Modern Physics Ques 16
- The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths $\lambda _1 / \lambda _2$ of the photons emitted in this process is
(a) $20 / 7$
(b) $27 / 5$
(c) $7 / 5$
(d) $9 / 7$
(Main 2019, 12 April II)
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Answer:
Correct Answer: 16.(a)
Solution:
Formula:
Wavelength Corresponding To Spectral Lines:
- Wavelength $\lambda$ of emitted photon as an electron transits from an initial energy level $n _i$ to some final energy level $n _f$ is given by Balmer’s formula,
$ \frac{1}{\lambda}=R (\frac{1}{n _f^{2}}-\frac{1}{n _i^{2}}) $
where, $R=$ Rydberg constant.
In transition from $n=4$ to $n=3$, we have
$ \begin{aligned} \frac{1}{\lambda _1} & =R (\frac{1}{3^{2}}-\frac{1}{4^{2}}) \\ & =R (\frac{7}{9 \times 16}) \quad …….(i) \end{aligned} $
In transition from $n=3$ to $n=2$, we have
$ \begin{aligned} \frac{1}{\lambda _2} & =R (\frac{1}{2^{2}}-\frac{1}{3^{2}}) \\ & =R (\frac{5}{9 \times 4}) \quad …….(ii) \end{aligned} $
So, from Eqs. (i) and (ii), the ratio of $\frac{\lambda _1}{\lambda _2}$ is
$ \frac{\lambda _1}{\lambda _2}=\frac{(\frac{9 \times 16}{7 R})}{(\frac{9 \times 4}{5 R})}=\frac{20}{7} $
BETA
