Calculate the ratio of the wavelength of the${\mathbf{K}}_{\mathbf{\alpha }}$ line for niobium (Nb) to that for gallium (Ga) .Take needed data from the periodic table of Appendix G.

The periodic table of Appendix G is given with elements niobium and gallium.

Moseley's law states that "the square root of the frequency of an atom's emitted x-ray is proportionate to its atomic number."

Formula:

According to the Moseley’s law, we can get the relation of frequency or wavelength to atomic number as:

$f_{\infty }{\left(Z-1\right)}^2$OR $\frac{c}{\lambda }(Z-1{)}^2$ ….. (1)

The atomic number of Niobium, ${\mathrm{Z}}_{\mathrm{Nb}}=41$

The atomic number of Gallium, ${\mathrm{Z}}_{\mathrm{Ga}}=31$

Using the relation of wavelength and atomic number, the ratio of the wavelengths of the line of the elements niobium and gallium is as follows:

$\frac{{\mathrm{\lambda }}_{\mathrm{Nb}}}{{\mathrm{\lambda }}_{\mathrm{Ga}}}=\frac{{\left({\mathrm{Z}}_{\mathrm{Ga}}-1\right)}^2}{{\left({\mathrm{Z}}_{\mathrm{Nb}}-1\right)}^2}$

Substitute known values in the above equation.

$$\begin{gathered} \frac{{\mathrm{\lambda }}_{\mathrm{Nb}}}{{\mathrm{\lambda }}_{\mathrm{Ga}}}=\frac{{\left(31-1\right)}^2}{{\left(41-1\right)}^2} \\ =\frac{{30}^2}{{40}^2} \\ =\frac{9}{16} \\ =0.563 \end{gathered}$$

Hence, the value of the required ratio of the wavelengths is 0.563 .