Doubly ionised helium atoms and hydrogen ions are accelerated from rest through the same potential drop. The ratio of the final velocities of the helium and the hydrogen ion is

Options

A. `1/2`
B. `2`
C. `1/sqrt(2)`
D. `sqrt(2)`

Answer

To find the ratio of the final velocities of doubly ionized helium atoms and hydrogen ions accelerated through the same potential difference, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges**:
- For doubly ionized helium (He²⁺), the charge \( Q_{He} = 2e \) (where \( e \) is the elementary charge).
- For hydrogen ion (H⁺), the charge \( Q_{H} = e \).

2. **Work Done by Electric Field**:
- The work done (or energy gained) when a charge is accelerated through a potential difference \( V \) is given by:
\[ W = Q \cdot V \]
- For helium:
\[ W_{He} = Q_{He} \cdot V = 2e \cdot V \]
- For hydrogen:
\[ W_{H} = Q_{H} \cdot V = e \cdot V \]

3. **Kinetic Energy Relation**:
- The work done on the particles is converted into kinetic energy. The kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2} m v^2 \]
- For helium:
\[ \frac{1}{2} m_{He} v_{He}^2 = 2eV \]
- For hydrogen:
\[ \frac{1}{2} m_{H} v_{H}^2 = eV \]

4. **Express Velocities**:
- Rearranging the kinetic energy equations gives:
\[ v_{He}^2 = \frac{4eV}{m_{He}} \]
\[ v_{H}^2 = \frac{2eV}{m_{H}} \]

5. **Finding the Ratio of Velocities**:
- To find the ratio \( \frac{v_{He}}{v_{H}} \):
\[ \frac{v_{He}^2}{v_{H}^2} = \frac{\frac{4eV}{m_{He}}}{\frac{2eV}{m_{H}}} = \frac{4}{2} \cdot \frac{m_{H}}{m_{He}} = 2 \cdot \frac{m_{H}}{m_{He}} \]
- Since the mass of helium \( m_{He} \) is approximately 4 times the mass of hydrogen \( m_{H} \) (i.e., \( m_{He} = 4m_{H} \)):
\[ \frac{v_{He}^2}{v_{H}^2} = 2 \cdot \frac{m_{H}}{4m_{H}} = \frac{2}{4} = \frac{1}{2} \]

6. **Final Velocity Ratio**:
- Taking the square root gives:
\[ \frac{v_{He}}{v_{H}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Conclusion:
The ratio of the final velocities of the doubly ionized helium atom to the hydrogen ion is:
\[ \frac{v_{He}}{v_{H}} = \frac{1}{\sqrt{2}} \]