The electron in a hydrogen atom first jumps from the third excited...

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths

\dfrac{{\lambda}_{1}}{{\lambda}_{2}}

of the photons emitted in this process is:

Options

A $\dfrac{7}{5}$
B $\dfrac{20}{7}$
C $\dfrac{9}{7}$
D $\dfrac{27}{5}$

Correct answer

B. $\dfrac{20}{7}$

Step-by-step solution

\dfrac{1}{{\lambda}_{1}} = R\left(\dfrac{1}{{3}^{2}} - \dfrac{1}{{4}^{2}}\right) = \dfrac{R \times 7}{144}

\dfrac{1}{{\lambda}_{2}} = R\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{3}^{2}}\right) = \dfrac{R \times 5}{36}

⇒ \dfrac{{\lambda}_{1}}{{\lambda}_{2}} = \dfrac{R \times 5}{36} \times \dfrac{144}{R \times 7} = \dfrac{20}{7}

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