MHT CET · Physics · Atomic Physics
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)
- A \(\frac{3 \mathrm{RC}}{27}\)
- B \(\frac{\mathrm{RC}}{25}\)
- C \(\frac{ \mathrm{8RC}}{9}\)
- D \(\frac{ \mathrm{5RC}}{36}\)
Answer & Solution
Correct Answer
(D) \(\frac{ \mathrm{5RC}}{36}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ \therefore \quad & \frac{1}{\lambda}=\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5}{36} \mathrm{R} \\ \therefore \quad & \mathrm{f}=\frac{\mathrm{c}}{\lambda}=\frac{5}{36} \mathrm{Rc}\end{array}\)
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