Q 12-12-086JEE MainJEE Main 2022 (25 Jul, Shift 2)Medium
A hydrogen atom comes from an excited state 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)
Answer: (B) $\sqrt{\dfrac{\lambda R}{\lambda R - 1}}$
$$\frac1\lambda = R\left(1 - \frac1{n^2}\right)\ \Rightarrow\ \frac1{n^2} = 1 - \frac{1}{\lambda R} = \frac{\lambda R - 1}{\lambda R}\ \Rightarrow\ n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$$
Solution by Sreeraj P, M.Sc Physics