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The energy of an electron present in Bohr’s second orbit of hydrogen atom is

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– 164 kJ mol^–1

– 328 J mol^–1

– 328 kJ mol^–1

–1312 J atom^–1

answer is B.

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Detailed Solution

$E = - \frac{1312}{n^{2}} k J / m o l f o r n = 2 E = - \frac{1312}{2^{2}} = - 328 k J / m o l$

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