An electron is emittied with a veloctiy of ` 5 xx 10 ^(6) ms^(-1)` . It is accelerated by an electric field in the direction of initial velocity at `3xx 10^(14) ms^(-2)` . If its final velocity is `7xx10^(6) ms^(-1)` calculation the distance coverd by the electron .

Answer

To solve the problem, we will use the kinematic equation that relates initial velocity (u), final velocity (v), acceleration (a), and distance (s):

\[ v^2 = u^2 + 2as \]

### Step 1: Identify the given values
- Initial velocity, \( u = 5 \times 10^6 \, \text{m/s} \)
- Final velocity, \( v = 7 \times 10^6 \, \text{m/s} \)
- Acceleration, \( a = 3 \times 10^{14} \, \text{m/s}^2 \)

### Step 2: Substitute the values into the equation
We need to rearrange the kinematic equation to solve for distance \( s \):

\[ s = \frac{v^2 - u^2}{2a} \]

### Step 3: Calculate \( v^2 \) and \( u^2 \)
- Calculate \( v^2 \):
\[ v^2 = (7 \times 10^6)^2 = 49 \times 10^{12} \, \text{m}^2/\text{s}^2 \]

- Calculate \( u^2 \):
\[ u^2 = (5 \times 10^6)^2 = 25 \times 10^{12} \, \text{m}^2/\text{s}^2 \]

### Step 4: Substitute \( v^2 \) and \( u^2 \) into the equation
Now substitute \( v^2 \) and \( u^2 \) back into the equation for \( s \):

\[ s = \frac{49 \times 10^{12} - 25 \times 10^{12}}{2 \times (3 \times 10^{14})} \]

### Step 5: Simplify the numerator
Calculate the numerator:

\[ 49 \times 10^{12} - 25 \times 10^{12} = 24 \times 10^{12} \]

### Step 6: Calculate \( s \)
Now substitute the values into the equation:

\[ s = \frac{24 \times 10^{12}}{6 \times 10^{14}} = \frac{24}{6} \times 10^{12 - 14} = 4 \times 10^{-2} \, \text{m} \]

### Final Answer
The distance covered by the electron is:

\[ s = 4 \times 10^{-2} \, \text{m} = 0.04 \, \text{m} \]

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