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NEET UG PHYSICS MCQ: ATOMS

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Application +4 | -1
Energy of electron (in eV\mathrm{eV}) in 2nd2^{\text {nd}} orbit of He+He^+ ion is
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Correct option: B
−13.6 eV-13.6 \mathrm{~eV}
Option B (−13.6 eV-13.6 \mathrm{~eV}) is correct because the energy of an electron in the nth orbit of a hydrogen-like atom is given by En=−13.6Z2n2eVE_n = -13.6 \frac{{Z^2}}{{n^2}} eV. For He+He^+, Z=2 and n=2, so E2=−13.62222=−13.6eVE_2 = -13.6 \frac{{2^2}}{{2^2}} = -13.6 eV.

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