The nucleus of helium atom contains two protons that are separated by distance `3.0xx10^(-15)m`. The magnitude of the electrostatic force that each proton exerts on the other is

Options

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

Explanation

To find the magnitude of the electrostatic force that each proton exerts on the other, we can use Coulomb's Law, which is given by the formula:

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

Where:
- \( F \) is the electrostatic force between the charges,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges (for protons, \( q = 1.6 \times 10^{-19} \, \text{C} \)),
- \( r \) is the distance between the charges.

### Step 1: Identify the values
- Charge of a proton, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Distance between the protons, \( r = 3.0 \times 10^{-15} \, \text{m} \)
- Permittivity of free space, \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)

### Step 2: Substitute the values into Coulomb's Law
Since both protons have the same charge, we can write:

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q^2}{r^2} \]

Substituting the known values:

\[ F = \frac{1}{4 \pi (8.85 \times 10^{-12})} \frac{(1.6 \times 10^{-19})^2}{(3.0 \times 10^{-15})^2} \]

### Step 3: Calculate \( q^2 \) and \( r^2 \)
Calculate \( q^2 \):

\[ q^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \]

Calculate \( r^2 \):

\[ r^2 = (3.0 \times 10^{-15})^2 = 9.0 \times 10^{-30} \, \text{m}^2 \]

### Step 4: Substitute \( q^2 \) and \( r^2 \) back into the equation
Now substituting \( q^2 \) and \( r^2 \) into the formula for \( F \):

\[ F = \frac{1}{4 \pi (8.85 \times 10^{-12})} \frac{2.56 \times 10^{-38}}{9.0 \times 10^{-30}} \]

### Step 5: Calculate the value of \( F \)
First, calculate the denominator:

\[ 4 \pi (8.85 \times 10^{-12}) \approx 1.11 \times 10^{-10} \]

Now substitute this back into the equation:

\[ F = \frac{1}{1.11 \times 10^{-10}} \cdot \frac{2.56 \times 10^{-38}}{9.0 \times 10^{-30}} \]

Calculating the fraction:

\[ \frac{2.56 \times 10^{-38}}{9.0 \times 10^{-30}} \approx 2.84 \times 10^{-9} \]

Now calculate \( F \):

\[ F \approx \frac{2.84 \times 10^{-9}}{1.11 \times 10^{-10}} \approx 25.6 \, \text{N} \]

### Final Answer
The magnitude of the electrostatic force that each proton exerts on the other is approximately \( 25.6 \, \text{N} \).

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