Calculate the nuclear density of `""_(26)Fe^(54)`. Given that the nuclear mass of `""_(26)Fe^(54)` is 53.9396 amu.

Answer

To calculate the nuclear density of \( _{26}^{54}Fe \), we will follow these steps:

### Step 1: Determine the mass number (A)
The mass number \( A \) of the iron isotope \( _{26}^{54}Fe \) is given as 54.

### Step 2: 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 \( R_0 \) is a constant approximately equal to \( 1.2 \times 10^{-15} \) m.

Substituting the values:
\[ R = 1.2 \times 10^{-15} \times (54)^{1/3} \]

Calculating \( (54)^{1/3} \):
\[ (54)^{1/3} \approx 3.78 \]
Now substituting this back into the equation for \( R \):
\[ R \approx 1.2 \times 10^{-15} \times 3.78 \approx 4.536 \times 10^{-15} \text{ m} \]

### Step 3: Convert nuclear mass from amu to kg
The nuclear mass of \( _{26}^{54}Fe \) is given as 53.9396 amu. To convert this to kilograms, we use the conversion factor:
\[ 1 \text{ amu} = 1.67 \times 10^{-27} \text{ kg} \]
Thus, the mass \( m \) in kg is:
\[ m = 53.9396 \times 1.67 \times 10^{-27} \approx 8.99 \times 10^{-26} \text{ kg} \]

### 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 value of \( R \):
\[ V = \frac{4}{3} \pi (4.536 \times 10^{-15})^3 \]
Calculating \( (4.536 \times 10^{-15})^3 \):
\[ (4.536 \times 10^{-15})^3 \approx 9.36 \times 10^{-44} \text{ m}^3 \]
Now substituting this back into the volume equation:
\[ V \approx \frac{4}{3} \pi (9.36 \times 10^{-44}) \approx 3.92 \times 10^{-43} \text{ m}^3 \]

### Step 5: Calculate the nuclear density
The nuclear density \( \rho \) can be calculated using the formula:
\[ \rho = \frac{m}{V} \]
Substituting the values of \( m \) and \( V \):
\[ \rho = \frac{8.99 \times 10^{-26}}{3.92 \times 10^{-43}} \approx 2.29 \times 10^{17} \text{ kg/m}^3 \]

### Final Answer
The nuclear density of \( _{26}^{54}Fe \) is approximately \( 2.29 \times 10^{17} \text{ kg/m}^3 \).

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