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)
Solution
<p>$\because$ ${1 \over \lambda } = R\left( {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right)$</p>
<p>$\Rightarrow {1 \over {\lambda R}} = 1 - {1 \over {{n^2}}}$</p>
<p>$$ \Rightarrow {1 \over {{n^2}}} = 1 - {1 \over {\lambda R}} = {{\lambda R - 1} \over {\lambda R}}$$</p>
<p>$\Rightarrow n = \sqrt {{{\lambda R} \over {\lambda R - 1}}}$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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