The nuclear radius of `._(8)O^(16)` is `3 xx10^(-15) m`. If an atomic mass unit is `1.67 xx 10^(-27) kg`, then the nuclear density is approximately.

Options

A. `2.35 xx 10^(17) g cm^(-3)`
B. `2.35 xx 10^(17) kg m^(-3)`
C. `2.35 xx 10^(17) g m^(-3)`
D. `2.35 xx 10^(17) kg mm^(-3)`

Explanation

To find the nuclear density of the oxygen-16 nucleus, we will follow these steps:

### Step 1: Calculate the mass of the nucleus
The mass of the nucleus can be calculated using the formula:
\[ \text{Mass} = A \times \text{(atomic mass unit)} \]
where \( A \) is the mass number of the nucleus. For oxygen-16, \( A = 16 \) and the atomic mass unit is given as \( 1.67 \times 10^{-27} \, \text{kg} \).

\[ \text{Mass} = 16 \times 1.67 \times 10^{-27} \, \text{kg} = 2.672 \times 10^{-26} \, \text{kg} \]

### Step 2: Calculate the volume of the nucleus
The volume \( V \) of a nucleus can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the nucleus. Given that the nuclear radius \( r = 3 \times 10^{-15} \, \text{m} \):

\[ V = \frac{4}{3} \pi (3 \times 10^{-15})^3 \]
Calculating \( (3 \times 10^{-15})^3 \):
\[ (3 \times 10^{-15})^3 = 27 \times 10^{-45} \, \text{m}^3 \]
Thus,
\[ V = \frac{4}{3} \pi (27 \times 10^{-45}) = 36 \pi \times 10^{-45} \, \text{m}^3 \]

### Step 3: Calculate the nuclear density
The nuclear density \( \rho \) can be calculated using the formula:
\[ \rho = \frac{\text{mass}}{\text{volume}} \]
Substituting the values we calculated:
\[ \rho = \frac{2.672 \times 10^{-26} \, \text{kg}}{36 \pi \times 10^{-45} \, \text{m}^3} \]

Calculating the denominator:
\[ 36 \pi \approx 113.097 \quad \text{(using } \pi \approx 3.14\text{)} \]
Thus,
\[ \rho \approx \frac{2.672 \times 10^{-26}}{113.097 \times 10^{-45}} \approx 2.36 \times 10^{17} \, \text{kg/m}^3 \]

### Conclusion
The nuclear density of the oxygen-16 nucleus is approximately \( 2.36 \times 10^{17} \, \text{kg/m}^3 \).

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