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In hydrogen atom, if an electron in the orbit with principal quantum number ' $n$ ' jumps to the first…
In hydrogen atom, if an electron in the orbit with principal quantum number ' $n$ ' jumps to the first excited state, the wavelength of the emitted photon is ' $\lambda$ '. Then the value of $\mathrm{n}$ is
(R-Rydberg constant)
$\sqrt{\frac{4 \lambda R}{\lambda R+4}}$ $\sqrt{\frac{4 \lambda R}{\lambda R-4}}$ $\sqrt{\frac{\lambda \mathrm{R}-4}{4 \lambda \mathrm{R}}}$ $\sqrt{\frac{\lambda \mathrm{R}+4}{4 \lambda \mathrm{R}}}$ Solution
(When electron jump from $\mathrm{n}_2=\mathrm{n}$ to first excited state
$\begin{aligned}
& \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]=\mathrm{R}\left[\frac{1}{4}-\frac{1}{\mathrm{n}^2}\right] \\
& \frac{1}{\lambda}=\frac{\mathrm{R}}{4}-\frac{\mathrm{R}}{\mathrm{n}^2} \Rightarrow \frac{\mathrm{R}}{\mathrm{n}^2}=\frac{\mathrm{R}}{4}-\frac{1}{\lambda} \Rightarrow \frac{\mathrm{R}}{\mathrm{n}^2}=\frac{\mathrm{R} \lambda-4}{4 \lambda} \\
& \Rightarrow \frac{\mathrm{n}^2}{\mathrm{R}}=\frac{4 \lambda}{\mathrm{R} \lambda-4} \Rightarrow \mathrm{n}=\sqrt{\frac{4 \lambda \mathrm{R}}{\mathrm{R} \lambda-4}}
\end{aligned}$
Asked in: AP EAMCET 2023 (18 May Shift 2)
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