Question

The X-ray wavelength of $L_{\alpha }$ line of platinum $(Z=78)$ is $1.30\dot{A}$. The X –ray wavelength of $L_{\alpha }$ line of Molybdenum (Z=42) is

• A. $5.41\dot{A}$
• B. $4.20\dot{A}$
• C. $2.70\dot{A}$
• D. $1.35\dot{A}$

Answer 1

The correct option is A $5.41\dot{A}$

The wave length of $L_{\alpha }$ line is given by
$\frac{1}{\lambda }=R(Z-7.4{)}^2(\frac{1}{2^2}-\frac{1}{3^2})\Rightarrow \lambda \propto \frac{1}{(Z-7.4{)}^2}$
$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{(Z_2-7.4{)}^2}{(Z_1-7.4{)}^2}\Rightarrow \frac{1.30}{{\lambda }_2}=\frac{(42-7.4{)}^2}{(78-7.4{)}^2}\Rightarrow {\lambda }_2=5.41\dot{A}$