Atoms & Nuclei

JEE Main level, numerical answer, medium

35 of 239

Question 35 of 239

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

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Answer: 2

E=13.6(1n12−1n12)\mathrm{E}=13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_1^2}\right)
2.86=13.6(1n2−152)2.86=13.6\left(\frac{1}{\mathrm{n}^2}-\frac{1}{5^2}\right)
1n2=0.21+12.5n2=4n=2\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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