`alpha`-particles of enegry `400 KeV` are boumbardel on nucleus of `._(82)Pb` . In scattering of `alpha`-particles, it minimum distance from nucleus will be

Options

A. 0.59 nm
B. 0.59 Å
C. 5.9 Å
D. 0.59 pm

Answer

To find the minimum distance of approach of an alpha particle bombarded on a lead nucleus, we can use the principle of conservation of energy. The kinetic energy of the alpha particle will be converted into potential energy when it comes closest to the nucleus.

### Step-by-Step Solution:

1. **Identify the Initial Kinetic Energy**:
The kinetic energy (KE) of the alpha particle is given as 400 keV.
\[ KE = 400 \, \text{keV} = 400 \times 10^3 \, \text{eV} \]

2. **Convert Kinetic Energy to Joules**:
We know that \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\). Therefore,
\[ KE = 400 \times 10^3 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 6.4 \times 10^{-14} \, \text{J} \]

3. **Use the Formula for Potential Energy**:
The potential energy (PE) at the minimum distance of approach (r) can be expressed as:
\[ PE = \frac{k \cdot Q_1 \cdot Q_2}{r} \]
where:
- \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\) (Coulomb's constant)
- \(Q_1\) is the charge of the lead nucleus, which is \(82e\) (where \(e = 1.6 \times 10^{-19} \, \text{C}\))
- \(Q_2\) is the charge of the alpha particle, which is \(2e\)

4. **Calculate the Charges**:
\[ Q_1 = 82 \cdot 1.6 \times 10^{-19} \, \text{C} = 1.312 \times 10^{-17} \, \text{C} \]
\[ Q_2 = 2 \cdot 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \]

5. **Substitute into the Potential Energy Formula**:
Set the kinetic energy equal to the potential energy:
\[ 6.4 \times 10^{-14} = \frac{(9 \times 10^9) \cdot (1.312 \times 10^{-17}) \cdot (3.2 \times 10^{-19})}{r} \]

6. **Calculate the Right Side**:
Calculate the product of the constants:
\[ 9 \times 10^9 \cdot 1.312 \times 10^{-17} \cdot 3.2 \times 10^{-19} = 3.78 \times 10^{-26} \]

7. **Rearranging to Solve for r**:
\[ r = \frac{3.78 \times 10^{-26}}{6.4 \times 10^{-14}} \approx 5.91 \times 10^{-13} \, \text{m} \]

8. **Convert to Picometers**:
Since \(1 \, \text{pm} = 10^{-12} \, \text{m}\),
\[ r \approx 0.591 \, \text{pm} = 5.91 \times 10^{-13} \, \text{m} \]

### Final Answer:
The minimum distance from the nucleus is approximately \(0.59 \, \text{pm}\).