When an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of…

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
  1. $\frac{20}{13} \lambda$
  2. $\frac{16}{25} \lambda$
  3. $\frac{9}{16} \lambda$
  4. $\frac{20}{7} \lambda$

Solution

$\begin{aligned} \frac{1}{\lambda} &=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{R \times 5}{36} \\ \frac{1}{\lambda^{\prime}} &=R\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)=R\left(\frac{1}{9}-\frac{1}{16}\right)=R \times \frac{7}{144} \\ \frac{\lambda^{\prime}}{\lambda} &=\frac{5}{36} \times \frac{144}{7}=\frac{20}{7} \\ \therefore \lambda^{\prime} &=\frac{20}{7} \lambda \end{aligned}$ ~

Asked in: MHT CET 2020 (13 Oct Shift 1)

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