The velocity of a particle at which the kinetic energy is eqyal to its rest mass energy is

Options

A. `((3c)/(2))`
B. `3(c)/(sqrt(2))`
C. `((3c)^(1//2))/(2)`
D. `(csqrt(3))/(2)`

Explanation

To find the velocity of a particle at which its kinetic energy is equal to its rest mass energy, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Rest Mass Energy**:
The rest mass energy (E₀) of a particle is given by the equation:
\[ E_0 = m_0 c^2 \]
where \( m_0 \) is the rest mass of the particle and \( c \) is the speed of light.

2. **Relativistic Kinetic Energy**:
The relativistic kinetic energy (K.E.) of a particle is given by:
\[ K.E. = \frac{m_0 c^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m_0 c^2 \]
Simplifying this, we get:
\[ K.E. = m_0 c^2 \left( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1 \right) \]

3. **Set Kinetic Energy Equal to Rest Mass Energy**:
We want to find the velocity \( v \) such that:
\[ K.E. = E_0 \]
Therefore, we set:
\[ m_0 c^2 \left( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1 \right) = m_0 c^2 \]

4. **Cancel \( m_0 c^2 \)**:
Assuming \( m_0 \neq 0 \), we can divide both sides by \( m_0 c^2 \):
\[ \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1 = 1 \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 2 \]

6. **Square Both Sides**:
Squaring both sides results in:
\[ \frac{1}{1 - \frac{v^2}{c^2}} = 4 \]

7. **Cross Multiply**:
Cross multiplying gives:
\[ 1 = 4 \left( 1 - \frac{v^2}{c^2} \right) \]
This simplifies to:
\[ 1 = 4 - 4 \frac{v^2}{c^2} \]

8. **Rearranging for \( v^2 \)**:
Rearranging gives:
\[ 4 \frac{v^2}{c^2} = 3 \]
Therefore:
\[ \frac{v^2}{c^2} = \frac{3}{4} \]

9. **Finding \( v \)**:
Taking the square root of both sides:
\[ v = c \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} c \]

### Final Answer:
The velocity of the particle at which the kinetic energy is equal to its rest mass energy is:
\[ v = \frac{\sqrt{3}}{2} c \]