An alpha particle (charge= +2e) is accelerated from rest through a potential difference of 500 volts. Its final kinetic energy is

Options

A. 50eV
B. 500eV
C. 1000eV
D. 2000eV

Answer

To solve the problem, we need to determine the final kinetic energy of an alpha particle after it has been accelerated through a potential difference of 500 volts.

### Step-by-Step Solution:

1. **Identify the Charge of the Alpha Particle**:
An alpha particle has a charge of +2e, where e (the elementary charge) is approximately \(1.6 \times 10^{-19}\) coulombs. Therefore, the charge of the alpha particle can be calculated as:
\[ q = 2e = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \]

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:
\[ KE = qV \]
where:
- \(KE\) is the kinetic energy in joules,
- \(q\) is the charge of the particle,
- \(V\) is the potential difference.

3. **Substitute the Values**:
Now, we substitute the values we have into the formula. The potential difference given is 500 volts:
\[ KE = (3.2 \times 10^{-19} \, \text{C}) \times (500 \, \text{V}) \]

4. **Calculate the Kinetic Energy**:
Performing the multiplication:
\[ KE = 3.2 \times 10^{-19} \times 500 = 1.6 \times 10^{-16} \, \text{J} \]

5. **Convert to Electron Volts**:
To express the kinetic energy in electron volts (eV), we use the conversion factor where \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ KE = \frac{1.6 \times 10^{-16} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 1000 \, \text{eV} \]

### Final Answer:
The final kinetic energy of the alpha particle is **1000 eV**.