An alpha particle of energy `5 MeV` is scattered through `180^(@)` by a found uramiam nucleus . The distance of closest approach is of the order of

Options

A. 1 Å
B. `10^(-10)` cm
C. `10^(-12)` cm
D. `10^(-15)` cm

Answer

To find the distance of closest approach of an alpha particle (with energy 5 MeV) when it is scattered by a uranium nucleus, we can use the principle of conservation of energy. Here's the step-by-step solution:

### Step 1: Identify the Charges
An alpha particle consists of 2 protons and 2 neutrons, giving it a charge of \( +2e \). The uranium nucleus (U-235) has 92 protons, giving it a charge of \( +92e \).

### Step 2: Write the Conservation of Energy Equation
Initially, the alpha particle has kinetic energy and no potential energy when it is far away. At the point of closest approach, all the kinetic energy is converted into potential energy. Thus, we can write:

\[ \text{Initial Kinetic Energy} + \text{Initial Potential Energy} = \text{Final Kinetic Energy} + \text{Final Potential Energy} \]

This simplifies to:

\[ KE_i + 0 = 0 + PE_f \]

Where:
- \( KE_i = 5 \, \text{MeV} = 5 \times 10^6 \, \text{eV} \)
- \( PE_f = \frac{k \cdot q_1 \cdot q_2}{D} \)

### Step 3: Substitute Values
Here, \( k \) (Coulomb's constant) is approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( q_1 = 2e \), and \( q_2 = 92e \). The potential energy at the distance of closest approach can be expressed as:

\[ PE_f = \frac{k \cdot (2e) \cdot (92e)}{D} \]

### Step 4: Convert Electron Charge to SI Units
The charge of an electron \( e \) is approximately \( 1.6 \times 10^{-19} \, \text{C} \). Thus:

\[ PE_f = \frac{9 \times 10^9 \cdot (2 \cdot 1.6 \times 10^{-19}) \cdot (92 \cdot 1.6 \times 10^{-19})}{D} \]

### Step 5: Set Up the Equation
Now we equate the initial kinetic energy to the potential energy at the closest approach:

\[ 5 \times 10^6 \, \text{eV} = \frac{9 \times 10^9 \cdot (2 \cdot 1.6 \times 10^{-19}) \cdot (92 \cdot 1.6 \times 10^{-19})}{D} \]

### Step 6: Solve for D
Rearranging gives:

\[ D = \frac{9 \times 10^9 \cdot (2 \cdot 1.6 \times 10^{-19}) \cdot (92 \cdot 1.6 \times 10^{-19})}{5 \times 10^6} \]

Calculating the numerator:

\[ = 9 \times 10^9 \cdot (2 \cdot 1.6 \times 10^{-19}) \cdot (92 \cdot 1.6 \times 10^{-19}) = 9 \times 10^9 \cdot 2 \cdot 92 \cdot (1.6 \times 10^{-19})^2 \]

Calculating \( (1.6 \times 10^{-19})^2 \):

\[ = 2.56 \times 10^{-38} \]

Now substituting back:

\[ D = \frac{9 \times 10^9 \cdot 2 \cdot 92 \cdot 2.56 \times 10^{-38}}{5 \times 10^6} \]

Calculating the values gives:

\[ D \approx 5.3 \times 10^{-14} \, \text{m} \]

### Step 7: Convert to Centimeters
To convert meters to centimeters:

\[ D \approx 5.3 \times 10^{-14} \, \text{m} = 5.3 \times 10^{-12} \, \text{cm} \]

### Conclusion
Thus, the distance of closest approach is of the order of \( 10^{-12} \, \text{cm} \).