An `O^(16)` nucleus is spherical and has a radius R and a volume `V=(4)/(3) pi R^(3)`.According to the empirical observation, the volume of the `._(54)X^(128)` nucleus assumed to be sphericla is `V'` and radius is `R'`. Then

Options

A. `V'=8V`
B. `V'=2V`
C. `R'=2R`
D. `R'=8R`

Explanation

To solve the problem, we need to analyze the relationship between the radius and volume of the nuclei based on their mass numbers.

### Step-by-Step Solution:

1. **Understanding the Volume of a Nucleus**:
The volume \( V \) of a spherical nucleus is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
where \( R \) is the radius of the nucleus.

2. **Relating Radius to Mass Number**:
According to empirical observations, the radius \( R \) of a nucleus can be expressed as:
\[ R = R_0 A^{1/3} \]
where \( R_0 \) is a constant and \( A \) is the mass number of the nucleus.

3. **Calculating the Radius for \( O^{16} \)**:
For the oxygen-16 nucleus (\( O^{16} \)), the mass number \( A = 16 \):
\[ R = R_0 (16)^{1/3} \]

4. **Calculating the Radius for \( X^{128} \)**:
For the nucleus \( X^{128} \) with mass number \( A = 128 \):
\[ R' = R_0 (128)^{1/3} \]

5. **Finding the Ratio of Radii**:
We can find the ratio of the radii \( R' \) and \( R \):
\[ \frac{R'}{R} = \frac{R_0 (128)^{1/3}}{R_0 (16)^{1/3}} = \frac{(128)^{1/3}}{(16)^{1/3}} = \left(\frac{128}{16}\right)^{1/3} = (8)^{1/3} = 2 \]
Therefore, we have:
\[ R' = 2R \]

6. **Finding the Volume Ratio**:
Now, we can find the ratio of the volumes \( V' \) and \( V \):
\[ V' = \frac{4}{3} \pi (R')^3 = \frac{4}{3} \pi (2R)^3 = \frac{4}{3} \pi (8R^3) = 8 \left(\frac{4}{3} \pi R^3\right) = 8V \]
Thus, we have:
\[ V' = 8V \]

### Conclusion:
From our calculations, we find that:
- The radius of nucleus \( X^{128} \) is \( R' = 2R \).
- The volume of nucleus \( X^{128} \) is \( V' = 8V \).

### Correct Options:
- \( V' = 8V \) (Option A)
- \( R' = 2R \) (Option C)