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

Options

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

Answer

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:

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

Where:
- \( R \) is the radius of the nucleus,
- \( R_0 \) is a constant approximately equal to \( 1.3 \, \text{Fermi} \) (or \( 1.3 \, \text{fm} \)),
- \( A \) 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 the radius**:
Substitute \( A \) into the formula:
\[ R = R_0 \times A^{1/3} \]

3. **Substitute the values**:
\[ R = 1.3 \, \text{fm} \times (64)^{1/3} \]

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

5. **Calculate the radius \( R \)**:
Substitute back into the equation:
\[ R = 1.3 \, \text{fm} \times 4 \]
\[ R = 5.2 \, \text{fm} \]

6. **Approximate the answer**:
The approximate size of the nucleus can be rounded to:
\[ R \approx 5 \, \text{fm} \]

### Final Answer:
The approximate size of the nucleus of \( _{28}^{64}Ni \) is \( 5 \, \text{fm} \).

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