When an alpha-particle is acceleration by a PD of 3 volt, its energy is

Options

A. 1eV
B. 5eV
C. 3eV
D. 6eV

Answer

To find the energy of an alpha particle when it is accelerated by a potential difference (PD) of 3 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. The charge of a proton is approximately \( e = 1.6 \times 10^{-19} \) coulombs. Since neutrons do not carry any charge, the total charge \( Q \) of the alpha particle is given by:
\[ Q = 2 \times e = 2 \times 1.6 \times 10^{-19} \text{ C} = 3.2 \times 10^{-19} \text{ C} \]

2. **Use the Formula for Energy**:
- The energy \( E \) gained by a charged particle when accelerated through a potential difference \( V \) is given by the formula:
\[ E = Q \times V \]
- Here, \( V = 3 \text{ volts} \).

3. **Substituting Values**:
- Substitute the charge of the alpha particle and the potential difference into the energy formula:
\[ E = (2e) \times V = (2 \times 1.6 \times 10^{-19} \text{ C}) \times 3 \text{ V} \]

4. **Calculate the Energy**:
- Now calculate the energy:
\[ E = (3.2 \times 10^{-19} \text{ C}) \times 3 \text{ V} = 9.6 \times 10^{-19} \text{ J} \]
- To convert this energy into electron volts (eV), we can use the conversion \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \):
\[ E = \frac{9.6 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} = 6 \text{ eV} \]

5. **Final Answer**:
- Therefore, the energy of the alpha particle when accelerated by a potential difference of 3 volts is:
\[ E = 6 \text{ eV} \]