MHT CET · Physics · Atomic Physics
When an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength \({\lambda} \lambda^{\prime}\). When it jumps from the fourth orbit to third orbit, the wavelength emitted by the photon will be
- A \(\frac{20}{13} \lambda\)
- B \(\frac{16}{25} \lambda\)
- C \(\frac{9}{16} \lambda\)
- D \(\frac{20}{7} \lambda\)
Answer & Solution
Correct Answer
(D) \(\frac{20}{7} \lambda\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{\lambda} =R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{R \times 5}{36} \)
\( \frac{1}{\lambda^{\prime}} =R\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)=R\left(\frac{1}{9}-\frac{1}{16}\right)=R ~\times\) \(\frac{7}{144} \)
\( \frac{\lambda^{\prime}}{\lambda} =\frac{5}{36} \times \frac{144}{7}=\frac{20}{7} \)
\( \therefore \lambda^{\prime} =\frac{20}{7} \lambda\)
\( \frac{1}{\lambda^{\prime}} =R\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)=R\left(\frac{1}{9}-\frac{1}{16}\right)=R ~\times\) \(\frac{7}{144} \)
\( \frac{\lambda^{\prime}}{\lambda} =\frac{5}{36} \times \frac{144}{7}=\frac{20}{7} \)
\( \therefore \lambda^{\prime} =\frac{20}{7} \lambda\)
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