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Chapter 7: Problem 185

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 \(A g 1 s\) 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.

Short Answer

Expert verified
The approximate effective nuclear charge ($Z_{\text{eff}}$) for the 1s electron in a silver (Ag) atom is 55.7, which is larger than its actual atomic number (Z) of 47. This difference is due to the influence of other electrons in the atom, as they cause electron-electron repulsion, making the effective nuclear charge experienced by the 1s electron higher than the actual nuclear charge. However, this is only an approximation, since the Bohr model is not fully accurate for multi-electron atoms like Ag.

Step by step solution

01

Recall the formula for ionization energy in the Bohr model

The ionization energy (E) for an electron in a hydrogen-like atom can be calculated using the relationship: \( E = \dfrac{Z_{\text{eff}}^2 \cdot 13.6\ \text{eV}}{n^2} \) Here, Z_eff is the effective nuclear charge, and n is the principal quantum number for the electron. Since we are dealing with the 1s electron, n equals 1.
02

Convert ionization energy from kJ/mol to eV

The given ionization energy is in kJ/mol. To use the formula above, we need to convert it to eV. We can use conversion factors: 1 eV = 96.485 kJ/mol and 1 mol = 6.022 x 10^23 atoms. \( 2.462 \times 10^{6} \ \frac{\text{kJ}}{\text{mol}} \cdot \dfrac{1 \ \text{eV}}{96.485 \ \text{kJ/mol}} \cdot \dfrac{1 \ \text{mol}}{6.022 \times 10^{23} \ \text{atoms}} = 4.222 \times 10^{4} \ \text{eV} \)
03

Rearrange the formula and solve for Z_eff

Now that we have the ionization energy in eV, we can rearrange and plug the values into our formula to solve for Z_eff: \( Z_{\text{eff}}^2 = \dfrac{E \cdot n^2}{13.6 \ \text{eV}} = \dfrac{4.222 \times 10^4 \ \text{eV}}{13.6 \ \text{eV}} \) \( Z_{\text{eff}}^2 = 3103.09 \) Solve for Z_eff: \( Z_{\text{eff}} = \sqrt{3103.09} = 55.7 \) Thus, the approximate effective nuclear charge (Z_eff) for the Ag 1s electron is 55.7. #b. How does Z_eff from part a compare to Z for Ag? Rationalize the relative numbers.
04

Compare Z_eff to Z for Ag

The atomic number of Ag (silver) is 47. Comparing Z_eff (55.7) to Z (47) shows that Z_eff is larger than the actual atomic number for Ag.
05

Rationalize the difference between Z_eff and Z

The higher Z_eff value for the Ag 1s electron compared to the actual atomic number is due to the influence of other electrons in the atom. In multi-electron atoms, electrons experience both nuclear charge from the nucleus and electron-electron repulsion. Consequently, the effective nuclear charge experienced by the 1s electron is higher than the actual nuclear charge. However, we should note that this is only an approximation since the Bohr model is not fully accurate for multi-electron atoms like Ag.

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