Question

The energy of a hydrogen atom in the ground state is -13.6eV. The energy of a He+ ion in the first excited state will be
(1) -6.8eV
(2) -13.6eV
(3) -27.2eV
(4) -54.4eV

Answer 1

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Answer 2

Explanation:

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$\huge\mathbb{\mid{\overline{\underline{Answer:-}}\mid}}$

$For \: helium, \\ \\ (atomic \: number) z = 2 \: and \: \\ \\ For \: first \: exited \: state \: n = 2$

$We \: know \leadsto \\ \\ Energy = E = -13.6 \times {(\frac{z}{n})}^{2} \\ \\ \implies E = -13.6 \times ({\frac{\cancel{2}}{\cancel{2}})}^{2} \\ \\ \boxed{\boxed{ \bold{ \implies E = -13.6 eV}}}$