Q 12-12-131JEE MainJEE Main 2025 (3 Apr, Shift 2)Easy
An electron in the hydrogen atom, initially in the fourth excited state, makes a transition to the $n^{\text{th}}$ energy state by emitting a photon of energy $2.86\ \text{eV}$. The integer value of $n$ will be ______.
Numerical value type. Enter your answer.
Answer: 2
The fourth excited state is $n = 5$.
$$13.6\left(\frac1{n^2} - \frac1{25}\right) = 2.86 \Rightarrow \frac1{n^2} = \frac{2.86}{13.6} + 0.04 = 0.21 + 0.04 = 0.25$$
$$n = 2$$
Solution by Sreeraj P, M.Sc Physics