Question

The $k_{\alpha }$ line from molybdenum (atomic number =42) has a wavelength of 0.7078 $\stackrel{o}{A}$ . The wavelength of $k_{\alpha }$ line of zinc (atomic number = 30) will be

• A. 1 $\stackrel{o}{A}$
• B. 1.3872 $\stackrel{o}{A}$
• C. 0.3541 $\stackrel{o}{A}$
• D. 0.5 $\stackrel{o}{A}$

Answer 1

Answer: (B)

$E\propto \frac{z^2}{n^2}$ and $E\propto \frac{1}{\lambda }$ $\therefore$ $\frac{Z^2}{n^2}=\frac{1}{\lambda }$ $\frac{Z_1^2}{Z_2^2}=\frac{{\lambda }_2}{{\lambda }_1}$
$\frac{{(42)}^2}{{(30)}^2}=\frac{{\lambda }^2}{0.7080}$ ${\lambda }_2=1.3872AA$