Question

An electron in hydrogen atom after absorbing energy photons can jump between energy states $n_1$ and $n_2(n_2>n_1)$. Then it may return to ground state after emitting six different wavelengths is emission spectrum. The energy of emitted photons is either equal to, less than or greater than the absorbed photons. Then $n_1$ and $n_2$ are

• A. $n_2=4,n_1=3$
• B. $n_2=5,n_1=3$
• C. $n_2=4,n_1=2$
• D. $n_2=4,n_1=1$

Answer 1

The correct option is A $n_2=4,n_1=3$

$\begin{array}{c}\text{Given-}\ast \text{electron can jump to ground state emitting}\\ \text{photons of six different wavelengths.}\\ \Rightarrow \frac{n_2(n_2-1)}{2}=6\\ \text{on solving, we get}{n}_2=4\end{array}$

$\begin{array}{c}\text{Now, we know that Energy of photon is given by-}\\ \begin{array}{c}E=13.6[\frac{1}{n_1^2}-\frac{1}{n_2^2}]eV\\ \Rightarrow E=13.6[\frac{1}{n_1^2}-\frac{1}{16}]eV\end{array}\end{array}$

$\begin{array}{c}\text{For minimum energy of the photon-}\\ \text{we have ,}{n}_1^2<16\text{for maximum value of}\left(n_1\right)\\ \Rightarrow n_1<4\\ \Rightarrow n_1=3\\ \Rightarrow \text{we get ,}{n}_1=3\text{\&}{n}_2=4\\ \Rightarrow \text{Hence , opt (a) is correct.}\end{array}$