The radius of the nucleus of `""_(8)O^(16)` is `3 x× 10^(–15)` m. Its density in `kg//m^(3)` will be about-

Options

A. `2.35 xx10^(-17)`
B. `2.35xx10^(14)`
C. `10^(14)`
D. `2.35xx10^(17)`

Answer

To find the density of the nucleus of the oxygen-16 isotope, we can follow these steps:

### Step 1: Understand the formula for density
Density (ρ) is defined as mass (m) divided by volume (V):
\[ \rho = \frac{m}{V} \]

### Step 2: Determine the mass of the nucleus
The mass of the oxygen-16 nucleus can be calculated using its atomic mass unit (amu). The atomic mass of oxygen-16 is 16 amu. We need to convert this mass into kilograms:
- 1 amu = \( 1.67 \times 10^{-27} \) kg
- Therefore, the mass of the oxygen-16 nucleus is:
\[ m = 16 \, \text{amu} \times 1.67 \times 10^{-27} \, \text{kg/amu} = 2.672 \times 10^{-26} \, \text{kg} \]

### Step 3: Calculate the volume of the nucleus
The volume (V) of a nucleus can be approximated 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 radius \( r = 3 \times 10^{-15} \) m, we can calculate the volume:
\[ V = \frac{4}{3} \pi (3 \times 10^{-15})^3 \]
Calculating \( (3 \times 10^{-15})^3 \):
\[ (3 \times 10^{-15})^3 = 27 \times 10^{-45} = 2.7 \times 10^{-44} \]
Now substituting back into the volume formula:
\[ V = \frac{4}{3} \pi (2.7 \times 10^{-44}) \]
Using \( \pi \approx 3.14 \):
\[ V \approx \frac{4}{3} \times 3.14 \times 2.7 \times 10^{-44} \approx 1.13 \times 10^{-43} \, \text{m}^3 \]

### Step 4: Calculate the density
Now we can find the density using the mass and volume calculated:
\[ \rho = \frac{m}{V} = \frac{2.672 \times 10^{-26} \, \text{kg}}{1.13 \times 10^{-43} \, \text{m}^3} \]
Calculating this gives:
\[ \rho \approx 2.36 \times 10^{17} \, \text{kg/m}^3 \]

### Final Answer
Thus, the density of the nucleus of oxygen-16 is approximately:
\[ \rho \approx 2.36 \times 10^{17} \, \text{kg/m}^3 \]