An electron in the hydrogen atom initially in the fourth excited state makes a transition to n th energy state by emitting a photon of energy 2.86 eV . The integer value of n will be .

Physics · Atomic Physics

JEE Main 2025 — 3 April, Evening Shift — Question 66

An electron in the hydrogen atom initially in the fourth excited state makes a transition to $\mathrm{n}^{\text {th }}$ energy state by emitting a photon of energy 2.86 eV . The integer value of $n$ will be $\qquad$ .

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$E_{n}=13.6\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

\begin{aligned} & 2.86=13.6\left|-\frac{1}{25}+\frac{1}{n^{2}}\right| \Rightarrow \\ & \frac{2.86}{13.6}+\frac{1}{25}=\frac{1}{n^{2}} \\ & \frac{1}{n^{2}}=0.25 \\ & n^{2}=4 \Rightarrow \\ & n=2 \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Evening Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Hydrogen Spectrum