The ionization energy for a \(1 s\) electron in a silver atom is \(2.462 \times 10^{6} \mathrm{kJ} / \mathrm{mol}\) a. Determine an approximate value for \(Z_{\text {eff }}\) for the Ag \(1 s\) electron. Assume the Bohr model applies to the 1 s electron. \(Z_{\mathrm{eff}}\) is the apparent nuclear charge experienced by the electrons. b. How does \(Z_{\text {eff }}\) from part a compare to \(Z\) for Ag? Rationalize the relative numbers.

To find the effective nuclear charge (Z_eff), we will use Bohr's model, which describes the behavior of electrons in hydrogen-like atoms. This model states that the ionization energy (IE) is related to the effective nuclear charge through the following formula: IE = \(\frac{Z_{eff}^2 * e^2}{2 * a_0 * n^2}\), where e is the elementary charge, a_0 is the Bohr radius, and n is the principal quantum number for the electron. The ionization energy for the given electron in the Ag atom is provided, and we know that the electron is in the 1s orbital, which means n = 1.

First, we need to isolate Z_eff in the equation. By multiplying both sides of the equation by \(\frac{2 * a_0 * n^2}{e^2}\), we get: Z_eff^2 = \(\frac{2 * IE * a_0 * n^2}{e^2}\) Now, by taking the square root of both sides of the equation, we get: Z_eff = \(\sqrt{\frac{2 * IE * a_0 * n^2}{e^2}}\)

Now that we have the equation for Z_eff, we can plug in the values for IE, a_0, and n. The given ionization energy for Ag is \(2.462 * 10^6 \frac{kJ}{mol}\), which we need to convert to Joules by multiplying by \(10^3\). The values for the other constants are: a_0 = \(5.2918 * 10^{-11} m\), e = \(1.6022 * 10^{-19} C\), n = 1. Substituting these values into the equation, we get: Z_eff = \(\sqrt{\frac{2 * 2.462 * 10^3 * 5.2918 * 10^{-11} * (1)^2}{(1.6022 * 10^{-19})^2}}\) Calculating this expression gives: Z_eff ≈ 46.72

The atomic number of Ag (silver) is 47. Our calculated value for Z_eff for the Ag 1s electron is approximately 46.72. These numbers are very close, which indicates that the 1s electron experiences an effective nuclear charge that is very similar to the actual nuclear charge of Ag. The small difference between Z and Z_eff can be rationalized by considering the electron shielding effect. In a multi-electron atom like Ag, inner electrons shield the outer electrons from the full nuclear charge. Since the 1s electron is the closest electron to the nucleus, it experiences minimal shielding from other electrons, resulting in a Z_eff value that is very close to Z. Now we have found an approximate value for Z_eff for the Ag 1s electron, and the result matches our expectations based on the shielding effect.