Answer
$\frac{\lambda_{Nb}}{\lambda_{Ga}}\approx0.563$
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 $(=4.96\times10^7\;Hz^{1/2})$
Substituting $\frac{c}{\lambda}$ for $f$ in Eq. $1$, we obtain
$\sqrt {\frac{c}{\lambda}}=C(Z-1)$
Therefore,
for a niobium (Nb) $(Z=41)$ target: $\sqrt {\frac{c}{\lambda_{Nb}}}=C(41-1)$ ..................$(2)$
and for a gallium (Ga) $(Z=31)$ target: $\sqrt {\frac{c}{\lambda_{Ga}}}=C(31-1)$ ..................$(3)$
Dividing Eq. $3$ by Eq. $2$, we obtain
$\sqrt {\frac{\lambda_{Nb}}{\lambda_{Ga}}}=\frac{31-1}{41-1}$
or, $\frac{\lambda_{Nb}}{\lambda_{Ga}}=\frac{30^2}{40^2}$
or, $\frac{\lambda_{Nb}}{\lambda_{Ga}}=\frac{9}{16}$
or, $\frac{\lambda_{Nb}}{\lambda_{Ga}}\approx0.563$
