The approximate size of the nucleus of `._28^64 Ni` is :

Options

A. 3 fm
B. 4 fm
C. 5 fm
D. 2 fm

Explanation

To find the approximate size of the nucleus of the isotope \( _{28}^{64}Ni \), we can use the formula for the radius of a nucleus given by:

\[ R = R_0 A^{1/3} \]

where:
- \( R \) is the radius of the nucleus,
- \( R_0 \) is a constant approximately equal to \( 1.3 \) femtometers (fm),
- \( A \) is the mass number of the nucleus.

### Step-by-Step Solution:

1. **Identify the Mass Number (A)**:
The mass number \( A \) for \( _{28}^{64}Ni \) is \( 64 \).

2. **Use the Formula for Nucleus Radius**:
Substitute \( A \) into the formula:
\[ R = R_0 A^{1/3} \]

3. **Substitute the Values**:
Here, \( R_0 = 1.3 \) fm and \( A = 64 \):
\[ R = 1.3 \times (64)^{1/3} \]

4. **Calculate \( (64)^{1/3} \)**:
The cube root of \( 64 \) is \( 4 \):
\[ (64)^{1/3} = 4 \]

5. **Calculate the Radius**:
Now substitute back into the equation:
\[ R = 1.3 \times 4 \]
\[ R = 5.2 \text{ fm} \]

6. **Conclusion**:
The approximate size of the nucleus of \( _{28}^{64}Ni \) is \( 5.2 \) femtometers.

### Final Answer:
The approximate size of the nucleus of \( _{28}^{64}Ni \) is **5.2 fm**.