Bohr's model can be used for hydrogen-like ions -ions that have only one electron, such as \(\mathrm{He}^{+}\) and \(\mathrm{Li}^{2+}\). (a) Why is the Bohr model applicable to \(\mathrm{He}^{+}\) ions but not to neutral He atoms? (b) The ground-state energies of \(\mathrm{H}, \mathrm{He}^{+},\) and \(\mathrm{Li}^{2+}\) are tabulated as follows: By examining these numbers, propose a relationship between the ground-state energy of hydrogen-like systems and the nuclear charge, \(Z\). (c) Use the relationship you derive in part (b) to predict the ground-state energy of the \(\mathrm{C}^{5+}\) ion.

The Bohr model is based on the assumption that an electron revolves in circular orbits around the nucleus with quantized energy levels. This model works well for hydrogen-like ions (ions with a single electron) because the force between the nucleus and the single electron can be easily analyzed. In the case of neutral helium atoms, there are two electrons in which both electrons mutually interact. This makes the analysis of forces far more complex than that of hydrogen-like ions, and the Bohr model is not capable of accurately describing this interaction. Thus, the Bohr model is applicable to \(\mathrm{He}^{+}\) ions but not to neutral He atoms.

We are given the ground-state energies of \(\mathrm{H}, \mathrm{He}^{+}\), and \(\mathrm{Li}^{2+}\) as follows: - \(\mathrm{H}\): \(E_{\mathrm{H}}=-13.6 \,\text{eV}\) - \(\mathrm{He}^{+}\): \(E_{\mathrm{He}^{+}}=-54.4 \,\text{eV}\) - \(\mathrm{Li}^{2+}\): \(E_{\mathrm{Li}^{2+}}=-122.4 \,\text{eV}\) We can notice that the ground-state energies are proportional to the square of the nuclear charge (Z). Let's write this relationship as \(E = kZ^2\), where k is a constant. We can express the ground-state energies as: - \(\mathrm{H}\): \(-13.6=-k(1)^2\) - \(\mathrm{He}^{+}\): \(-54.4=-k(2)^2\) - \(\mathrm{Li}^{2+}\): \(-122.4=-k(3)^2\) By comparing the expression for \(\mathrm{H}\) with the others, we find that \(k=13.6\). Thus, the relationship between the ground-state energy of hydrogen-like systems and the nuclear charge is given by: \(E = -13.6Z^2\).

Now, we will use the relationship found in part (b) to predict the ground-state energy of \(\mathrm{C}^{5+}\) ion. The nuclear charge \(Z\) for \(\mathrm{C}^{5+}\) ion is 6. Applying the relationship, we get: \(E = -13.6(6)^2 = -13.6(36) = -489.6 \,\text{eV}\) Hence, the predicted ground-state energy of the \(\mathrm{C}^{5+}\) ion is \(-489.6 \,\text{eV}\).