Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm.

Solution:

The radius of the $n^{\text {th }}$ orbit of hydrogen-like particles is given by,

$r=\frac{0.529 n^{2}}{Z} A$

$r=\frac{52.9 n^{2}}{Z} \mathrm{pm}$

For radius $\left(r_{1}\right)=1.3225 \mathrm{~nm}$

$=1.32225 \times 10^{-9} \mathrm{~m}$

$=1322.25 \times 10^{-12} \mathrm{~m}$

$=1322.25 \mathrm{pm}$

$n_{1}^{2}=\frac{r_{1} Z}{52.9}$

$n_{1}^{2}=\frac{1322.25 Z}{52.9}$

Similarly,

$n_{2}^{2}=\frac{211.6 Z}{52.9}$

$\frac{n_{1}^{2}}{n_{2}^{2}}=\frac{1322.5}{211.6}$

$\frac{n_{1}^{2}}{n_{2}^{2}}=6.25$

$\frac{n_{1}}{n_{2}}=2.5$

$\frac{n_{1}}{n_{2}}=\frac{25}{10}=\frac{5}{2}$

$\Rightarrow n_{1}=5$ and $n_{2}=2$

Thus, the transition is from the $5^{\text {th }}$ orbit to the $2^{\text {nd }}$ orbit. It belongs to the Balmer series.

Wave number $(\bar{v})$ for the transition is given by,

$1.097 \times 10^{7} \mathrm{~m}^{-1}\left(\frac{1}{2^{2}}-\frac{1}{5^{2}}\right)$

$=1.097 \times 10^{7} \mathrm{~m}^{-1}\left(\frac{21}{100}\right)$

$=2.303 \times 10^{6} \mathrm{~m}^{-1}$

$\therefore$ Wavelength $(\lambda)$ associated with the emission transition is given by,

$\lambda=\frac{1}{\bar{v}}$

$=\frac{1}{2.303 \times 10^{6} \mathrm{~m}^{-1}}$

$=0.434 \times 10^{-6} \mathrm{~m}$

$\lambda=434 \mathrm{~nm}$

This transition belongs to Balmer series and comes in the visible region of the spectrum.