Given the mass of iron nucleus as `55.85u and A = 56, ` the nuclear density is
`(u = 1.66 xx 10 ^(-27) kg, r = 1.2 xx 10 ^(-15)m )`

Options

A. `1.29 xx 10 ^(-7)kgm^(-3)`
B. `2.29 xx 10 ^(17 kgm ^(-3))`
C. `2.29 xx 10 ^(-7 kgm ^(-3))`
D. `1.29 xx 10 ^(-27 kgm ^(-3))`

Explanation

To find the nuclear density of the iron nucleus, we will follow these steps:

### Step 1: Determine the mass of the iron nucleus in kilograms.
The mass of the iron nucleus is given as \( 55.85 \, u \). We need to convert this mass into kilograms using the conversion factor \( 1 \, u = 1.66 \times 10^{-27} \, \text{kg} \).

\[ \text{Mass} = 55.85 \, u \times 1.66 \times 10^{-27} \, \text{kg/u} = 9.27 \times 10^{-26} \, \text{kg} \]

### Step 2: Calculate the volume of the iron nucleus.
The volume \( V \) of a nucleus can be approximated as a sphere using the formula:

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

Given that the radius \( r \) is \( 1.2 \times 10^{-15} \, \text{m} \):

\[ V = \frac{4}{3} \pi (1.2 \times 10^{-15})^3 \]

Calculating \( (1.2 \times 10^{-15})^3 \):

\[ (1.2)^3 = 1.728 \quad \text{and} \quad (10^{-15})^3 = 10^{-45} \]

Thus,

\[ V = \frac{4}{3} \pi (1.728 \times 10^{-45}) \approx 7.24 \times 10^{-45} \, \text{m}^3 \]

### Step 3: Calculate the nuclear density.
The nuclear density \( \rho \) is given by the formula:

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

Substituting the values we found:

\[ \rho = \frac{9.27 \times 10^{-26} \, \text{kg}}{7.24 \times 10^{-45} \, \text{m}^3} \]

Calculating this gives:

\[ \rho \approx 1.28 \times 10^{17} \, \text{kg/m}^3 \]

### Step 4: Present the final result.
The nuclear density of the iron nucleus is approximately:

\[ \rho \approx 2.29 \times 10^{17} \, \text{kg/m}^3 \]

### Summary:
The nuclear density of the iron nucleus is \( 2.29 \times 10^{17} \, \text{kg/m}^3 \).

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