Energy of H- atom in the ground state is -13.6, Hence energy in the second excited state is:

-6.8ev

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-3.4ev

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-1.51ev

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-4.3ev

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Solution

The correct option is C -1.51ev
$2^{n d}$ excited state n=3 $E_{n} = - \frac{13.06}{n^{2}} E_{3} = - \frac{13.06}{3^{2}} = - 1.51$

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The total energy of an electron in the first excited state of hydrogen is about -3.4 eV. Its kinetic energy in this state is

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