A proton a deutron and an `alpha` particle are accelerated through potentials of `V, 2V` and `4V` respectively. Their velocity will bear a ratio

Options

A. `1:1:1`
B. `1:sqrt2:1`
C. `sqrt2:1:1`
D. `1:1:sqrt2`

Explanation

To find the ratio of velocities of a proton, deuteron, and alpha particle when they are accelerated through potentials of \( V, 2V, \) and \( 4V \) respectively, we can follow these steps:

### Step 1: Understand the relationship between potential energy and kinetic energy
When a charged particle is accelerated through a potential difference \( V \), the change in potential energy is converted into kinetic energy. The relationship can be expressed as:
\[ \Delta PE = q \Delta V \]
\[ KE = \frac{1}{2} mv^2 \]
Equating these gives:
\[ \frac{1}{2} mv^2 = qV \]

### Step 2: Rearranging the equation for velocity
From the equation above, we can solve for the velocity \( v \):
\[ v = \sqrt{\frac{2qV}{m}} \]

### Step 3: Determine the charge and mass for each particle
- **Proton**: Charge \( q = e \), Mass \( m = m \)
- **Deuteron**: Charge \( q = e \), Mass \( m = 2m \)
- **Alpha particle**: Charge \( q = 2e \), Mass \( m = 4m \)

### Step 4: Calculate the velocity for each particle
1. **Velocity of Proton** (accelerated through potential \( V \)):
\[ v_p = \sqrt{\frac{2eV}{m}} \]

2. **Velocity of Deuteron** (accelerated through potential \( 2V \)):
\[ v_d = \sqrt{\frac{2e(2V)}{2m}} = \sqrt{\frac{4eV}{2m}} = \sqrt{\frac{2eV}{m}} \]

3. **Velocity of Alpha Particle** (accelerated through potential \( 4V \)):
\[ v_{\alpha} = \sqrt{\frac{2(2e)(4V)}{4m}} = \sqrt{\frac{16eV}{4m}} = \sqrt{\frac{4eV}{m}} \]

### Step 5: Establish the ratio of velocities
Now we can write the ratio of the velocities:
\[ \text{Ratio} = v_p : v_d : v_{\alpha} = \sqrt{\frac{2eV}{m}} : \sqrt{\frac{2eV}{m}} : \sqrt{\frac{4eV}{m}} \]
This simplifies to:
\[ = 1 : 1 : 2 \]

### Final Answer
The ratio of the velocities of the proton, deuteron, and alpha particle is:
\[ 1 : 1 : 2 \]