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 : ( R: Rydberg constant)

\sqrt\frac\lambda R^2\lambda R-1

Atoms and Nuclei PYQ — JEE Main 2022 · 25 Jul · Shift 2 · Q62 with Solution

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Correct answer: B

When an electron in hydrogen falls from an excited state $n$ to the ground state $1$ , the emitted photon wavelength $\lambda$ satisfies

\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)

So,

\frac{1}{\lambda}=R\left(1-\frac{1}{n^2}\right)

Divide by $R$ :

\frac{1}{\lambda R}=1-\frac{1}{n^2}

Hence,

\frac{1}{n^2}=1-\frac{1}{\lambda R}

Take LCM:

\frac{1}{n^2}=\frac{\lambda R-1}{\lambda R}

Therefore,

n^2=\frac{\lambda R}{\lambda R-1}

and

n=\sqrt{\frac{\lambda R}{\lambda R-1}}

This matches Option B:

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

A: $\sqrt{\frac{\lambda R}{\lambda-1}}$ has inconsistent form.
C: $\sqrt{\frac{\lambda}{\lambda R-1}}$ misses a factor of $R$ in the numerator.
D: $\sqrt{\frac{\lambda R^2}{\lambda R-1}}$ has an extra factor of $R$ .

So only B is correct.