A cathod rays 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/(ev) `
C. `sqrt(mev)`
D. `sqrt((4eV)/(m))`

Answer

To find the speed of a cathode ray particle accelerated from rest through a potential difference of \( V \) volts, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The cathode ray particle (which is usually an electron) starts from rest. Therefore, its initial kinetic energy is zero.

2. **Work Done on the Particle**:
- When the particle is accelerated through a potential difference \( V \), the work done on the particle can be expressed as:
\[ W = qV \]
- Here, \( q \) is the charge of the particle. For an electron, \( q = e \) (the elementary charge).

3. **Kinetic Energy Relation**:
- The work done on the particle is converted into kinetic energy. Thus, we can write:
\[ qV = \text{Kinetic Energy} \]
- The kinetic energy (KE) of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
- Where \( m \) is the mass of the particle and \( v \) is its final speed.

4. **Setting Up the Equation**:
- Equating the work done to the kinetic energy, we have:
\[ qV = \frac{1}{2} mv^2 \]
- Substituting \( q = e \):
\[ eV = \frac{1}{2} mv^2 \]

5. **Solving for Speed \( v \)**:
- Rearranging the equation to solve for \( v \):
\[ mv^2 = 2eV \]
\[ v^2 = \frac{2eV}{m} \]
\[ v = \sqrt{\frac{2eV}{m}} \]

### Final Result:
The speed of the cathode ray particle after being accelerated through a potential difference \( V \) is given by:
\[ v = \sqrt{\frac{2eV}{m}} \]