Question

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

• A. ${\lambda }_1={\lambda }_2={\lambda }_3$
• B. ${\lambda }_3=\sqrt{{\lambda }_1\cdot {\lambda }_2}$
• C. ${\lambda }_1<{\lambda }_2<{\lambda }_3$
• D. Both(1) & (2)

Answer 1

Answer: (D)

Value of $\lambda$ here depends on atomic number. Since $z$ is same for all species hence ${\lambda }_1={\lambda }_2={\lambda }_3$