Verified Solution • 2025–26 Session moderate

QJEEM22Ph93c3: JEE Main 2022 Physics - Class 12 JEEM22Ph93c3 Solution: Hydrogen atom from excited state comes to the ground state by…

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

Multiple Choice Options

  • A λRλR−1\sqrt{\frac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1}}
  • B λRλ−1\sqrt{\frac{\lambda \mathrm{R}}{\lambda-1}}
  • C λλR−1\sqrt{\frac{\lambda}{\lambda \mathrm{R}-1}}
  • D λR2λR−1\sqrt{\frac{\lambda R^{2}}{\lambda R-1}}

Step-by-Step Verified Solution

∵\because 1λ=R(112−1n2){1 \over \lambda } = R\left( {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right)

⇒1λR=1−1n2\Rightarrow {1 \over {\lambda R}} = 1 - {1 \over {{n^2}}}

⇒1n2=1−1λR=λR−1λR\Rightarrow {1 \over {{n^2}}} = 1 - {1 \over {\lambda R}} = {{\lambda R - 1} \over {\lambda R}}

⇒n=λRλR−1\Rightarrow n = \sqrt {{{\lambda R} \over {\lambda R - 1}}}

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