An electron is accelerated through a potential difference of `200` volts. If `e//m` for the electron be `1.6 xx 10^(11)` coulomb/kg, the velocity acquired by the electron will be

Options

A. `8xx10^(6)ms^(-1)`
B. `8xx10^(5)ms^(-1)`
C. `5.9xx10^(6) ms^(-1)`
D. `5.9xx10^(5) ms^(-1)`

Explanation

To find the velocity acquired by an electron accelerated through a potential difference of 200 volts, we can use the relationship between potential energy and kinetic energy. Here’s a step-by-step solution:

### Step 1: Understand the relationship between potential energy and kinetic energy
When an electron is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field. The potential energy gained by the electron is given by:
\[ PE = eV \]
where \( e \) is the charge of the electron and \( V \) is the potential difference.

### Step 2: Write the expression for kinetic energy
The kinetic energy (KE) acquired by the electron is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron and \( v \) is the velocity acquired.

### Step 3: Set the potential energy equal to kinetic energy
Since the potential energy gained is converted into kinetic energy, we can set these two equations equal to each other:
\[ eV = \frac{1}{2} mv^2 \]

### Step 4: Rearrange the equation to solve for velocity \( v \)
Rearranging the equation gives us:
\[ v^2 = \frac{2eV}{m} \]
Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{2eV}{m}} \]

### Step 5: Substitute the given values
We know:
- The potential difference \( V = 200 \) volts
- The charge-to-mass ratio \( \frac{e}{m} = 1.6 \times 10^{11} \) coulomb/kg

We can express \( e \) in terms of \( \frac{e}{m} \):
\[ e = \frac{e}{m} \cdot m \]
However, we can directly substitute \( \frac{e}{m} \) into the equation:
\[ v = \sqrt{2 \cdot \frac{e}{m} \cdot V} \]
Substituting the values:
\[ v = \sqrt{2 \cdot (1.6 \times 10^{11}) \cdot (200)} \]

### Step 6: Calculate the velocity
Calculating the value inside the square root:
\[ v = \sqrt{2 \cdot 1.6 \times 10^{11} \cdot 200} \]
\[ = \sqrt{6.4 \times 10^{13}} \]
\[ = 8 \times 10^{6} \text{ m/s} \]

### Final Answer
Thus, the velocity acquired by the electron is:
\[ v = 8 \times 10^{6} \text{ m/s} \]

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