The order of magnitude of the density of nuclear matter is=

Options

A. `10^(4) kg//m^(3)`
B. `10^(17) kg//m^(3)`
C. `10^(27) kg//m^(3)`
D. `10^(34) kg//m^(3)`

Answer

To find the order of magnitude of the density of nuclear matter, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Nuclear Matter**:
- Nuclear matter refers to the matter that makes up the nucleus of an atom, which consists of protons and neutrons.

2. **Density Formula**:
- The density (\( \rho \)) of nuclear matter can be calculated using the formula:
\[ \rho = \frac{\text{mass of nucleus}}{\text{volume of nucleus}} \]

3. **Mass of the Nucleus**:
- The mass of the nucleus can be approximated as:
\[ \text{mass} = A \times m_u \]
where \( A \) is the atomic mass number (total number of protons and neutrons) and \( m_u \) is the atomic mass unit, approximately \( 1.67 \times 10^{-27} \) kg.

4. **Volume of the Nucleus**:
- The volume of a nucleus (assuming it is spherical) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- The radius \( r \) can be estimated using the formula:
\[ r = r_0 A^{1/3} \]
where \( r_0 \) is a constant approximately equal to \( 1.1 \times 10^{-15} \) m.

5. **Substituting the Volume**:
- Substituting the expression for \( r \) into the volume formula:
\[ V = \frac{4}{3} \pi (r_0 A^{1/3})^3 = \frac{4}{3} \pi r_0^3 A \]

6. **Calculating Density**:
- Now substituting the mass and volume into the density formula:
\[ \rho = \frac{A \times m_u}{\frac{4}{3} \pi r_0^3 A} \]
- The \( A \) cancels out:
\[ \rho = \frac{m_u}{\frac{4}{3} \pi r_0^3} \]

7. **Plugging in Values**:
- Plugging in the values:
\[ \rho = \frac{1.67 \times 10^{-27}}{\frac{4}{3} \pi (1.1 \times 10^{-15})^3} \]
- Calculate \( \frac{4}{3} \pi (1.1 \times 10^{-15})^3 \):
\[ \frac{4}{3} \pi (1.1 \times 10^{-15})^3 \approx 5.58 \times 10^{-46} \text{ m}^3 \]

8. **Final Calculation**:
- Now calculating the density:
\[ \rho \approx \frac{1.67 \times 10^{-27}}{5.58 \times 10^{-46}} \approx 3 \times 10^{17} \text{ kg/m}^3 \]

9. **Conclusion**:
- Therefore, the order of magnitude of the density of nuclear matter is approximately \( 10^{17} \text{ kg/m}^3 \).

### Final Answer:
The order of magnitude of the density of nuclear matter is \( 10^{17} \text{ kg/m}^3 \).

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