=C(Z-1)$ ..................$(2)$
and the atom having atomic number $Z^{\prime}$: $\sqrt {\frac{E^{\prime}}{h}}=C(Z^{\prime}-1)$ ..................$(3)$
Dividing Eq. $2$ by Eq. $3$, we obtain
$\sqrt {\frac{E}{E^{\prime}}}=\frac{Z-1}{Z^{\prime}-1}$
or, $\frac{E}{E^{\prime}}=\frac{(Z-1)^2}{(Z^{\prime}-1)^2}$
Therefore, the ratio of the photon energies due to $K_{\alpha}$ transitions in two atoms whose atomic numbers are $Z$ and $Z^{\prime}$ is $\frac{(Z-1)^2}{(Z^{\prime}-1)^2}$
