A proton and an `alpha` particle accellerated throught the same potential differenc e enter a region of uniform magnetic field normally if the radius of the proton orbit is 10 cm ten radius of `alpha` particle is

Options

A. 10 cm
B. `10sqrt2 cm`
C. 20 cm
D. `5sqrt2 cm`

Answer

To solve the problem, we need to find the radius of the orbit of an alpha particle when both a proton and an alpha particle are accelerated through the same potential difference and enter a uniform magnetic field. We know the radius of the proton's orbit is 10 cm.

### Step-by-Step Solution:

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

2. **Kinetic energy and potential difference:**
The kinetic energy \( KE \) gained by the particle when accelerated through a potential difference \( V \) is:
\[ KE = qV \]
This can also be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Setting these equal gives:
\[ qV = \frac{1}{2} mv^2 \]
Rearranging for \( v \):
\[ v = \sqrt{\frac{2qV}{m}} \]

3. **Substituting velocity into the radius formula:**
Substitute \( v \) into the radius formula:
\[ R = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{\sqrt{2m qV}}{qB} \]
Simplifying gives:
\[ R = \frac{\sqrt{2mV}}{B\sqrt{q}} \]

4. **Comparing the radii of the proton and alpha particle:**
Let \( R_p \) be the radius of the proton and \( R_\alpha \) be the radius of the alpha particle. From the above formula, we have:
\[ R_p = \frac{\sqrt{2m_pV}}{B\sqrt{q_p}} \quad \text{and} \quad R_\alpha = \frac{\sqrt{2m_\alpha V}}{B\sqrt{q_\alpha}} \]
Taking the ratio:
\[ \frac{R_p}{R_\alpha} = \frac{\sqrt{m_p}}{\sqrt{q_p}} \cdot \frac{\sqrt{q_\alpha}}{\sqrt{m_\alpha}} \]

5. **Substituting known values:**
- For the proton:
- Mass \( m_p = m \)
- Charge \( q_p = e \)
- For the alpha particle:
- Mass \( m_\alpha = 4m \) (since an alpha particle consists of 2 protons and 2 neutrons)
- Charge \( q_\alpha = 2e \)

Substituting these values gives:
\[ \frac{R_p}{R_\alpha} = \frac{\sqrt{m}}{\sqrt{e}} \cdot \frac{\sqrt{2e}}{\sqrt{4m}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \]
Rearranging gives:
\[ R_\alpha = R_p \cdot 2 = R_p \cdot \sqrt{2} \]

6. **Calculating the radius of the alpha particle:**
Given \( R_p = 10 \, \text{cm} \):
\[ R_\alpha = 10 \sqrt{2} \, \text{cm} \]

### Final Answer:
The radius of the alpha particle is \( 10\sqrt{2} \, \text{cm} \).