Question

The wavelengths of $K_{\alpha }X$ -rays for lead isotopes $P{b}^{208},P{b}^{206}$ and $P{b}^{204}$ are ${\lambda }_1,{\lambda }_2$ and ${\lambda }_3$ respectively. Then

• A. ${\lambda }_2=\sqrt{{\lambda }_1{\lambda }_3}$
• B. ${\lambda }_2={\lambda }_1+{\lambda }_3$
• C. ${\lambda }_2={\lambda }_1{\lambda }_3$
• D. ${\lambda }_2=\frac{{\lambda }_1}{{\lambda }_3}$

Answer 1

Answer: (A)

Wavelengths of the $K_{\alpha }$ lines for given isotopes of lead (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 $P{b}^{208},P{b}^{206}$ and $P{b}^{204}$ have different atomic masses, $Z$ will be same for them i.e. $82$ .
$\therefore \frac{1}{{\lambda }_1}=R(82-1{)}^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4}R(81{)}^2$
$\frac{1}{{\lambda }_2}=\frac{3}{4}R\left(81^2\right)$ and $\frac{1}{{\lambda }_3}=\frac{3}{4}R\left(81^2\right)$
$\Rightarrow {\left(\frac{1}{{\lambda }_2}\right)}^2=\frac{1}{{\lambda }_1}\times \frac{1}{{\lambda }_3}$
$\Rightarrow {\lambda }_2=\sqrt{{\lambda }_1{\lambda }_3}$