Question

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

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

Answer 1

The correct option is A $5.41\stackrel{\circ }{A}$

The wave length of $L_{\alpha }$ line is given by $\frac{1}{\lambda }=R(z-7.4{)}^2(\frac{1}{2^2}-\frac{1}{3^2})\Rightarrow \lambda \infty \frac{1}{(z-7.4{)}^2}\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{(z_2-7.4{)}^2}{(z_1-7.4{)}^2}\Rightarrow \frac{1.30}{{\lambda }_2}=\frac{(42-7.4{)}^2}{(78-7.4{)}^2}\Rightarrow {\lambda }_2=5.41\stackrel{\circ }{A}$.