Nuclear Matter Density Calculation: A Detailed Explanation

This solution provides a step-by-step guide to calculating the approximate order of magnitude of nuclear matter density. We will use the given formula for the radius of an atomic nucleus, $R = R_0 A^{1/3}$, along with the provided average mass of a nucleon.

Density Explained: Concepts of Nuclear Matter

Nuclear matter density refers to how tightly packed the matter is inside an atomic nucleus. A key finding in nuclear physics is that this density is almost the same for all atomic nuclei, regardless of how many protons and neutrons they contain (represented by the mass number, $A$). This constant density is a fundamental property of nuclear matter.

Step-by-Step Calculation for Nuclear Density

To find the density, we need to know the total mass of the nucleus and the total volume it occupies. We'll assume the nucleus is shaped like a sphere.

1. Density Formula Basics

The fundamental definition of density ($\rho$) is mass ($M$) divided by volume ($V$):

$ \rho = \frac{M}{V} $

2. Nucleus Mass Calculation

An atomic nucleus is made up of protons and neutrons, collectively called nucleons. If a nucleus has a mass number $A$, it means it contains $A$ nucleons in total. Given the average mass of a single nucleon ($m_{nucleon}$), the total mass ($M$) of the nucleus is:

$ M = A \times m_{nucleon} $

We are given that the average mass of a single nucleon is approximately $m_{nucleon} \approx 1.67 \times 10^{-27}$ kg.

3. Nucleus Volume Calculation

Assuming the nucleus is a sphere, its volume ($V$) is calculated using the standard formula:

$ V = \frac{4}{3} \pi R^3 $

The problem provides a formula for the nuclear radius: $R = R_0 A^{1/3}$, where $R_0$ is the Fermi radius constant ($R_0 \approx 1.2 \times 10^{-15}$ m). Let's substitute this into the volume formula:

$ V = \frac{4}{3} \pi (R_0 A^{1/3})^3 $

When we cube the term $(R_0 A^{1/3})$, we get $R_0^3 (A^{1/3})^3 = R_0^3 A$. So, the volume becomes:

$ V = \frac{4}{3} \pi R_0^3 A $

4. Deriving the Constant Nuclear Density Formula

Now, let's put the expressions for Mass ($M$) and Volume ($V$) into the density formula ($\rho = M/V$):

$ \rho = \frac{A \times m_{nucleon}}{\frac{4}{3} \pi R_0^3 A} $

Observe that the mass number $A$ appears in both the numerator and the denominator. It cancels out:

$ \rho = \frac{m_{nucleon}}{\frac{4}{3} \pi R_0^3} $

This simplified formula shows that the nuclear density ($\rho$) depends only on the mass of a nucleon ($m_{nucleon}$) and the constant $R_0$, and it does not depend on the size of the nucleus ($A$). This is why nuclear density is approximately constant.

5. Numerical Calculation of Nuclear Density

Let's substitute the given numerical values to find the density:

• Fermi radius constant: $R_0 \approx 1.2 \times 10^{-15}$ m
• Average mass of a nucleon: $m_{nucleon} \approx 1.67 \times 10^{-27}$ kg
• We'll use $\pi \approx 3.14$.

First, calculate $R_0^3$:

$ R_0^3 = (1.2 \times 10^{-15} \text{ m})^3 $

$ R_0^3 = (1.2)^3 \times (10^{-15})^3 \text{ m}^3 = 1.728 \times 10^{-45} \text{ m}^3 $

Next, calculate the term $\frac{4}{3} \pi R_0^3$, which represents the effective volume occupied per nucleon:

$ \frac{4}{3} \pi R_0^3 \approx \frac{4}{3} \times 3.14 \times (1.728 \times 10^{-45} \text{ m}^3) $

$ \frac{4}{3} \pi R_0^3 \approx 4.189 \times 1.728 \times 10^{-45} \text{ m}^3 $

$ \frac{4}{3} \pi R_0^3 \approx 7.240 \times 10^{-45} \text{ m}^3 $

Now, compute the density $\rho$ using the simplified formula:

$ \rho \approx \frac{1.67 \times 10^{-27} \text{ kg}}{7.240 \times 10^{-45} \text{ m}^3} $

To perform the division, we divide the numerical parts and subtract the exponents:

$ \rho \approx \left( \frac{1.67}{7.240} \right) \times 10^{-27 - (-45)} \text{ kg/m}^3 $

$ \rho \approx 0.2306 \times 10^{18} \text{ kg/m}^3 $

Order of Magnitude Determination

The question asks for the approximate order of magnitude of the nuclear matter density. To find this, we express our calculated density in standard scientific notation, which has the form $a \times 10^n$, where $a$ is a number between 1 and 10 ($1 \le a < 10$).

Our calculated density is:

$ \rho \approx 0.2306 \times 10^{18} \text{ kg/m}^3 $

To convert this into standard scientific notation, we adjust the decimal point:

$ \rho \approx 2.306 \times 10^{17} \text{ kg/m}^3 $

The exponent of 10 in this notation is 17. Thus, the approximate order of magnitude of the nuclear matter density is $10^{17}$ kg/m$^3$.

Summary of Result

The calculation shows that the approximate order of magnitude of nuclear matter density is $10^{17} \text{ kg/m}^3$. This matches the first option provided in the question.