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

Answer

To solve the problem, we will follow these steps:

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

\[ \text{Kinetic Energy (KE)} = 6 \, \text{MeV} = 6 \times 10^6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]

Calculating this gives:

\[ \text{KE} = 6 \times 10^6 \times 1.6 \times 10^{-19} = 9.6 \times 10^{-13} \, \text{J} \]

### Step 2: Use Conservation of Energy
According to the conservation of energy, the kinetic energy of the alpha particle will be converted into potential energy when it is at the closest approach to the nucleus. The potential energy (PE) between two charges is given by:

\[ PE = \frac{k \cdot Q_1 \cdot Q_2}{R} \]

Where:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) (Coulomb's constant)
- \( Q_1 = 100 \, e \) (charge of the nucleus)
- \( Q_2 = 2 \, e \) (charge of the alpha particle)
- \( R \) is the distance of closest approach (size of the nucleus)

### Step 3: Set Kinetic Energy Equal to Potential Energy
Setting the kinetic energy equal to the potential energy, we have:

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

### Step 4: Simplify the Equation
Substituting the values into the equation:

\[ 9.6 \times 10^{-13} = \frac{9 \times 10^9 \cdot (100 \cdot 2.56 \times 10^{-38})}{R} \]

This simplifies to:

\[ 9.6 \times 10^{-13} = \frac{2.56 \times 10^{-28} \cdot 9 \times 10^9}{R} \]

### Step 5: Solve for R
Rearranging the equation to solve for \( R \):

\[ R = \frac{2.56 \times 10^{-28} \cdot 9 \times 10^9}{9.6 \times 10^{-13}} \]

Calculating the right side:

\[ R = \frac{2.304 \times 10^{-18}}{9.6 \times 10^{-13}} = 2.4 \times 10^{-6} \, \text{m} \]

### Step 6: Convert to Appropriate Units
Since we are looking for the size of the nucleus, we can express this in femtometers (1 fm = \( 10^{-15} \) m):

\[ R = 2.4 \times 10^{-6} \, \text{m} = 2400 \, \text{fm} \]

### Final Answer
The size of the nucleus is approximately \( 2.4 \, \text{fm} \).

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