An `alpha`-particle and a proton are fired through the same magnetic field which is perpendicular to their velocity vectors. The `alpha`-partcles and the proton move such that radius of curvature of their paths is same. Find the ratio of their de Broglie wavelengths.

Options

A. `2:0`
B. `3:4`
C. `5:7`
D. `1:2`

Explanation

To solve the problem, we need to find the ratio of the de Broglie wavelengths of an alpha particle and a proton, given that they are moving in the same magnetic field and have the same radius of curvature in their paths.

### Step-by-Step Solution:

1. **Understanding the Forces**:
The force experienced by a charged particle moving in a magnetic field is given by:
\[ F = qvB \sin \theta \]
Since the magnetic field is perpendicular to the velocity, \(\sin \theta = 1\). Therefore, the force simplifies to:
\[ F = qvB \]

2. **Centripetal Force**:
This magnetic force acts as the centripetal force required to keep the particle in circular motion. Thus, we can equate the magnetic force to the centripetal force:
\[ qvB = \frac{mv^2}{r} \]
Rearranging this gives us:
\[ mv = qBr \]

3. **De Broglie Wavelength**:
The de Broglie wavelength (\(\lambda\)) of a particle is given by:
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \]
where \(p\) is the momentum of the particle.

4. **Finding the Ratio of Wavelengths**:
We can express the de Broglie wavelengths for the alpha particle (\(\lambda_\alpha\)) and the proton (\(\lambda_p\)):
\[ \lambda_\alpha = \frac{h}{m_\alpha v_\alpha} \quad \text{and} \quad \lambda_p = \frac{h}{m_p v_p} \]
To find the ratio of the wavelengths:
\[ \frac{\lambda_\alpha}{\lambda_p} = \frac{m_p v_p}{m_\alpha v_\alpha} \]

5. **Using the Radius of Curvature**:
Since the radius of curvature for both particles is the same (\(r_\alpha = r_p\)), we can substitute \(mv\) from step 2:
\[ \frac{m_p v_p}{m_\alpha v_\alpha} = \frac{q_p r_p}{q_\alpha r_\alpha} \]
Given that \(r_p = r_\alpha\), this simplifies to:
\[ \frac{m_p v_p}{m_\alpha v_\alpha} = \frac{q_p}{q_\alpha} \]

6. **Charge of Particles**:
The charge of a proton (\(q_p\)) is \(e\) and the charge of an alpha particle (\(q_\alpha\)) is \(2e\):
\[ \frac{q_p}{q_\alpha} = \frac{e}{2e} = \frac{1}{2} \]

7. **Final Ratio of Wavelengths**:
Thus, we have:
\[ \frac{\lambda_\alpha}{\lambda_p} = \frac{1}{2} \]
Therefore, the ratio of the de Broglie wavelengths is:
\[ \lambda_\alpha : \lambda_p = 1 : 2 \]

### Conclusion:
The ratio of the de Broglie wavelengths of the alpha particle to the proton is \(1:2\).