An electron moves past a gold atom's nucleus, which comprises 79 protons and 118 neutrons. At a specific instant, the electron is 7.5 × 10^−9 m from the nucleus. (a) What is the magnitude of the electric force the nucleus exerts on the electron?

To calculate the magnitude of the electric force that the gold atom nucleus exerts on the electron, we can use Coulomb's Law. Coulomb's Law states that the electric force (F) between two charged particles is directly proportional to the product of the charges (q₁ and q₂) and inversely proportional to the square of the distance (r) between them. The formula is expressed as:

\[ F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \]

where \( k \) is Coulomb's constant (\( 8.9875 \times 10^9 \) N·m²/C²), \( q_1 \) and \( q_2 \) are the charges of the two particles, and \( r \) is the distance between the charges.

The charge of one proton is \( +e \), where \( e \) is the elementary charge (\( 1.602 \times 10^{-19} \) C). The charge of the gold nucleus will be \( +79e \), because there are 79 protons and the neutrons do not contribute to the charge. The charge of the electron is \( -e \).

Substituting the known values into the Coulomb's Law formula, we get:

\[ F = (8.9875 \times 10^9 \text{ N·m}^2/\text{C}^2) \cdot \frac{|79 \cdot e \cdot (-e)|}{(7.5 \times 10^{-9} \text{ m})^2} \] \[ F = (8.9875 \times 10^9 \text{ N·m}^2/\text{C}^2) \cdot \frac{79 \cdot (1.602 \times 10^{-19} \text{ C})^2}{(7.5 \times 10^{-9} \text{ m})^2} \]

Now perform the calculation:

\[ F ≈ (8.9875 \times 10^9) \cdot \frac{79 \cdot (1.602 \times 10^{-19})^2}{(7.5 \times 10^{-9})^2} \] \[ F ≈ (8.9875 \times 10^9) \cdot \frac{79 \cdot (2.56 \times 10^{-38})}{5.625 \times 10^{-17}} \] \[ F ≈ (8.9875 \times 10^9) \cdot \frac{202.24 \times 10^{-38}}{5.625 \times 10^{-17}} \] \[ F ≈ (8.9875 \times 10^9) \cdot 35.927 \times 10^{-21} \] \[ F ≈ 3.228 \times 10^{-11} \text{ N} \]

The magnitude of electric force the gold atom nucleus exerts on the electron is approximately \( 3.228 \times 10^{-11} \) newtons.