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JEE Main 202225 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
25 July
Shift
Evening Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2022 · 25 July · Evening Shift · Physics · Atomic Physics

Hydrogen atom from excited state comes to the ground by emitting a...

Hydrogen atom from excited state comes to the ground by emitting a photon of wavelength λ\lambda. The value of principal quantum number nn of the excited state will be :

( R:R: Rydberg constant)

Options

  1. A\lambdaRλ−1\sqrt{\dfrac{\lambdaR}{\lambda - 1}}
  2. B\lambdaR\lambdaR−1\sqrt{\dfrac{\lambdaR}{\lambdaR - 1}}
  3. Cλ\lambdaR−1\sqrt{\dfrac{\lambda}{\lambdaR - 1}}
  4. DλR2\lambdaR−1\sqrt{\dfrac{\lambda{R}^{2}}{\lambdaR - 1}}

Correct answer

B. \lambdaR\lambdaR−1\sqrt{\dfrac{\lambdaR}{\lambdaR - 1}}

Step-by-step solution

When an electron jumps from a nth{n}^{th} orbit to ground state, using Rydberg formula,

1λ=R(1n12−1n22)\dfrac{1}{\lambda} = R\left(\dfrac{1}{{{n}_{1}}^{2}} - \dfrac{1}{{{n}_{2}}^{2}}\right)

In this case, n1=1{n}_{1} = 1 and n2=n{n}_{2} = n. Therefore,

⇒1λ=R(112−1n2)⇒n=\lambdaR\lambdaR−1⇒ \dfrac{1}{\lambda} = R\left(\dfrac{1}{{1}^{2}} - \dfrac{1}{{n}^{2}}\right) ⇒ n = \sqrt{\dfrac{\lambdaR}{\lambdaR - 1}}

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