Question

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

Answer 1

The radius of $n$th orbit of H-like species is given by the expression $\frac{0.529n^2}{Z}\mathring{A}$ or $\frac{52.9n^2}{Z}$ pm.
$r_1=1.3225$ nm $=1322.5$ pm $=\frac{52.9n_1^2}{Z}$pm $......\left(1\right)$
$r_2=211.6$ pm $=\frac{52.9n_2^2}{Z}$ pm $......\left(2\right)$
Dividing $\left(1\right)$ by $\left(2\right)$, we get
$\frac{r_1}{r_2}=\frac{1322.5}{211.6}=\frac{n_1^2}{n_2^2}$
$\frac{n_1^2}{n_2^2}=6.25$ $⟹\frac{n_1}{n_2}=2.5$
$n_2=5$, $n_1=2$.

The transition is in Balmer series.
$\overline{\nu }=1.097\times {10}^7(\frac{1}{2^2}-\frac{1}{5^2})=1.097\times {10}^7\times 0.21m^{-1}$
$\lambda =\frac{1}{v}=\frac{1}{23.037\times {10}^5}=434\times {10}^{-9}$ m or $434$ nm
The transition lies in the visible region.