Nuclear Density of Iron Nucleus: A Detailed Solution

This problem asks us to calculate the nuclear density of an iron nucleus. We are given the mass of the iron nucleus in atomic mass units (u) and its mass number (A). Nuclear density is a fundamental property of atomic nuclei, representing the mass per unit volume of the nucleus.

Iron Nucleus: Understanding Nuclear Density

The nuclear density of an atom's nucleus is remarkably constant across most atomic nuclei, indicating that nuclear matter is nearly incompressible. This problem specifically focuses on an iron nucleus with a given mass and mass number. To find the nuclear density, we need to determine the mass of the nucleus in kilograms and its volume in cubic meters.

Formulas for Nuclear Density Calculation

To solve this problem, we will use the following key formulas and constants:

• Mass Conversion: The atomic mass unit (u) needs to be converted to kilograms (kg). \(1 \text{ u} = 1.6605 \times 10^{-27} \text{ kg}\)
• Nuclear Radius Formula: The radius (R) of a nucleus is related to its mass number (A) by the empirical formula: \(R = R_0 A^{1/3}\) Where \(R_0\) is the Fermi constant, approximately \(1.2 \times 10^{-15} \text{ m}\). This constant represents the radius of a single nucleon (proton or neutron).
• Volume of a Sphere: Nuclei are generally approximated as spherical. The volume (V) of a sphere is given by: \(V = \frac{4}{3} \pi R^3\) Where \(\pi \approx 3.14159\).
• Nuclear Density Formula: Density (\(\rho\)) is defined as mass (M) divided by volume (V): \(\rho = \frac{M}{V}\)

Detailed Calculation for Iron Nucleus Density

Let's apply the formulas step-by-step to calculate the nuclear density of the iron nucleus.

1. Iron Nucleus Mass Conversion to Kilograms

Given the mass of the iron nucleus \(M_{\text{iron}} = 55.85 \text{ u}\).

Convert this mass into kilograms using the conversion factor \(1 \text{ u} = 1.6605 \times 10^{-27} \text{ kg}\):

\(M = 55.85 \text{ u} \times (1.6605 \times 10^{-27} \text{ kg/u})\)

\(M \approx 92.731425 \times 10^{-27} \text{ kg}\)

\(M \approx 9.273 \times 10^{-26} \text{ kg}\)

2. Iron Nucleus Radius Calculation

The mass number of the iron nucleus is given as \(A = 56\). Using the nuclear radius formula \(R = R_0 A^{1/3}\) with \(R_0 = 1.2 \times 10^{-15} \text{ m}\):

\(R = (1.2 \times 10^{-15} \text{ m}) \times (56)^{1/3}\)

First, calculate \((56)^{1/3}\):

\((56)^{1/3} \approx 3.82587\)

Now, calculate the radius R:

\(R = (1.2 \times 10^{-15} \text{ m}) \times 3.82587\)

\(R \approx 4.591044 \times 10^{-15} \text{ m}\)

\(R \approx 4.59 \times 10^{-15} \text{ m}\)

3. Iron Nucleus Volume Calculation

Using the calculated radius R and the formula for the volume of a sphere \(V = \frac{4}{3} \pi R^3\):

\(V = \frac{4}{3} \times 3.14159 \times (4.591044 \times 10^{-15} \text{ m})^3\)

\(V = \frac{4}{3} \times 3.14159 \times (4.591044)^3 \times (10^{-15})^3 \text{ m}^3\)

\(V = \frac{4}{3} \times 3.14159 \times 96.611 \times 10^{-45} \text{ m}^3\)

\(V \approx 404.65 \times 10^{-45} \text{ m}^3\)

\(V \approx 4.0465 \times 10^{-43} \text{ m}^3\)

4. Nuclear Density Calculation for Iron

Now, we can calculate the nuclear density using the mass in kilograms and the volume in cubic meters using the formula \(\rho = \frac{M}{V}\):

\(\rho = \frac{9.273 \times 10^{-26} \text{ kg}}{4.0465 \times 10^{-43} \text{ m}^3}\)

\(\rho = \frac{9.273}{4.0465} \times 10^{(-26 - (-43))} \text{ kg/m}^3\)

\(\rho = 2.2915 \times 10^{17} \text{ kg/m}^3\)

Rounding to two decimal places, the nuclear density is:

\(\rho \approx 2.29 \times 10^{17} \text{ kg/m}^3\)

Summary of Nuclear Density Result

The calculated nuclear density for the iron nucleus is approximately \(2.29 \times 10^{17} \text{ kg/m}^3\).

Parameter	Value
Mass of Iron Nucleus (M)	\(55.85 \text{ u}\)
Mass of Iron Nucleus (M) in kg	\(9.273 \times 10^{-26} \text{ kg}\)
Mass Number (A)	\(56\)
Fermi Constant (\(R_0\))	\(1.2 \times 10^{-15} \text{ m}\)
Radius of Iron Nucleus (R)	\(4.59 \times 10^{-15} \text{ m}\)
Volume of Iron Nucleus (V)	\(4.0465 \times 10^{-43} \text{ m}^3\)
Nuclear Density (\(\rho\))	\(2.29 \times 10^{17} \text{ kg/m}^3\)