A deuteron of kinetic energy `50` keV is desvribing 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`

Explanation

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 kinetic energy of 50 keV.

### Step-by-Step Solution:

1. **Understand the relationship between radius, mass, charge, and kinetic energy in a magnetic field:**
The radius \( r \) of the circular motion of a charged particle in a magnetic field is given by the formula:
\[ r = \frac{mv}{qB} \]
where \( m \) is the mass of the particle, \( v \) is its velocity, \( q \) is the charge, and \( B \) is the magnetic field strength.

2. **Relate kinetic energy to velocity:**
The kinetic energy \( KE \) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
From this, we can express the velocity \( v \) in terms of kinetic energy:
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

3. **Substituting velocity into the radius formula:**
Substituting \( v \) into the radius formula gives:
\[ r = \frac{m \sqrt{\frac{2 \cdot KE}{m}}}{qB} = \frac{\sqrt{2m \cdot KE}}{qB} \]

4. **Setting up the equations for the deuteron and proton:**
For the deuteron (mass \( m_d \) and kinetic energy \( KE_d = 50 \) keV):
\[ r = \frac{\sqrt{2m_d \cdot KE_d}}{qB} \]
For the proton (mass \( m_p \) and kinetic energy \( KE_p \)):
\[ r = \frac{\sqrt{2m_p \cdot KE_p}}{qB} \]

5. **Equating the radii:**
Since both particles are in the same magnetic field and have the same radius:
\[ \frac{\sqrt{2m_d \cdot KE_d}}{qB} = \frac{\sqrt{2m_p \cdot KE_p}}{qB} \]
This simplifies to:
\[ \sqrt{m_d \cdot KE_d} = \sqrt{m_p \cdot KE_p} \]

6. **Squaring both sides:**
Squaring both sides gives:
\[ m_d \cdot KE_d = m_p \cdot KE_p \]

7. **Substituting known values:**
The mass of the deuteron \( m_d \) is approximately \( 2m_p \) (since a deuteron consists of one proton and one neutron). Thus:
\[ 2m_p \cdot 50 \text{ keV} = m_p \cdot KE_p \]

8. **Solving for \( KE_p \):**
Dividing both sides by \( m_p \):
\[ 2 \cdot 50 \text{ keV} = KE_p \]
Therefore:
\[ KE_p = 100 \text{ keV} \]

### Final Answer:
The kinetic energy of the proton is \( 100 \) keV.

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