A proton and a deuteron both having the same kinetic energy, enter perpendicularly into a uniform magnetic field `B`. For motion of proton and deuteron on circular path or radius `R_(p)` and `R_(d)` respectively, the correct statement is

Options

A. `R_(d)=sqrt(2)R _(p)`
B. `R_(d)=R_9p)//sqrt(2)`
C. `R_(d)=R_(p)`
D. `R_(d)=2R_(p)`

Answer

To solve the problem, we need to analyze the motion of a proton and a deuteron in a magnetic field when both have the same kinetic energy. We will derive the relationship between their radii of circular motion in the magnetic field.

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

### Step 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:
\[ v = \sqrt{\frac{2 KE}{m}} \]

### Step 3: Substitute velocity into the radius formula
Substituting the expression for \( v \) into the radius formula gives:
\[ R = \frac{m \sqrt{\frac{2 KE}{m}}}{qB} = \frac{\sqrt{2m \cdot KE}}{qB} \]

### Step 4: Apply the formula for both particles
Let’s denote:
- \( m_p \) and \( m_d \) as the masses of the proton and deuteron respectively,
- \( q_p \) and \( q_d \) as the charges of the proton and deuteron respectively (both have the same charge \( e \)),
- \( KE \) as the common kinetic energy for both particles.

For the proton:
\[ R_p = \frac{\sqrt{2m_p \cdot KE}}{q_p B} \]

For the deuteron (which has a mass approximately twice that of the proton):
\[ R_d = \frac{\sqrt{2m_d \cdot KE}}{q_d B} = \frac{\sqrt{2(2m_p) \cdot KE}}{e B} = \frac{\sqrt{4m_p \cdot KE}}{e B} = 2 \cdot \frac{\sqrt{2m_p \cdot KE}}{e B} = 2R_p \]

### Step 5: Final relationship
Thus, we find that:
\[ R_d = 2R_p \]

### Conclusion
The correct statement is that the radius of the deuteron \( R_d \) is equal to \( 2R_p \). Therefore, the correct answer is:
\[ R_d = 2R_p \]