An electron from various excited states of hydrogen atom emit… | JEE Main 2018

Q107JEE Main 2018Difficult

Bohr Model: Orbits, Speed & Angular Momentum

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 nth state and the ground state respectively. Let $\Lambda_n$ be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)

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Answer: (a)

$2\pi r_n = n\lambda_n$ with $r_n \propto n^2$ , so $\lambda_n \propto n$ , i.e. $\lambda_n = n\lambda_g$ .
Photon: $\frac{1}{\Lambda_n} = R\left(1 - \frac{1}{n^2}\right) \Rightarrow \Lambda_n = \frac{1}{R}\left(1 - \frac{1}{n^2}\right)^{-1}$ .
For large $n$ : $\Lambda_n \approx \frac{1}{R}\left(1 + \frac{1}{n^2}\right) = \frac{1}{R} + \frac{\lambda_g^2}{R}\cdot\frac{1}{\lambda_n^2}$
So the answer is (a) $\Lambda_n \approx A + \frac{B}{\lambda_n^2}$ .

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