Physics - Modern Physics Question with Solution | TestHub

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 . $\begin{array}{rl}& \therefore \frac{1}{{\lambda }_1}=R(82-1{)}^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4}R(81{)}^2\\ & \frac{1}{{\lambda }_2}=\frac{3}{4}R\left(81^2\right)\text{ and }\frac{1}{{\lambda }_3}=\frac{3}{4}R\left(81^2\right)\\ & \Rightarrow {\left(\frac{1}{{\lambda }_2}\right)}^2=\frac{1}{{\lambda }_1}\times \frac{1}{{\lambda }_3}\Rightarrow {\lambda }_2=\sqrt{{\lambda }_1{\lambda }_3}\end{array}$