Two electrons each moving with a velocity of `10^(6) ms^(-1)` are released towards eachother. What will be the closest distance of approach between them ?

Options

A. `1.53 xx 10^(-8)` m
B. `2.53 xx 10^(-10)` m
C. `2.53 xx 10^(-6)` m
D. zero

Answer

To find the closest distance of approach between two electrons moving towards each other with a velocity of \(10^6 \, \text{m/s}\), we can use the principle of conservation of energy. Here's a step-by-step solution:

### Step 1: Understand the Initial Conditions
Both electrons are moving towards each other with the same velocity \(v = 10^6 \, \text{m/s}\). Since they are both negatively charged, they will repel each other due to electrostatic force.

### Step 2: Initial Kinetic Energy
The initial kinetic energy (\(KE_i\)) of each electron can be calculated using the formula:
\[ KE = \frac{1}{2}mv^2 \]
where \(m\) is the mass of an electron and \(v\) is the velocity.

The mass of an electron \(m = 9.1 \times 10^{-31} \, \text{kg}\).

Thus, the total initial kinetic energy of the two electrons is:
\[ KE_i = 2 \times \frac{1}{2}mv^2 = mv^2 \]

### Step 3: Calculate Initial Kinetic Energy
Substituting the values:
\[ KE_i = 9.1 \times 10^{-31} \, \text{kg} \times (10^6 \, \text{m/s})^2 \]
\[ KE_i = 9.1 \times 10^{-31} \times 10^{12} = 9.1 \times 10^{-19} \, \text{J} \]

### Step 4: Final Potential Energy
As the electrons approach each other, they will come to a stop at the closest distance of approach \(r\). At this point, all the kinetic energy will be converted into potential energy (\(PE\)) due to electrostatic repulsion.

The potential energy between two charges is given by:
\[ PE = \frac{k \cdot |q_1 \cdot q_2|}{r} \]
where \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\) (Coulomb's constant) and \(q_1 = q_2 = -e\) (charge of an electron, \(e = 1.6 \times 10^{-19} \, \text{C}\)).

Thus,
\[ PE = \frac{k \cdot e^2}{r} \]

### Step 5: Apply Conservation of Energy
According to the conservation of energy:
\[ KE_i = PE \]
\[ 9.1 \times 10^{-19} \, \text{J} = \frac{(9 \times 10^9) \cdot (1.6 \times 10^{-19})^2}{r} \]

### Step 6: Solve for \(r\)
Rearranging the equation to solve for \(r\):
\[ r = \frac{(9 \times 10^9) \cdot (1.6 \times 10^{-19})^2}{9.1 \times 10^{-19}} \]

Calculating the right side:
\[ r = \frac{(9 \times 10^9) \cdot (2.56 \times 10^{-38})}{9.1 \times 10^{-19}} \]
\[ r = \frac{23.04 \times 10^{-29}}{9.1 \times 10^{-19}} \approx 2.53 \times 10^{-10} \, \text{m} \]

### Conclusion
The closest distance of approach between the two electrons is approximately:
\[ r \approx 2.53 \times 10^{-10} \, \text{m} \]