Find the density of `._(6)^(12)C` nucleus. Take atomic mass of `._(6)^(12)C` as `12.00 am u` Take `R_(0) =1.2 xx10^(-15) m` .

Explanation

To find the density of the \( _{6}^{12}C \) nucleus, we will follow these steps:

### Step 1: Calculate the mass of the \( _{6}^{12}C \) nucleus.
The atomic mass of the \( _{6}^{12}C \) nucleus is given as \( 12 \, \text{amu} \). We need to convert this mass into kilograms.

\[ \text{Mass} = 12 \, \text{amu} \times 1.66 \times 10^{-27} \, \text{kg/amu} = 12 \times 1.66 \times 10^{-27} \, \text{kg} \]

### Step 2: Calculate the volume of the \( _{6}^{12}C \) nucleus.
The volume \( V \) of a nucleus can be calculated using the formula:

\[ V = \frac{4}{3} \pi R^3 \]

where \( R \) is the radius of the nucleus. The radius \( R \) can be calculated using the formula:

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

where \( R_0 = 1.2 \times 10^{-15} \, \text{m} \) and \( A = 12 \) (the mass number of carbon).

Calculating the radius:

\[ R = 1.2 \times 10^{-15} \, \text{m} \times 12^{1/3} \]

Calculating \( 12^{1/3} \):

\[ 12^{1/3} \approx 2.29 \]

Now substituting this value back into the equation for \( R \):

\[ R \approx 1.2 \times 10^{-15} \, \text{m} \times 2.29 \approx 2.748 \times 10^{-15} \, \text{m} \]

Now substituting \( R \) into the volume formula:

\[ V = \frac{4}{3} \pi (2.748 \times 10^{-15})^3 \]

Calculating \( V \):

\[ V \approx \frac{4}{3} \pi (2.748 \times 10^{-15})^3 \approx 8.66 \times 10^{-45} \, \text{m}^3 \]

### Step 3: Calculate the density of the \( _{6}^{12}C \) nucleus.
Density \( \rho \) is given by the formula:

\[ \rho = \frac{\text{mass}}{\text{volume}} \]

Substituting the values we have calculated:

\[ \rho = \frac{12 \times 1.66 \times 10^{-27}}{8.66 \times 10^{-45}} \]

Calculating \( \rho \):

\[ \rho \approx \frac{1.992 \times 10^{-26}}{8.66 \times 10^{-45}} \approx 2.30 \times 10^{17} \, \text{kg/m}^3 \]

### Final Answer:
The density of the \( _{6}^{12}C \) nucleus is approximately \( 2.30 \times 10^{17} \, \text{kg/m}^3 \).

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