Ionization Energy From Bohr Model

Question

What is the minimum energy required in electron volts to completely remove an electron from the n=2 orbit of a He^+ ion? (Given: Z=2 for helium, ionization energy of H in ground state = 13.6 eV)

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The energy of an electron in a hydrogen-like ion is given by E_n = -13.6 × Z^2 / n^2 eV, where Z is the atomic number and n is the principal quantum number. For He^+ (Z=2) in the n=2 orbit: E_2 = -13.6 × (2)^2 / (2)^2 = -13.6 × 4/4 = -13.6 eV. The ionization energy is the energy needed to remove the electron to \infty (E = 0), so the minimum energy required is the magnitude of E_2, which is 13.6 eV. Option 27.2 eV corresponds to He^+ in n=1 (ground state): -13.6 × 4/1 = -54.4 eV, so from n=1 the ionization energy is 54.4 eV, not 27.2 eV. Option 54.4 eV is the ionization energy from the ground state (n=1) of He^+. Option 6.04 eV is incorrect; it would correspond to n=3 in H, not He^+ in n=2. This problem tests the extension of Bohr's formula to hydrogen-like ions, a key JEE topic. Plausibility check: substituting back, E_2(He^+) = -13.6 × 4/4 = -13.6 eV, so 13.6 eV is required to reach E=0 ✓.

Ionization Energy From Bohr Model

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