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.

Answer 1

The radius of the $n^{th}$ orbit of hydrogen-like particles is given by,
$r=\frac{0.529n^2}{Z}Ar=\frac{52.9n^2}{Z}pm$
For radius ( $r_1$ ) = 1.3225 nm
= 1.32225 × ${10}^{–9}$ m
= 1322.25 × ${10}^{–12}$ m
= 1322.25 pm
$n_1^2=\frac{r_{1}Z}{52.9}{n}_1^2=\frac{1322.25Z}{52.9}Similarly,n_2^2=\frac{211.6Z}{52.9}\frac{n_1^2}{n_2^2}=\frac{1322.5}{211.6}\frac{n_1^2}{n_2^2}=6.25\frac{n_1}{n_2}=2.5\frac{n_1}{n_2}=\frac{25}{10}=\frac{5}{2}$
⇒ $n_1$ = 5 and $n_2$ = 2
Thus, the transition is from the 5th orbit to the 2nd orbit. It belongs to the Balmer series. wave number $\left(\overline{v}\right)$ for the transition is given by,
$1.097\times {10}^7(\frac{1}{2^2}-\frac{1}{5^2})m^{-1}=1.097\times {10}^7{m}^{-1}\left(\frac{21}{100}\right)=2.303\times {10}^6{m}^{-1}$
Wavelength (λ) associated with the emission transition is given by,
$\lambda =\frac{1}{\overline{v}}=\frac{1}{2.303\times {10}^6{m}^{-1}}=0.434\times {10}^{-6}m\lambda =434nm$