An alpha particle is accelerated from rest through a potential difference of 100 volt. Its final kinetic energy is

Options

A. 100 eV
B. 1000 eV
C. 200 eV
D. 400 eV

Answer

To find the final kinetic energy of an alpha particle that has been accelerated from rest through a potential difference of 100 volts, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Charge of the Alpha Particle**:
An alpha particle consists of 2 protons and 2 neutrons. Therefore, it has a charge of \( +2e \), where \( e \) is the elementary charge (\( e \approx 1.6 \times 10^{-19} \) coulombs).

2. **Use the Formula for Kinetic Energy**:
The kinetic energy (KE) gained by a charged particle when it is accelerated through a potential difference (V) is given by the formula:
\[ KE = Q \cdot V \]
where \( Q \) is the charge of the particle and \( V \) is the potential difference.

3. **Substitute the Values**:
- The charge \( Q \) of the alpha particle is \( 2e = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \).
- The potential difference \( V \) is given as 100 volts.

Now substitute these values into the kinetic energy formula:
\[ KE = (3.2 \times 10^{-19} \, \text{C}) \times (100 \, \text{V}) = 3.2 \times 10^{-17} \, \text{J} \]

4. **Convert Joules to Electron Volts**:
To convert the kinetic energy from joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ KE = \frac{3.2 \times 10^{-17} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 200 \, \text{eV} \]

5. **Final Answer**:
The final kinetic energy of the alpha particle after being accelerated through a potential difference of 100 volts is:
\[ \text{KE} = 200 \, \text{eV} \]