An electron moving with velocity of `5 xx 10^(4)` m/s. When it enters into an electric field, it gains an acceleration of `1 xx 10^(15)`. In how much time, velocity of electron will become double of initial velocity?

Options

A. `5xx10^(-11) s`
B. `4xx10^(-11)s`
C. `3xx10^(-11)s`
D. `2xx10^(-11)s`

Explanation

To solve the problem step by step, we will use the kinematic equation that relates initial velocity, final velocity, acceleration, and time.

### Step 1: Identify the given values
- Initial velocity (u) = \(5 \times 10^4 \, \text{m/s}\)
- Acceleration (a) = \(1 \times 10^{15} \, \text{m/s}^2\)
- We need to find the time (t) when the final velocity (V) becomes double the initial velocity.

### Step 2: Determine the final velocity
The final velocity (V) when it becomes double the initial velocity can be calculated as:
\[ V = 2u = 2 \times (5 \times 10^4) = 10 \times 10^4 = 1 \times 10^5 \, \text{m/s} \]

### Step 3: Use the kinematic equation
We will use the kinematic equation:
\[ V = u + at \]
Substituting the known values:
\[ 1 \times 10^5 = 5 \times 10^4 + (1 \times 10^{15})t \]

### Step 4: Rearrange the equation to solve for time (t)
Rearranging the equation gives:
\[ 1 \times 10^5 - 5 \times 10^4 = (1 \times 10^{15})t \]
\[ 5 \times 10^4 = (1 \times 10^{15})t \]

### Step 5: Solve for time (t)
Now, divide both sides by \(1 \times 10^{15}\):
\[ t = \frac{5 \times 10^4}{1 \times 10^{15}} = 5 \times 10^{-11} \, \text{seconds} \]

### Final Answer
The time required for the velocity of the electron to become double its initial velocity is:
\[ t = 5 \times 10^{-11} \, \text{seconds} \]

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