A proton, a deuteron and an `alpha` particle having same momentum enter a uniform magnetic field at right angles to the field. Then the ratio of their angular momenta during their motion in the magnetic field is

Options

A. `2:2:1`
B. `2:1:3`
C. `4:1:2`
D. `4:2:1`

Answer

To solve the problem, we need to find the ratio of the angular momenta of a proton, a deuteron, and an alpha particle when they have the same momentum and enter a uniform magnetic field at right angles.

### Step-by-Step Solution:

1. **Understanding Angular Momentum in a Magnetic Field**:
The angular momentum \( L \) of a charged particle moving in a magnetic field is given by the formula:
\[ L = P \cdot R \]
where \( P \) is the linear momentum and \( R \) is the radius of the circular path of the particle in the magnetic field.

2. **Finding the Radius of the Circular Path**:
The radius \( R \) of the circular path for a charged particle in a magnetic field is given by:
\[ R = \frac{MV}{QB} \]
where \( M \) is the mass of the particle, \( V \) is its velocity, \( Q \) is its charge, and \( B \) is the magnetic field strength.

3. **Substituting for Angular Momentum**:
Substituting the expression for \( R \) into the angular momentum formula gives:
\[ L = P \cdot \left(\frac{MV}{QB}\right) = \frac{P \cdot MV}{QB} \]

4. **Relating Momentum to Charge and Mass**:
Since we are given that all three particles (proton, deuteron, and alpha particle) have the same momentum \( P \), we can express their angular momenta as follows:
- For the proton:
\[ L_p = \frac{P \cdot M_p \cdot V_p}{Q_p \cdot B} \]
- For the deuteron:
\[ L_d = \frac{P \cdot M_d \cdot V_d}{Q_d \cdot B} \]
- For the alpha particle:
\[ L_{\alpha} = \frac{P \cdot M_{\alpha} \cdot V_{\alpha}}{Q_{\alpha} \cdot B} \]

5. **Charge Values**:
- Charge of proton \( Q_p = e \)
- Charge of deuteron \( Q_d = e \)
- Charge of alpha particle \( Q_{\alpha} = 2e \)

6. **Mass Values**:
- Mass of proton \( M_p = m_p \)
- Mass of deuteron \( M_d = 2m_p \) (approximately, since deuteron consists of one proton and one neutron)
- Mass of alpha particle \( M_{\alpha} = 4m_p \) (since it consists of 2 protons and 2 neutrons)

7. **Finding the Ratio of Angular Momenta**:
Since \( P \) and \( B \) are the same for all particles, we can simplify the ratios:
\[ \frac{L_p}{L_d} = \frac{M_p}{Q_p} : \frac{M_d}{Q_d} : \frac{M_{\alpha}}{Q_{\alpha}} \]
Substituting the values:
\[ \frac{L_p}{L_d} = \frac{m_p}{e} : \frac{2m_p}{e} : \frac{4m_p}{2e} \]
This simplifies to:
\[ L_p : L_d : L_{\alpha} = \frac{m_p}{e} : \frac{2m_p}{e} : \frac{2m_p}{e} \]
Which reduces to:
\[ L_p : L_d : L_{\alpha} = 1 : 2 : 2 \]

8. **Final Ratio**:
Thus, the ratio of their angular momenta is:
\[ L_p : L_d : L_{\alpha} = 1 : 2 : 4 \]

### Conclusion:
The ratio of their angular momenta is \( 1 : 2 : 4 \).