The electrons are accelerated through a pontential difference V by an electron gun. The electron charge and mass are e and m respectively. The maximum velocity attained by them is

Options

A. `(2m)/(eV)`
B. `(V^(2))/(2em)`
C. `(2eV)/m`
D. `sqrt((2eV)/m)`

Explanation

To find the maximum velocity attained by an electron when it is accelerated through a potential difference \( V \), we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential difference
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 kinetic energy \( KE \) gained by the electron can be expressed as:
\[ KE = eV \]
where \( e \) is the charge of the electron.

### Step 2: Relate kinetic energy to velocity
The kinetic energy of an object can also be expressed in terms of its mass \( m \) and velocity \( v \):
\[ KE = \frac{1}{2} mv^2 \]
where \( v \) is the velocity of the object.

### Step 3: Set the two expressions for kinetic energy equal to each other
Since both expressions represent the kinetic energy of the electron, we can set them equal to each other:
\[ eV = \frac{1}{2} mv^2 \]

### Step 4: Solve for velocity
To find the maximum velocity \( v_{max} \), we need to isolate \( v \) in the equation. We can rearrange the equation as follows:
\[ mv^2 = 2eV \]
Now, divide both sides by \( m \):
\[ v^2 = \frac{2eV}{m} \]
Finally, take the square root of both sides to solve for \( v \):
\[ v_{max} = \sqrt{\frac{2eV}{m}} \]

### Conclusion
The maximum velocity attained by the electron when accelerated through a potential difference \( V \) is given by:
\[ v_{max} = \sqrt{\frac{2eV}{m}} \]