Wavelengths of the $K_\alpha$ lines for given isotopes of lead $(\mathrm{Pb})$ can be given by a general expression
$\frac{1}{\lambda}=R(Z-1)^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$
where $R=$ Rydberg's constant, $Z=$ atomic number of the isotopes. Though $\mathrm{Pb}^{208}, \mathrm{~Pb}^{206}$ and $\mathrm{Pb}^{204}$ have different atomic masses, $Z$ will be same for them i.e. 82 .
$\begin{aligned} & \therefore \frac{1}{\lambda_1}=R(82-1)^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4} R(81)^2 \\ & \frac{1}{\lambda_2}=\frac{3}{4} R\left(81^2\right) \text { and } \frac{1}{\lambda_3}=\frac{3}{4} R\left(81^2\right) \\ & \Rightarrow\left(\frac{1}{\lambda_2}\right)^2=\frac{1}{\lambda_1} \times \frac{1}{\lambda_3} \Rightarrow \lambda_2=\sqrt{\lambda_1 \lambda_3}\end{aligned}$