Question: Q. 5. A monochromatic radiation of wavelength $975{ }^{\circ}$ excites the hydrogen atom from its ground state to a higher state. How many different spectral lines are possible in the resulting spectrum? Which transition corresponds to the longest wavelength amongst them.
[SQP 2018]
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Solution:
Ans. Energy corresponding to the giyen wavelength $\mathrm{E}($ in $\mathrm{eV})=\frac{12400}{\lambda(\text { in } \AA)}=12.71 \mathrm{eV}$
The excited state :
$$ \begin{aligned} E_{n}-E_{1} & =12.71 \ \frac{-13.6}{n^{2}}+13.6 & =12.71 \end{aligned} $$
$$ \therefore \quad n=3.9 \approx 4 $$
Total no. of spectral lines emitted : $\frac{n(n-1)}{2}=6$
Longest wavelength will correspond to the transition from $n=1$ to $n=4$
[CBSE Marking Scheme, 2018]
