A deutron of kinetic energy `50` keV is describing a circular orbit of radius `0.5` meter in a plane perpendicular to magnetic field `vecB`. The kinetic energy of the proton that describes a circular orbit of radius `0.5` meter in the same plane with the same `vecB` is

Options

A. 25 KeV
B. 50 KeV
C. 200 KeV
D. 100 KeV

Answer

To solve the problem, we need to find the kinetic energy of a proton that describes a circular orbit of radius 0.5 meters in the same magnetic field as a deuteron with a given kinetic energy of 50 keV.

### Step-by-Step Solution:

1. **Understanding the Given Data:**
- Kinetic energy of the deuteron (K_d) = 50 keV
- Radius of the circular orbit (r) = 0.5 m
- Magnetic field (B) is the same for both particles.

2. **Kinetic Energy Formula:**
The kinetic energy (K) of a charged particle moving in a magnetic field can be expressed as:
\[ K = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the particle and \(v\) is its velocity.

3. **Centripetal Force and Magnetic Force:**
The centripetal force required to keep the particle in circular motion is provided by the magnetic force. Thus, we have:
\[ \frac{mv^2}{r} = qvB \]
where \(q\) is the charge of the particle.

4. **Solving for Velocity (v):**
Rearranging the above equation gives:
\[ mv = qBr \implies v = \frac{qBr}{m} \]

5. **Substituting Velocity into Kinetic Energy:**
Substituting \(v\) back into the kinetic energy formula:
\[ K = \frac{1}{2} m \left(\frac{qBr}{m}\right)^2 = \frac{1}{2} \frac{q^2B^2r^2}{m} \]

6. **Kinetic Energy for Deuteron:**
For the deuteron (mass \(m_d = 2m_p\) and charge \(q_d = q_p\)):
\[ K_d = \frac{1}{2} \frac{q_d^2B^2r^2}{m_d} = \frac{1}{2} \frac{q_p^2B^2r^2}{2m_p} = \frac{1}{4} \frac{q_p^2B^2r^2}{m_p} \]
Given \(K_d = 50 \text{ keV}\), we can express this as:
\[ K_d = \frac{1}{4} K_p \]

7. **Kinetic Energy for Proton:**
From the relationship derived:
\[ K_p = 4K_d = 4 \times 50 \text{ keV} = 200 \text{ keV} \]

8. **Final Answer:**
The kinetic energy of the proton is \(200\) keV.