A particle is given a velocity `(v_(e ))/(sqrt(3))` in a vertically upward direction from the surface of the earth, where `v_(e )` is the escape velocity from the surface of the earth. Let the radius of the earth be R. The height of the particle above the surface of the earth at the instant it comes to rest is :

Options

A. 3R
B. 2R
C. `(3R)/(4)`
D. `(R )/(2)`

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The initial kinetic energy of the particle will be converted into gravitational potential energy as it rises until it comes to rest.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The particle is given an initial velocity \( v = \frac{v_e}{\sqrt{3}} \), where \( v_e \) is the escape velocity.
- The radius of the Earth is \( R \).

2. **Calculate Initial Kinetic Energy (KE)**:
- The initial kinetic energy (KE) of the particle can be calculated using the formula:
\[ KE_i = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{v_e}{\sqrt{3}}\right)^2 = \frac{1}{2} m \frac{v_e^2}{3} \]

3. **Calculate Initial Potential Energy (PE)**:
- The initial potential energy (PE) at the surface of the Earth is given by:
\[ PE_i = -\frac{GMm}{R} \]
- Here, \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the particle.

4. **Final Conditions**:
- At the maximum height \( h \), the particle comes to rest, so its final kinetic energy (KE) is:
\[ KE_f = 0 \]
- The final potential energy (PE) at height \( h \) is:
\[ PE_f = -\frac{GMm}{R + h} \]

5. **Apply Conservation of Mechanical Energy**:
- According to the conservation of energy:
\[ KE_i + PE_i = KE_f + PE_f \]
- Substituting the values we have:
\[ \frac{1}{2} m \frac{v_e^2}{3} - \frac{GMm}{R} = 0 - \frac{GMm}{R + h} \]

6. **Simplify the Equation**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} \frac{v_e^2}{3} - \frac{GM}{R} = -\frac{GM}{R + h} \]
- Rearranging gives:
\[ \frac{1}{2} \frac{v_e^2}{3} = \frac{GM}{R} - \frac{GM}{R + h} \]

7. **Substitute \( v_e^2 \)**:
- We know that \( v_e^2 = \frac{2GM}{R} \). Substitute this into the equation:
\[ \frac{1}{2} \frac{2GM}{R \cdot 3} = \frac{GM}{R} - \frac{GM}{R + h} \]
- Simplifying gives:
\[ \frac{GM}{3R} = \frac{GM}{R} - \frac{GM}{R + h} \]

8. **Clear the GM**:
- Dividing through by \( GM \) (assuming \( GM \neq 0 \)):
\[ \frac{1}{3R} = \frac{1}{R} - \frac{1}{R + h} \]

9. **Combine the Terms**:
- Rearranging gives:
\[ \frac{1}{3R} = \frac{(R + h) - R}{R(R + h)} = \frac{h}{R(R + h)} \]

10. **Cross Multiply**:
- Cross multiplying gives:
\[ h = \frac{R(R + h)}{3} \]

11. **Solve for \( h \)**:
- Rearranging gives:
\[ 3h = R + h \implies 2h = R \implies h = \frac{R}{2} \]

### Final Answer:
The height of the particle above the surface of the Earth at the instant it comes to rest is:
\[ h = \frac{R}{2} \]