A proton and an antiproton come close to each other in vacuum such that the distance between them is 10 cm. Consider the potential energy to be zero at infinity. The velocity at this distance will be

Options

A. 1.17 m/s
B. 2.3 m/s
C. 3.0 m/s
D. 23 m/s

Explanation

To find the velocity of a proton and an antiproton when they are 10 cm apart, we can use the principles of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the system
We have a proton (charge \( +e \)) and an antiproton (charge \( -e \)). The distance between them is given as \( r = 10 \, \text{cm} = 0.1 \, \text{m} \). The potential energy is considered to be zero at infinity.

### Step 2: Write the expression for potential energy
The potential energy \( U \) between two point charges is given by the formula:
\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]
where:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) (Coulomb's constant),
- \( q_1 = +e = 1.6 \times 10^{-19} \, \text{C} \) (charge of the proton),
- \( q_2 = -e = -1.6 \times 10^{-19} \, \text{C} \) (charge of the antiproton),
- \( r = 0.1 \, \text{m} \).

### Step 3: Calculate the potential energy at 10 cm
Substituting the values into the potential energy formula:
\[ U = \frac{(9 \times 10^9) \cdot (1.6 \times 10^{-19}) \cdot (-1.6 \times 10^{-19})}{0.1} \]
Calculating this gives:
\[ U = \frac{-9 \times 10^9 \cdot (2.56 \times 10^{-38})}{0.1} = -2.56 \times 10^{-28} \, \text{J} \]

### Step 4: Apply conservation of energy
Initially, the kinetic energy \( K_i \) is zero (as they are at rest), and the potential energy \( U_i \) is zero at infinity. At a distance of 10 cm, the potential energy is \( U_f \) and the kinetic energy is \( K_f \).

Using conservation of energy:
\[ K_i + U_i = K_f + U_f \]
\[ 0 + 0 = K_f + (-2.56 \times 10^{-28}) \]
Thus,
\[ K_f = 2.56 \times 10^{-28} \, \text{J} \]

### Step 5: Relate kinetic energy to velocity
The kinetic energy of a proton is given by:
\[ K = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the proton \( m = 1.67 \times 10^{-27} \, \text{kg} \).

Setting \( K_f = \frac{1}{2} mv^2 \):
\[ 2.56 \times 10^{-28} = \frac{1}{2} (1.67 \times 10^{-27}) v^2 \]

### Step 6: Solve for velocity \( v \)
Rearranging gives:
\[ v^2 = \frac{2 \cdot 2.56 \times 10^{-28}}{1.67 \times 10^{-27}} \]
Calculating this:
\[ v^2 = \frac{5.12 \times 10^{-28}}{1.67 \times 10^{-27}} = 0.307 \, \text{m}^2/\text{s}^2 \]
Taking the square root:
\[ v = \sqrt{0.307} \approx 0.554 \, \text{m/s} \]

### Step 7: Final answer
Thus, the velocity of the proton and antiproton when they are 10 cm apart is approximately:
\[ v \approx 0.554 \, \text{m/s} \]