Question

The $ {{k}_{\alpha }} $ line from molybdenum (atomic number =42) has a wavelength of 0.7078 $ \overset{o}{\mathop{\text{A}}}\, $ . The wavelength of $ {{k}_{\alpha }} $ line of zinc (atomic number = 30) will be

• A. 1 $ \overset{o}{\mathop{\text{A}}}\, $
• B. 1.3872 $ \overset{o}{\mathop{\text{A}}}\, $
• C. 0.3541 $ \overset{o}{\mathop{\text{A}}}\, $
• D. 0.5 $ \overset{o}{\mathop{\text{A}}}\, $

Answer 1

Answer: (B)

$ E\propto \frac{{{z}^{2}}}{{{n}^{2}}} $ and $ E\propto \frac{1}{\lambda } $ $ \therefore $ $ \frac{{{Z}^{2}}}{{{n}^{2}}}=\frac{1}{\lambda } $ $ \frac{Z_{1}^{2}}{Z_{2}^{2}}=\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}} $
$ \frac{{{(42)}^{2}}}{{{(30)}^{2}}}=\frac{{{\lambda }^{2}}}{0.7080} $ $ {{\lambda }_{2}}=1.3872{AA} $