Which of the following atoms has the lowest ionization energy?

The correct answer is Option (4) → \( {}_{1}^{1}\mathrm{H} \)

All four are hydrogen-like (one-electron) species. By the Bohr model, the energy of the electron in the ground state (\( n = 1 \)) of a nucleus of atomic number \( Z \) is \( -13.6\,Z^{2} \) eV, so the ionization energy is

    \( \displaystyle E_{\mathrm{ion}} = 13.6\,Z^{2}\,\mathrm{eV} \)

    \( \mathrm{Be^{+++}}\ (Z = 4): 13.6 \times 16 = 217.6\,\mathrm{eV} \)
    \( \mathrm{Li^{++}}\ (Z = 3): 13.6 \times 9 = 122.4\,\mathrm{eV} \)
    \( \mathrm{He^{+}}\ (Z = 2): 13.6 \times 4 = 54.4\,\mathrm{eV} \)
    \( \mathrm{H}\ (Z = 1): 13.6 \times 1 = 13.6\,\mathrm{eV} \)

The ionization energy grows as \( Z^{2} \), so the species with the smallest nuclear charge, hydrogen, has the lowest ionization energy.