MHT CET · Physics · Atomic Physics
If an electron in hydrogen atom jumps from \(3^{\text {rd }}\) orbit to \(2^{\text {nd }}\) orbit, it emits a photon of wavelength ' \(\lambda\) '. When it jumps from \(4^{\text {th }}\) orbit to \(3^{\text {rd }}\) orbit, the corresponding wavelength of the photon will be
- A \(\frac{20}{13} \lambda\)
- B \(\frac{20}{7} \lambda\)
- C \(\frac{9}{16} \lambda\)
- D \(\frac{16}{25} \lambda\)
Answer & Solution
Correct Answer
(B) \(\frac{20}{7} \lambda\)
Step-by-step Solution
Detailed explanation
Rydberg's relation is given by
\(\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \mathrm{Z}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
For hydrogen, \(\mathrm{Z}=1\).
Given, for \(3^{\text {rd }}\) to \(2^{\text {nd }}\) orbit transition

And for \(4^{\text {th }}\) to \(3^{\text {rd }}\) orbit transition

Taking ratio of eqn (1) \& (2)
\(\begin{aligned}
& \frac{\lambda_n}{\lambda}=\frac{5}{36} \times \frac{(16 \times 9)}{7}=\frac{20}{7} \\
& \Rightarrow \lambda_n=\frac{20}{7} \lambda
\end{aligned}\)
\(\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \mathrm{Z}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
For hydrogen, \(\mathrm{Z}=1\).
Given, for \(3^{\text {rd }}\) to \(2^{\text {nd }}\) orbit transition

And for \(4^{\text {th }}\) to \(3^{\text {rd }}\) orbit transition

Taking ratio of eqn (1) \& (2)
\(\begin{aligned}
& \frac{\lambda_n}{\lambda}=\frac{5}{36} \times \frac{(16 \times 9)}{7}=\frac{20}{7} \\
& \Rightarrow \lambda_n=\frac{20}{7} \lambda
\end{aligned}\)
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