Question: Q. 5. Calculate the ratio of the frequencies of the radiation emitted due to transition of the electron in a hydrogen atom from its (i) second permitted energy level to the first level and (ii) highest permitted energy level to the second permitted level.
A&E [Comptt. I, II, III 2018]
Show Answer
Solution:
Ans. Formulae
(i) Frequency of first case
$1 / 2$
(ii) Frequency of second case
Ratio
We have,
(i)
(ii)
$$ \begin{aligned} h v & =E_{f}-E_{i} \ & =\frac{E_{0}}{n_{f}^{2}}-\frac{E_{0}}{n_{i}^{2}} \ h v_{1} & =E_{0}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=E_{0} \times \frac{3}{4} \ h v_{2} & =E_{0}\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)=E_{0} \times \frac{1}{4} \end{aligned} $$
$$ \begin{align*} & =\frac{E_{0}}{n_{f}^{2}}-\frac{E_{0}}{n_{i}^{2}} \ h v_{1} & =E_{0}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=E_{0} \times \frac{3}{4} \ h v_{2} & =E_{0}\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)=E_{0} \times \frac{1}{4} \ \therefore \quad \frac{v_{1}}{v_{2}} & =3 \end{align*} $$
