Question

An electron in $nth$ excited state in a hydrogen atom comes down to first excited state by emitting ten different wavelengths. Find the value of $n$ (an integer).

Answer 1

First excited state means $n=2$ orbit and thus $nth$ excited state means $(n+1)th$ orbit.

As the electron comes down to $n_1=2$ orbit, thus number of specteral line is given by- $N=\frac{[n_2-n_1]\left[\right(n_2-n_1)+1]}{2}$

$\therefore 10=\frac{[n_2-2]\left[\right(n_2-2)+1]}{2}⟹n_2=6$ orbit

Thus $n_2=n+1⟹n=5$

Thus the transition talkes place from $5th$ excited state OR $6th$ orbit.