The most common kind of iron nucleus has a mass number of `56`. Find the approximate density of the nucleus.

Options

A. `2.29xx10^16 "kg m"^(-3)`
B. `2.29xx10^17 "kg m"^(-3)`
C. `2.29xx10^18 "kg m"^(-3)`
D. `2.29xx10^15 "kg m"^(-3)`

Answer

To find the approximate density of the iron nucleus with a mass number of 56, we can follow these steps:

### Step 1: Determine the radius of the nucleus
The radius \( R \) of a nucleus can be estimated using the formula:
\[ R = R_0 \cdot A^{1/3} \]
where \( R_0 \) is a constant approximately equal to \( 1.1 \times 10^{-15} \) meters (or 1.1 Fermi), and \( A \) is the mass number (in this case, 56).

### Step 2: Calculate the radius
Substituting the values:
\[ R = 1.1 \times 10^{-15} \cdot 56^{1/3} \]
Calculating \( 56^{1/3} \):
\[ 56^{1/3} \approx 3.83 \]
Now substituting back:
\[ R \approx 1.1 \times 10^{-15} \cdot 3.83 \approx 4.22 \times 10^{-15} \text{ meters} \]

### Step 3: Calculate the volume of the nucleus
The volume \( V \) of a spherical nucleus is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
Substituting the radius we found:
\[ V = \frac{4}{3} \pi (4.22 \times 10^{-15})^3 \]

### Step 4: Calculate \( R^3 \)
Calculating \( (4.22 \times 10^{-15})^3 \):
\[ (4.22 \times 10^{-15})^3 \approx 7.52 \times 10^{-44} \text{ cubic meters} \]
Now substituting this into the volume formula:
\[ V \approx \frac{4}{3} \pi (7.52 \times 10^{-44}) \approx 3.15 \times 10^{-43} \text{ cubic meters} \]

### Step 5: Calculate the mass of the nucleus
The mass \( m \) of the nucleus can be calculated using the mass number:
\[ m = A \cdot m_p \]
where \( m_p \) is the mass of a proton (approximately \( 1.67 \times 10^{-27} \) kg). Since the mass number is 56:
\[ m = 56 \cdot 1.67 \times 10^{-27} \approx 9.35 \times 10^{-26} \text{ kg} \]

### Step 6: Calculate the density of the nucleus
Density \( \rho \) is given by:
\[ \rho = \frac{m}{V} \]
Substituting the mass and volume:
\[ \rho \approx \frac{9.35 \times 10^{-26}}{3.15 \times 10^{-43}} \approx 2.97 \times 10^{17} \text{ kg/m}^3 \]

### Conclusion
The approximate density of the iron nucleus is \( \rho \approx 2.97 \times 10^{17} \text{ kg/m}^3 \).