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 $1.3225nm$ and ends at $211.6pm$?

• A. $434nm$, Balmer
• B. $434pm$, Paschen
• C. $545pm$, Pfund
• D. $600nm$, Lyman

Answer 1

Answer: (A)

Radius of $n^{th}$ orbit
$=\frac{0.529\times n^2}{Z}\mathring{A}=\frac{52.9\times n^2}{Z}pm$
Radius $\left(r_2\right)=1.3225nm=1322.5pm=\frac{52.9n_2^2}{Z}$
Radius $\left(r_1\right)=211.6pm=\frac{52.9n_1^2}{Z}$
$\therefore \frac{r_2}{r_1}=\frac{1322.5}{211.6}=\frac{n_2^2}{n_1^2}$
$\Rightarrow 6.25=\frac{n_2^2}{n_1^2}={\left(\frac{n_2}{n_1}\right)}^{{}^2}$
$\Rightarrow \frac{n_2}{n_1}=\sqrt{6.25}$
$=2.5$
$\therefore n_1=2$, $n_2=5$ thus, the transition is from $5^{th}$ orbit to $2^{nd}$ orbit. It belongs to Balmer series.
$\bar{v}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
$\Rightarrow \bar{v}=1.097\times 10^7\left(\frac{1}{2^2}-\frac{1}{5^2}\right)$
$=1.097\times 10^7\left(\frac{1}{4}-\frac{1}{25}\right)$
$=1.097\times 10^7\times \frac{21}{100}$
$\lambda =\frac{1}{\bar{v}}=\frac{100}{1.097\times 21\times 10^7}m$
$=4.34\times 10^{-7}m=434\times 10^{-9}m$
$\lambda =434nm$
Thus, it lies in the visible region.