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In hydrogen atom, when electron jumps from second to first orbit, then energy emitted is

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None of these

– 27.2 eV

– 13.6 eV

– 6.8 eV

answer is D.

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Detailed Solution

$E_{n_{1} \to n_{2}} = - 13.6 [\frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}}]; n_{1} = 2 & n_{2} = 1 \Rightarrow E_{II} \to E_{I} = - 13.6 \times \frac{3}{4} = - 10.2 eV$

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