Modern Physics Ques 178
- An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let $\lambda _n, \lambda _g$ be the de-Broglie wavelength of the electron in the $n$th state and the ground state, respectively. Let $\Lambda _n$ be the wavelength of the emitted photon in the transition from the $n$th state to the ground state. For large $n,(A, B$ are constants)
(2018 main)
(a) $\Lambda _n^{2} \approx \lambda$
(b) $\Lambda _n \approx A+\frac{B}{\lambda _n^{2}}$
(c) $\Lambda _n \approx A+B \lambda _n^{2}$
(d) $\Lambda _n^{2} \approx A+B \lambda _n^{2}$
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Answer:
Correct Answer: 178.(b)
Solution:
Formula:
Wavelength Corresponding To Spectral Lines:
- $2 \pi r=n \lambda _n$
$ \begin{aligned} & \lambda _n=\frac{2 \pi r}{n}=\frac{2 \pi r _0 n^{2}}{n}=2 \pi r _0 n \quad …….(i) \\ & \frac{1}{\Lambda _n}=R \{\frac{1}{1^{2}}-\frac{1}{n^{2}}\} \end{aligned} $
$ \begin{aligned} & \Lambda _n=\frac{1}{R} \{1+\frac{1}{n^{2}-1}\} \\ & \Lambda _n=\frac{1}{R} \{1+\frac{1}{n^{2}}\} \quad(n \gg 1)\quad …….(ii) \end{aligned} $
From Eqs. (i) and (ii),
$ \Lambda _n=\frac{1}{R} \{1+\frac{4 \pi^{2} r _0^{2}}{\lambda _n^{2}}\}=A+\frac{B}{\lambda _n^{2}} $
BETA
