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 can follow these steps:

### Step 1: Understand the formula for nuclear density
The nuclear density \( \rho \) is defined as the mass of the nucleus divided by its volume. The formula can be expressed as:
\[ \rho = \frac{m}{V} \]
where \( m \) is the mass of the nucleus and \( V \) is the volume of the nucleus.

### Step 2: Convert the mass from atomic mass units (amu) to kilograms (kg)
Given that the atomic mass of \( _{6}^{12}C \) is \( 12 \, \text{amu} \), we need to convert this to kg using the conversion factor:
\[ 1 \, \text{amu} = 1.66 \times 10^{-27} \, \text{kg} \]
Thus, the mass in kg is:
\[ m = 12 \, \text{amu} \times 1.66 \times 10^{-27} \, \text{kg/amu} = 1.992 \times 10^{-26} \, \text{kg} \]

### Step 3: Calculate the radius of the nucleus
The radius \( R \) of the nucleus can be calculated using the formula:
\[ R = R_0 A^{1/3} \]
where \( R_0 = 1.2 \times 10^{-15} \, \text{m} \) and \( A \) is the mass number (which is 12 for \( _{6}^{12}C \)). Thus, we have:
\[ R = 1.2 \times 10^{-15} \, \text{m} \times 12^{1/3} \]
Calculating \( 12^{1/3} \) gives approximately \( 2.29 \), so:
\[ R \approx 1.2 \times 10^{-15} \, \text{m} \times 2.29 \approx 2.748 \times 10^{-15} \, \text{m} \]

### Step 4: Calculate the volume of the nucleus
The volume \( V \) of the nucleus can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi R^3 \]
Substituting the radius we found:
\[ V = \frac{4}{3} \pi (2.748 \times 10^{-15})^3 \]
Calculating \( (2.748 \times 10^{-15})^3 \) gives approximately \( 2.073 \times 10^{-44} \, \text{m}^3 \), so:
\[ V \approx \frac{4}{3} \pi \times 2.073 \times 10^{-44} \approx 8.694 \times 10^{-44} \, \text{m}^3 \]

### Step 5: Calculate the nuclear density
Now we can substitute the mass and volume into the density formula:
\[ \rho = \frac{m}{V} = \frac{1.992 \times 10^{-26} \, \text{kg}}{8.694 \times 10^{-44} \, \text{m}^3} \]
Calculating this gives:
\[ \rho \approx 2.287 \times 10^{17} \, \text{kg/m}^3 \]

### Final Answer
Thus, the density of the \( _{6}^{12}C \) nucleus is approximately:
\[ \rho \approx 2.4 \times 10^{17} \, \text{kg/m}^3 \]