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

The correct option is A ${\lambda }_2=\sqrt{{\lambda }_1{\lambda }_3}$

Wavelengths of $K_{\alpha }$ lines for given isotopes of lead can be given by a general expression

$\frac{1}{\lambda }=R(Z-1{)}^2(\frac{1}{1^2}-\frac{1}{2^2})$

Where, $R$ is Rydberg's constant, $Z$ is atomic number of the isotopes ($Z$ will be same for all the three isotopes, i.e $82$)

$\frac{1}{{\lambda }_1}=R(82-1{)}^2(\frac{1}{1^2}-\frac{1}{2^2})=\frac{3}{4}R(81{)}^2$

Similarly, $\frac{1}{{\lambda }_2}=\frac{1}{{\lambda }_3}=\frac{3}{4}R(81{)}^2$

$\Rightarrow (\frac{1}{{\lambda }_2}{)}^2=\frac{1}{{\lambda }_1}\times \frac{1}{{\lambda }_3}$

$\Rightarrow {\lambda }_2=\sqrt{{\lambda }_1{\lambda }_3}$