(A) Find the radius of nucleus of an atom having atomic mass number equal to `125. ("Take" R_(0)=1.3xx10^(-15) m)` (B) Find the distance of closest approach when an `alpha` particle is projected towards the nucleus of silver atom having speed `v`. (mass of `alpha` particle `=m_(alpha)`, atomic number pf `Ag=47`)

Explanation

### Step-by-Step Solution

#### Part (A): Finding the Radius of the Nucleus

1. **Identify the formula for the radius of a nucleus**:
The radius \( R \) of a nucleus can be calculated using the formula:
\[ R = R_0 A^{1/3} \]
where \( R_0 \) is a constant (given as \( 1.3 \times 10^{-15} \) m) and \( A \) is the atomic mass number.

2. **Substitute the values into the formula**:
Given that the atomic mass number \( A = 125 \):
\[ R = 1.3 \times 10^{-15} \, \text{m} \times (125)^{1/3} \]

3. **Calculate \( (125)^{1/3} \)**:
\[ (125)^{1/3} = 5 \]

4. **Calculate the radius**:
\[ R = 1.3 \times 10^{-15} \, \text{m} \times 5 = 6.5 \times 10^{-15} \, \text{m} \]

#### Final Result for Part (A):
The radius of the nucleus of the atom with atomic mass number 125 is:
\[ R = 6.5 \times 10^{-15} \, \text{m} \]

---

#### Part (B): Finding the Distance of Closest Approach

1. **Identify the formula for the distance of closest approach**:
The distance of closest approach \( d \) for an alpha particle approaching a nucleus can be calculated using the formula:
\[ d = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{E} \]
where \( q_1 \) and \( q_2 \) are the charges of the alpha particle and the nucleus, and \( E \) is the kinetic energy of the alpha particle.

2. **Identify the charges**:
The charge of the alpha particle \( q_1 = 2e \) (since it has 2 protons) and the charge of the silver nucleus \( q_2 = Ze = 47e \) (where \( Z = 47 \) is the atomic number of silver).

3. **Substitute the charges into the formula**:
\[ d = \frac{1}{4 \pi \epsilon_0} \frac{(2e)(47e)}{E} \]

4. **Express the kinetic energy \( E \)**:
The kinetic energy \( E \) of the alpha particle can be expressed as:
\[ E = \frac{1}{2} m_{\alpha} v^2 \]

5. **Substitute \( E \) into the distance formula**:
\[ d = \frac{1}{4 \pi \epsilon_0} \frac{(2e)(47e)}{\frac{1}{2} m_{\alpha} v^2} \]

6. **Simplify the expression**:
\[ d = \frac{4 \cdot 47 e^2}{4 \pi \epsilon_0 \cdot \frac{1}{2} m_{\alpha} v^2} = \frac{4 \cdot 47 e^2}{2 \pi \epsilon_0 m_{\alpha} v^2} \]

7. **Plug in the value of \( e \) and \( \epsilon_0 \)**:
The value of \( e \) (elementary charge) is \( 1.6 \times 10^{-19} \, \text{C} \) and \( \epsilon_0 \) (permittivity of free space) is \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

8. **Final expression for distance**:
\[ d = \frac{4 \cdot 47 (1.6 \times 10^{-19})^2}{2 \pi (8.85 \times 10^{-12}) m_{\alpha} v^2} \]

#### Final Result for Part (B):
The distance of closest approach when an alpha particle is projected towards the nucleus of a silver atom is:
\[ d = \frac{4 \cdot 47 (1.6 \times 10^{-19})^2}{2 \pi (8.85 \times 10^{-12}) m_{\alpha} v^2} \]

---