If an electron in a hydrogen atom jumps from an orbit of level $\mathrm{n}=3$ to orbit of level $\mathrm{n}=2$, then the emitted radiation frequency is (where $\mathrm{R}=$ Rydberg's constant, $\mathrm{C}=$ Velocity of light)

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Aicharya · Accepted Answer

$\begin{array}{ll} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}ight] \ 	herefore \quad & \frac{1}{\lambda}=\left[\frac{1}{2^2}-\frac{1}{3^2}ight]=\frac{5}{36} \mathrm{R} \ 	herefore \quad & \mathrm{f}=\frac{\mathrm{c}}{\lambda}=\frac{5}{36} \mathrm{Rc}\end{array}$

If an electron in a hydrogen atom jumps from an orbit of level $\mathrm{n}=3$ to orbit of level…

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