Question

The wavelength of $K_{\alpha} X$-rays for lead isotopes $Pb ^{208}, Pb ^{206}, Pb ^{204}$ 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 $\lambda_{1}=\lambda_{2}=\lambda_{3}$