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

Question:

Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

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.