The linear momentum of an electron, initially at rest, accelerated through a potential difference of 100 V is 

Options

A. `9.1 xx 10^(-24) kg ms^(-1)`
B. `6.5 xx 10^(-24) kg ms^(-1)`
C. `5.4 xx 10^(-24) kg ms^(-1)`
D. `1.6 xx 10^(-24) kg ms^(-1)`

Explanation

To solve the problem of finding the linear momentum of an electron that has been accelerated through a potential difference of 100 V, we can follow these steps:

### Step 1: Calculate the kinetic energy of the electron
When an electron is accelerated through a potential difference (V), it gains kinetic energy (KE) equal to the work done on it by the electric field. The kinetic energy gained by the electron can be calculated using the formula:

\[ KE = eV \]

where:
- \( e \) is the charge of the electron (\( e \approx 1.6 \times 10^{-19} \) coulombs),
- \( V \) is the potential difference (100 V in this case).

Substituting the values:

\[ KE = (1.6 \times 10^{-19} \, \text{C})(100 \, \text{V}) = 1.6 \times 10^{-17} \, \text{J} \]

### Step 2: Relate kinetic energy to momentum
The kinetic energy of an electron can also be expressed in terms of its momentum (p) using the formula:

\[ KE = \frac{p^2}{2m} \]

where:
- \( m \) is the mass of the electron (\( m \approx 9.11 \times 10^{-31} \) kg).

We can rearrange this formula to solve for momentum:

\[ p = \sqrt{2m \cdot KE} \]

### Step 3: Substitute the values into the momentum formula
Now we can substitute the values of \( KE \) and \( m \) into the momentum formula:

\[ p = \sqrt{2 \cdot (9.11 \times 10^{-31} \, \text{kg}) \cdot (1.6 \times 10^{-17} \, \text{J})} \]

Calculating this step-by-step:

1. Calculate \( 2m \):
\[ 2m = 2 \cdot (9.11 \times 10^{-31}) = 1.822 \times 10^{-30} \, \text{kg} \]

2. Multiply by \( KE \):
\[ 2m \cdot KE = (1.822 \times 10^{-30}) \cdot (1.6 \times 10^{-17}) = 2.9152 \times 10^{-47} \, \text{kg} \cdot \text{J} \]

3. Take the square root:
\[ p = \sqrt{2.9152 \times 10^{-47}} \approx 5.4 \times 10^{-24} \, \text{kg m/s} \]

### Final Answer
Thus, the linear momentum of the electron after being accelerated through a potential difference of 100 V is approximately:

\[ p \approx 5.4 \times 10^{-24} \, \text{kg m/s} \]