The ground state energy of hydrogen atom is $- 13.6$ eV. The energy of second excited state $H e^{+} i o n$ in eV is:

- 6.04

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- 27.2

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- 54.4

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- 3.4

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Solution

The correct option is A $- 6.04$
$(E)_{n}^{t h} = (E_{G N D})_{H}^{.} \frac{Z^{2}}{n^{2}}$

$E_{3 r d} (H e^{+}) = (- 13.6 e V) \times \frac{2^{2}}{3^{2}} = - 6.04 e V$

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Q. The energy of the ground electronic state of hydrogen atom is -13.6 eV. The energy of the first excited stae will be :

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The ground state energy of hydrogen atom is −13.6 eV. The energy of second excited state He+ion in eV is:

The correct option is A −6.04(E)thn=(EGND).HZ2n2 E3rd(He+)=(−13.6eV)×2232=−6.04eV

The ground state energy of hydrogen atom is $- 13.6$ eV. The energy of second excited state $H e^{+} i o n$ in eV is:

The correct option is A $- 6.04$
$(E)_{n}^{t h} = (E_{G N D})_{H}^{.} \frac{Z^{2}}{n^{2}}$

$E_{3 r d} (H e^{+}) = (- 13.6 e V) \times \frac{2^{2}}{3^{2}} = - 6.04 e V$