A electron beam particle is accelerated from rest through a potential difference of V volt. The speed of the particle is

Options

A. `sqrt((2eV)/(m))`
B. `(m)/(e V)`
C. `sqrt(me V)`
D. `sqrt((4e V)/(m))`.

Explanation

To find the speed of an electron beam particle that is accelerated from rest through a potential difference of \( V \) volts, we can use the principles of energy conservation. Here’s a step-by-step solution:

### Step 1: Understand the energy conversion
When the electron is accelerated through a potential difference \( V \), the electrical potential energy is converted into kinetic energy. The work done on the electron by the electric field is given by the equation:
\[ W = Q \cdot V \]
where \( Q \) is the charge of the electron and \( V \) is the potential difference.

### Step 2: Relate work done to kinetic energy
The work done on the electron is equal to its change in kinetic energy. Since the electron starts from rest, its initial kinetic energy is zero. Therefore, the work done can be expressed as:
\[ W = \text{K.E.} = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its final speed.

### Step 3: Set the equations equal
Equating the work done to the kinetic energy, we have:
\[ Q \cdot V = \frac{1}{2} mv^2 \]

### Step 4: Substitute the charge of the electron
The charge of an electron \( Q \) is equal to \( e \) (approximately \( 1.6 \times 10^{-19} \) coulombs). Thus, we can rewrite the equation as:
\[ eV = \frac{1}{2} mv^2 \]

### Step 5: Solve for speed \( v \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{2eV}{m} \]
Taking the square root of both sides gives:
\[ v = \sqrt{\frac{2eV}{m}} \]

### Final Result
Thus, the speed of the electron beam particle after being accelerated through a potential difference \( V \) is:
\[ v = \sqrt{\frac{2eV}{m}} \]