In Rutherford experiment `alpha` – particles are scattered by nucleus having change `100 e^(-)` Initial kinetic energy of `alpha` - particles is 6 MeV . The size of the nucleus is

Options

A. `10^(-14) m`
B. `3xx10^(-14)m`
C. `10^(-13)m`
D. `10^(-16)m`

Answer

To solve the problem of finding the size of the nucleus in the Rutherford experiment, we will use the relationship between kinetic energy and potential energy of the alpha particles as they approach the nucleus.

### Step 1: Convert Kinetic Energy to Joules
The initial kinetic energy of the alpha particles is given as 6 MeV. We need to convert this energy into Joules.

\[ \text{K.E.} = 6 \text{ MeV} = 6 \times 10^6 \text{ eV} \]

Using the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ Joules}\):

\[ \text{K.E.} = 6 \times 10^6 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 9.6 \times 10^{-13} \text{ J} \]

### Step 2: Set Up the Potential Energy Equation
The potential energy (U) between two charged particles can be expressed as:

\[ U = \frac{k \cdot Q_1 \cdot Q_2}{r} \]

Where:
- \(k\) is Coulomb's constant (\(9 \times 10^9 \text{ N m}^2/\text{C}^2\))
- \(Q_1\) is the charge of the nucleus (\(100 \text{ e} = 100 \times 1.6 \times 10^{-19} \text{ C}\))
- \(Q_2\) is the charge of the alpha particle (\(2 \text{ e} = 2 \times 1.6 \times 10^{-19} \text{ C}\))
- \(r\) is the distance (size of the nucleus) we want to find.

### Step 3: Equate Kinetic Energy and Potential Energy
At the point of closest approach, the kinetic energy of the alpha particle is converted into potential energy:

\[ \text{K.E.} = U \]

Substituting the expressions we have:

\[ 9.6 \times 10^{-13} \text{ J} = \frac{(9 \times 10^9) \cdot (100 \times 1.6 \times 10^{-19}) \cdot (2 \times 1.6 \times 10^{-19})}{r} \]

### Step 4: Solve for r
Now we can rearrange the equation to solve for \(r\):

\[ r = \frac{(9 \times 10^9) \cdot (100 \times 1.6 \times 10^{-19}) \cdot (2 \times 1.6 \times 10^{-19})}{9.6 \times 10^{-13}} \]

Calculating the numerator:

\[ = 9 \times 10^9 \cdot 100 \cdot 1.6 \times 10^{-19} \cdot 2 \cdot 1.6 \times 10^{-19} \]
\[ = 9 \times 10^9 \cdot 100 \cdot 3.2 \times 10^{-38} \]
\[ = 2880 \times 10^{-29} = 2.88 \times 10^{-26} \]

Now substituting back into the equation for \(r\):

\[ r = \frac{2.88 \times 10^{-26}}{9.6 \times 10^{-13}} = 3 \times 10^{-14} \text{ m} \]

### Conclusion
The size of the nucleus is approximately:

\[ r \approx 3 \times 10^{-14} \text{ m} \]