Question 167: Atoms & Nuclei

Question

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength $\lambda$. The value of principal quantum number '$n$' of the excited state will be : ($\mathrm{R}:$ Rydberg constant)

Accepted Answer

$\because$ ${1 \over \lambda } = R\left( {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right)$
$ \Rightarrow {1 \over {\lambda R}} = 1 - {1 \over {{n^2}}}$
$ \Rightarrow {1 \over {{n^2}}} = 1 - {1 \over {\lambda R}} = {{\lambda R - 1} \over {\lambda R}}$
$ \Rightarrow n = \sqrt {{{\lambda R} \over {\lambda R - 1}}} $

Answer

$\sqrt{\frac{\lambda \mathrm{R}}{\lambda-1}}$

Answer

$\sqrt{\frac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1}}$

Answer

$\sqrt{\frac{\lambda}{\lambda \mathrm{R}-1}}$

Answer

$\sqrt{\frac{\lambda R^{2}}{\lambda R-1}}$

Atoms & Nuclei

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength \lambda. The value of principal quantum number 'n' of the excited state will be : (\mathrm{R}: Rydberg constant)

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