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.

Question:

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, $\lambda_{1} / \lambda_{2}$, of the photons emitted in this process is :

  1. (1) $20 / 7$

  2. (2) $27 / 5$

  3. (3) $7 / 5$

  4. (4) $9 / 7$


Correct Option: 1,

Solution:

(1) $\frac{1}=\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)$

$=\frac{7 R}{16 \times 9}$

And $\frac{1}{\lambda_{2}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$

$=\frac{5 R}{36}$

Now $\frac{\lambda_{1}}{\lambda_{2}}=\frac{(5 R / 36)}{7 R /(16 \times 9)}=\frac{20}{7}$

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