An alpha particle is acceleration through a potential difference of `10^(6)` volt. Its kinetic energy will be

Options

A. 1MeV
B. 2MeV
C. 4MeV
D. 8MeV

Explanation

To find the kinetic energy of an alpha particle accelerated through a potential difference of \(10^6\) volts, we can use the relationship between electric potential energy and kinetic energy.

### 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 \(e\) (the charge of a proton) is approximately \(1.6 \times 10^{-19}\) coulombs. Thus, the charge of the alpha particle is:
\[ q = 2e = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \]

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:
\[ KE = qV \]
Here, \(V\) is the potential difference through which the particle is accelerated.

3. **Substitute the Values:**
Given that the potential difference \(V = 10^6\) volts, we can substitute the values into the equation:
\[ KE = (3.2 \times 10^{-19} \, \text{C})(10^6 \, \text{V}) \]

4. **Calculate the Kinetic Energy:**
\[ KE = 3.2 \times 10^{-19} \times 10^6 = 3.2 \times 10^{-13} \, \text{J} \]

5. **Convert Joules to Mega Electron Volts:**
To convert joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ KE = \frac{3.2 \times 10^{-13} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 2 \times 10^6 \, \text{eV} = 2 \, \text{MeV} \]

### Final Answer:
The kinetic energy of the alpha particle is \(2 \, \text{MeV}\).

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