Q. The $X$-ray wavelength of $L_{\alpha}$ line of platinum $(Z = 78)$ is $1.30\mathring{\mathrm{A}}.$ The $X$-ray wavelength of $L_{\alpha}$ line of Molybdenum $(Z = 42)$ is
A) $5.41\ \mathring{\mathrm{A}}$
B) $4.20\ \mathring{\mathrm{A}}$
C) $2.70\ \mathring{\mathrm{A}}$
D) $1.35\ \mathring{\mathrm{A}}$
Explanation
(a) The wave length of $L_{\alpha}$ line is given by
$\frac{1}{\lambda} = R(z - 7.4)^{2}\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)$
$\Rightarrow \lambda \propto \frac{1}{(Z - 7.4)^{2}}$
$\Rightarrow \frac{\lambda_{1}}{\lambda_{2}} = \frac{\left( z_{2} - 7.4 \right)^{2}}{\left( z_{1} - 7.4 \right)^{2}} \Rightarrow \frac{1.30}{\lambda_{2}} = \frac{(42 - 7.4)^{2}}{(78 - 7.4)^{2}}$
$\Rightarrow \lambda_{2} = 5.41\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.