An `alpha`-particle is accelerated through a.p.d of `10^(6)` volt the `K.E.` of particle will be

Options

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

Answer

To find the kinetic energy (K.E.) of an alpha particle that has been accelerated through a potential difference (p.d.) of \(10^6\) volts, we can use the formula for kinetic energy in terms of charge and potential difference:

### Step-by-Step Solution:

1. **Identify 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). Hence, the charge \(Q\) 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 gained by a charged particle when it is accelerated through a potential difference \(V\) is given by:
\[ K.E. = Q \cdot V \]
Here, \(V = 10^6\) volts.

3. **Substitute the Values into the Formula**:
Now, substituting the values of \(Q\) and \(V\) into the formula:
\[ K.E. = (3.2 \times 10^{-19} \text{ C}) \cdot (10^6 \text{ V}) = 3.2 \times 10^{-13} \text{ J} \]

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

### Final Answer:
The kinetic energy of the alpha particle after being accelerated through a potential difference of \(10^6\) volts is \(2 \text{ MeV}\).

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