JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2025
- Easy
- Numerical
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 ____.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 2
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_1^2}\right)\)
\(2.86=13.6\left(\frac{1}{\mathrm{n}^2}-\frac{1}{5^2}\right)\)
\(\begin{aligned} & \frac{1}{\mathrm{n}^2}=0.21+\frac{1}{2.5} \\ & \mathrm{n}^2=4 \\ & \mathrm{n}=2\end{aligned}\)
\(2.86=13.6\left(\frac{1}{\mathrm{n}^2}-\frac{1}{5^2}\right)\)
\(\begin{aligned} & \frac{1}{\mathrm{n}^2}=0.21+\frac{1}{2.5} \\ & \mathrm{n}^2=4 \\ & \mathrm{n}=2\end{aligned}\)
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