Problem 61 - Physics for Scientists and Engineers with Modern Physics

Question

The ionization (binding) energy of the outermost electron in boron is \(8.26 \mathrm{eV}\). \((a)\) Use the Bohr model to estimate the "effective charge," \(Z_{\text {eff }}\), seen by this electron. (b) Estimate the average orbital radius.

Step-by-Step Solution

Answer

(a) The effective charge is approximately 1.33. (b) The average orbital radius is about 1.59 angstroms.

Step 1: Understanding Ionization Energy

Ionization energy is the energy required to remove an electron from an atom in its gaseous state. For boron, its outermost electron has an ionization energy of 8.26 eV.

Step 2: Applying the Bohr Model

The Bohr model relates the ionization energy to the effective nuclear charge seen by the electron. The formula is given by \(E_n = - 13.6 \frac{Z_{ ext{eff}}^2}{n^2}\), where \(E_n\) is the ionization energy, \(Z_{\text{eff}}\) is the effective charge, and \(n\) is the principal quantum number. Since we're dealing with the outermost electron of boron (in the 2p orbital), \(n = 2\).

Step 3: Solving for Effective Nuclear Charge

Given that the ionization energy \(E_n = 8.26 \text{ eV}\), and \(n = 2\), we solve for \(Z_{\text{eff}}\) using the equation \(8.26 = 13.6 \frac{Z_{\text{eff}}^2}{4}\). Rearranging and solving, we find \(Z_{\text{eff}} = \sqrt{\frac{8.26 \times 4}{13.6}} \approx 1.33\).

Step 4: Estimating Average Orbital Radius

The average orbital radius in the Bohr model is given by \( r = \frac{n^2 a_0}{Z_{\text{eff}}}\), where \(a_0 = 0.529 \) angstroms is the Bohr radius. For boron, \(n = 2\) and \(Z_{\text{eff}} \approx 1.33\), hence \( r = \frac{2^2 \times 0.529}{1.33} \) angstroms, which calculates to approximately 1.59 angstroms.

Key Concepts

Ionization Energy

Ionization energy is a fundamental concept in chemistry and physics. It refers to the energy needed to remove an electron from an atom or ion. This concept helps us understand an element's reactivity and how tightly an atom holds onto its electrons. For boron, the outermost electron requires 8.26 eV to be removed. This value is important as it indicates the energy level stability of the electron.

In the Bohr model, ionization energy is used to estimate the effective nuclear charge, which affects the electron's attraction to the nucleus. Higher ionization energies imply stronger nuclear-electron attractions. Understanding ionization energy also aids in predicting how an atom will bond with others. It is a key player in determining the chemical behavior of elements."

Effective Nuclear Charge

Effective nuclear charge (\(Z_\text{eff}\)) is the net positive charge experienced by an electron in a multi-electron atom. This charge impacts an electron's energy level and how tightly it is bound to the nucleus. \(Z_\text{eff}\) is often less than the total nuclear charge due to electron shielding and repulsion.

In the Bohr model, the formula \[E_n = -13.6 \frac{Z_\text{eff}^2}{n^2}\] helps us calculate the effective nuclear charge. For boron, solving \[8.26 = 13.6 \frac{Z_\text{eff}^2}{4}\] gives \(Z_\text{eff} \approx 1.33\). This suggests the outer electron "feels" a reduced charge due to shielding by inner electrons.

Understanding effective nuclear charge is vital for predicting elemental properties like size, ionization energy, and electronegativity. It shapes how atoms interact in chemical reactions and influences periodic trends."

Average Orbital Radius

The average orbital radius of an electron in an atom shows us the typical distance of the electron from the nucleus. In the Bohr model, this is calculated using the formula:\[ r = \frac{n^2 a_0}{Z_\text{eff}} \]where \(a_0\) is the Bohr radius (0.529 angstroms). This helps us understand the scale of the electron's orbit and indirectly, its energy level.

For boron, using \(n = 2\) and \(Z_\text{eff} \approx 1.33\), the orbital radius is roughly 1.59 angstroms. The orbital radius affects the atom's size and chemical properties like bond length and energy transitions within the atom. It provides a visual of how spread out the electron cloud is around the nucleus, which is critical in understanding atomic interactions in molecules."