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 ____.

Correct answer: 2. E=13.6 (1/(n_1^2)-1/(n_1^2) ) 2.86=13.6 (1/(n^2)-1/(5^2) ) 1/(n^2)=0.21+1/2.5 n^2=4 n=2

Q53 / 75 · JEE Main 2025 (3 Apr Shift 2) · Physics

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 ____.

Correct answer: 2

Solution

$\mathrm{E}=13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_1^2}\right)$
$2.86=13.6\left(\frac{1}{\mathrm{n}^2}-\frac{1}{5^2}\right)$
$\begin{aligned} & \frac{1}{\mathrm{n}^2}=0.21+\frac{1}{2.5} \\ & \mathrm{n}^2=4 \\ & \mathrm{n}=2\end{aligned}$

Hence, the correct answer is 2.

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