Question

The energy corresponding to one of the lines in the Paschen series for $H$-atom is $18.16\times {10}^{-20}J$. Find the quantum numbers for the transition which produce this line.

Answer 1

For paschen series of H-atom,$n_1$=3

$\frac{1}{\lambda }$=$R{Z}^2$$[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$ put $n_2=n$

$E=\frac{hc}{\lambda }$

given,$E=18.16\times {10}^{-20}J$

$\Rightarrow \frac{1}{\lambda }=913571.794m^{-1}$

$R=109678{cm}^{-1}$

$\Rightarrow 913572.794=\frac{109678}{cm}$${\left(1\right)}^2[\frac{1}{3^2}-\frac{1}{n^2}]$

$\frac{1}{100}\times 8.3296=\frac{1}{9}-\frac{1}{n^2}$

$\frac{1}{n^2}$ $=\left[\frac{1}{9}\right]-\left(8.3296\right)\left(\frac{1}{100}\right)$

$\Rightarrow n^2=35.95155$

n$\cong$6

principle quantum number,n=6