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If the potential energy of a H-atom in the ground state be zero then its potential energy in the first excited state will be

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10.2 eV

20.4 eV

23.8 eV

27.2 Ev

answer is B.

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Detailed Solution

Given -27.2 is taken is zero,

$So - 27.2 + 27.2 = 0$

$In 2^{nd} orbit, E = - 3.4 + 27.2$

$E = 20.4 eV$

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