Question

Energy corresponding to one of the lines in Paschen-series for $H-atom$ is $18.16\times {10}^{-20}J$. Find the possible quantum numbers for the $e^{-}$ when it is present in the initial energy state.

Answer 1

$△E=R_H(\frac{1}{n_i^2}-\frac{1}{n_f^2})$

for Paschen series,$n_i^2=3$; $n_f^2=n=?$

$18.16\times {+10}^{-20}=+2.18\times {10}^{-18}(\frac{1}{3^2}-\frac{1}{n^2})$

$\frac{{18.16\times 10}^{-20}}{{2.18\times 10}^{-18}}=\frac{n^2-9}{{9n}^2}$

${9n}^2\left({8.33\times 10}^{-2}\right)=n^2-90.0833\times {9n}^2=n^2-90.7497n^2-n^2=-9-0.2503n^2=-9n=\frac{3}{\sqrt{0.25}}=\frac{3}{0.5}\therefore n=6$

possible quantum numbers $=4,5,6$.