The wavelength of the characteristics X-ray ${\mathrm{K}}_{\mathrm{\alpha }}$ line emitted from zinc (Z = 30) is 1.415 $\text{Å}$. Find the wavelength of the ${\mathrm{K}}_{\mathrm{\alpha }}$ line (in $\text{Å}$) emitted from molybdenum (Z = 42)

According to Moseley's law, the frequency for K series is given by

$\begin{array}{l}\nu \propto (\mathrm{Z}-1{)}^2\\ \Rightarrow \frac{\mathrm{c}}{\mathrm{\lambda }}\propto (\mathrm{Z}-1{)}^2\\ \Rightarrow \frac{1}{\mathrm{\lambda }}=\mathrm{k}(\mathrm{Z}-1{)}^2 ...\left(\mathrm{i}\right)\end{array}$

where k is a constant. Let $\mathrm{\lambda }\text{'}$ be the wavelength of ${\mathrm{K}}_{\mathrm{\alpha }}$ line emitted from molybdenum, then 

$\frac{1}{{\mathrm{\lambda }}^{\mathrm{\prime }}}=\mathrm{k}{\left({\mathrm{Z}}^{\mathrm{\prime }}-1\right)}^2 ...\left(\mathrm{ii}\right)$

Dividing (i) and (ii) we get

${\mathrm{\lambda }}^{\mathrm{\prime }}={\left(\frac{\mathrm{Z}-1}{{\mathrm{Z}}^{\mathrm{\prime }}-1}\right)}^2\mathrm{\lambda }\Rightarrow {\mathrm{\lambda }}^{\mathrm{\prime }}={\left(\frac{30-1}{42-1}\right)}^2\times 1.415\text{Å}=0.708\text{Å}$