Question

The nucleus of helium atom contains two protons that are separated by a distance $3.0 \times 10^{-15}$ m. The magnitude of the electrostatic force that each proton exerts on the other is

• A. 20.6 N
• B. 25.6 N
• C. 15.6 N
• D. 12.6 N

Hint:

When setting up calculations involving nuclear dimensions, look for numbers that cancel nicely. Squaring the distance $3.0 \times 10^{-15}$ gives a $9.0$ in the denominator, which completely cancels out Coulomb's constant ($9.0 \times 10^9$). This leaves you to calculate only the square of the proton's base charge: $1.6^2 = 2.56$, making it easy to identify 25.6 N as the matching choice.

Answer 1

The Correct Option is B

Concept: According to Coulomb's Law, the magnitude of the electrostatic force ($F$) acting between two stationary point charges, $q_1$ and $q_2$, separated by a distance $r$ in vacuum or air is given by the formula: \[ F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \] Where:
• $k$ is Coulomb's constant, approximately equal to $9.0 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2$.
• $q_1$ and $q_2$ are the magnitudes of the charges.
• $r$ is the separation distance between the centers of the two charges.

Step 1: Identify the given values from the problem.

• Charge of a single proton ($q_1 = q_2 = e$) $\approx 1.6 \times 10^{-19}\text{ C}$
• Separation distance ($r$) = $3.0 \times 10^{-15}\text{ m}$ (often referred to as $3 \text{ fm}$ or femtometers)
• Electrostatic constant ($k$) = $9.0 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$

Step 2: Substitute the values into Coulomb's law equation.
\[ F = (9.0 \times 10^9) \times \frac{(1.6 \times 10^{-19}) \times (1.6 \times 10^{-19})}{(3.0 \times 10^{-15})^2} \] First, let's square the terms in the denominator: \[ (3.0 \times 10^{-15})^2 = 9.0 \times 10^{-30}\text{ m}^2 \] Now, write out the complete expression with the simplified denominator: \[ F = \frac{9.0 \times 10^9 \times 1.6 \times 1.6 \times 10^{-38}}{9.0 \times 10^{-30}} \]

Step 3: Simplify the exponents and numbers to calculate the final force.
We can cancel out the factor of $9.0$ from both the numerator and the denominator: \[ F = \frac{\cancel{9.0} \times 10^9 \times 2.56 \times 10^{-38}}{\cancel{9.0} \times 10^{-30}} \] Combine the powers of 10 in the numerator: \[ 10^9 \times 10^{-38} = 10^{-29} \] This leaves us with: \[ F = \frac{2.56 \times 10^{-29}}{10^{-30}} \] Using exponent rules to divide power bases ($\frac{10^{-29}}{10^{-30}} = 10^{-29 - (-30)} = 10^1 = 10$): \[ F = 2.56 \times 10 = 25.6\text{ N} \] Therefore, the magnitude of the electrostatic repulsive force that each proton exerts on the other is exactly $25.6\text{ N}$.