The energy of a hydrogen atom in the ground state is 13.6 eV. The energy of a He $^{+}$ ion in the first excited state will be:

-6.8 eV

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-13.6 eV

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-27.2 eV

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-54.4 eV

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Solution

The correct option is D -13.6 eV
Energy of electron in n th orbit with atomic number Z for H-like ion is:
$E_{n} = - \frac{13.6 \times Z^{2}}{n^{2}}$

For $H e^{+}$ ion, for first excited state $n = 2$ and atomic number $Z = 2$

Now,

$E_{2} = - \frac{13.6 \times 2^{2}}{2^{2}} = - 13.6 e V$

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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 energy of a hydrogen atom in the ground state is 13.6 eV. The energy of a He+ ion in the first excited state will be:

The correct option is D -13.6 eVEnergy of electron in n th orbit with atomic number Z for H-like ion is:En=−13.6×Z2n2For He+ ion, for first excited state n=2 and atomic number Z=2Now, E2=−13.6×2222=−13.6 eV

The energy of a hydrogen atom in the ground state is 13.6 eV. The energy of a He $^{+}$ ion in the first excited state will be:

The correct option is D -13.6 eV
Energy of electron in n th orbit with atomic number Z for H-like ion is:
$E_{n} = - \frac{13.6 \times Z^{2}}{n^{2}}$

For $H e^{+}$ ion, for first excited state $n = 2$ and atomic number $Z = 2$

Now,

$E_{2} = - \frac{13.6 \times 2^{2}}{2^{2}} = - 13.6 e V$