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Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelength λ₁ : λ₂ emitted in the two cases is

1. 7/5
2. 27/20
3. 27/5
4. 20/7

Correct answer: 27/20

Solution

The energy difference between levels in a hydrogen atom is given by the formula E = 13.6 * (1/n₁² - 1/n₂²). Using this, the wavelengths are inversely proportional to the energy differences. For the transitions 4→3 and 3→2, the ratio of wavelengths λ₁:λ₂ is calculated as 27/20.

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