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 of finding the ratio of de Broglie wavelengths of an alpha particle and a proton moving in the same magnetic field with the same radius of curvature, we can follow these steps:

### Step 1: Understand the relationship between radius of curvature and momentum
The radius of curvature (r) of a charged particle moving in a magnetic field (B) 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: Set up the equation for both particles
For the alpha particle:
\[ r_{\alpha} = \frac{m_{\alpha} v_{\alpha}}{q_{\alpha} B} \]

For the proton:
\[ r_{p} = \frac{m_{p} v_{p}}{q_{p} B} \]

Since the radius of curvature is the same for both particles, we can equate the two equations:
\[ \frac{m_{\alpha} v_{\alpha}}{q_{\alpha}} = \frac{m_{p} v_{p}}{q_{p}} \]

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

Thus, for the alpha particle:
\[ \lambda_{\alpha} = \frac{h}{m_{\alpha} v_{\alpha}} \]

And for the proton:
\[ \lambda_{p} = \frac{h}{m_{p} v_{p}} \]

### Step 4: Find the ratio of the de Broglie wavelengths
Now we can find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{m_{p} v_{p}}{m_{\alpha} v_{\alpha}} \]

### Step 5: Substitute the expressions from the radius of curvature
From the earlier step, we have:
\[ \frac{m_{\alpha} v_{\alpha}}{q_{\alpha}} = \frac{m_{p} v_{p}}{q_{p}} \]

Rearranging gives:
\[ \frac{m_{\alpha} v_{\alpha}}{m_{p} v_{p}} = \frac{q_{\alpha}}{q_{p}} \]

### Step 6: Substitute the charge values
The charge of the proton \( q_{p} = e \) and the charge of the alpha particle \( q_{\alpha} = 2e \). Thus:
\[ \frac{q_{\alpha}}{q_{p}} = \frac{2e}{e} = 2 \]

### Step 7: Substitute back into the ratio of wavelengths
Now substituting this back into our ratio of wavelengths:
\[ \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{m_{p} v_{p}}{m_{\alpha} v_{\alpha}} = \frac{1}{2} \]

### Conclusion
Thus, the ratio of the de Broglie wavelengths of the alpha particle to the proton is:
\[ \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{1}{2} \]

### Final Answer
The ratio of their de Broglie wavelengths is \( \frac{1}{2} \).

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