Given that the mass of carbon nucleus as 11.011 amu and A = 11, find the nuclear density.

Answer

To find the nuclear density of the carbon nucleus, we can follow these steps:

### Step 1: Convert the mass of the carbon nucleus from amu to kg
The mass of the carbon nucleus is given as 11.011 amu. We know that 1 amu is approximately equal to \(1.66 \times 10^{-27}\) kg.

\[ m = 11.011 \, \text{amu} \times 1.66 \times 10^{-27} \, \text{kg/amu} \]
\[ m = 11.011 \times 1.66 \times 10^{-27} \, \text{kg} \approx 1.83 \times 10^{-26} \, \text{kg} \]

### Step 2: Determine the radius of the nucleus
We are given that the radius of the carbon nucleus is \(r = 2.7 \times 10^{-15}\) m.

### Step 3: Calculate the volume of the nucleus
The nucleus is approximately spherical, so we can use the formula for the volume of a sphere:

\[ V = \frac{4}{3} \pi r^3 \]
Substituting the value of \(r\):

\[ V = \frac{4}{3} \pi (2.7 \times 10^{-15})^3 \]
Calculating \(r^3\):

\[ (2.7 \times 10^{-15})^3 = 19.713 \times 10^{-45} \, \text{m}^3 \]
Now substituting this back into the volume formula:

\[ V = \frac{4}{3} \pi (19.713 \times 10^{-45}) \approx 8.26 \times 10^{-44} \, \text{m}^3 \]

### Step 4: Calculate the nuclear density
The nuclear density \(\rho\) is given by the formula:

\[ \rho = \frac{m}{V} \]
Substituting the values we have calculated:

\[ \rho = \frac{1.83 \times 10^{-26} \, \text{kg}}{8.26 \times 10^{-44} \, \text{m}^3} \]
Calculating this gives:

\[ \rho \approx 2.21 \times 10^{17} \, \text{kg/m}^3 \]

### Final Answer
The nuclear density of the carbon nucleus is approximately \(2.21 \times 10^{17} \, \text{kg/m}^3\).

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