What would be the wavelength and name of series respectively for the emission transition for H -atom

Question

What would be the wavelength and name of series respectively for the emission transition for H -atom if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm ?\

Professor PC Thomas Classes · Answer

(a) : Radius of $n^{	ext {th }}$ orbit $=\frac{0.529 	imes n^{2}}{Z} \AA=\frac{52.9 	imes n^{2}}{Z} \mathrm{pm}$ 
Radius $\left(r_{2}ight)=1.3225 \mathrm{~nm}=1322.5 \mathrm{pm}=\frac{52.9 n_{2}^{2}}{Z}$\ Radius $\left(r_{1}ight)=211.6 \mathrm{pm}=\frac{52.9 n_{1}^{2}}{Z} \quad 	herefore \frac{r_{2}}{r_{1}}=\frac{1322.5}{211.6}=\frac{n_{2}^{2}}{n_{1}^{2}}$\ $\Rightarrow \quad 6.25=\frac{n_{2}^{2}}{n_{1}^{2}}=\left(\frac{n_{2}}{n_{1}}ight)^{2} \Rightarrow \frac{n_{2}}{n_{1}}=\sqrt{6.25}=2.5$\ $	herefore \quad n_{1}=2, n_{2}=5$ thus, the transition is from $5^{	ext {th }}$ orbit to $2^{	ext {nd }}$ orbit. It belongs to Balmer series.\ $\bar{v}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}ight) \Rightarrow \bar{v}=1.097 	imes 10^{7}\left(\frac{1}{2^{2}}-\frac{1}{5^{2}}ight)$\ $=1.097 	imes 10^{7}\left(\frac{1}{4}-\frac{1}{25}ight)=1.097 	imes 10^{7} 	imes \frac{21}{100}$\ $\lambda=\frac{1}{\bar{v}}=\frac{100}{1.097 	imes 21 	imes 10^{7}} \mathrm{~m}=4.34 	imes 10^{-7} \mathrm{~m}=434 	imes 10^{-9} \mathrm{~m}$\ $\lambda=434 \mathrm{~nm}$\