Problem 3 - College Physics

Question

Density of the nucleus. (a) Using the empirical formula for the radius of a nucleus, show that the volume of a nucleus is directly proportional to its nucleon number \(A .\) (b) Give a reasonable argument concluding that the mass \(m\) of a nucleus of nucleon number \(A\) is approximately \(m=m_{\mathrm{p}} A,\) where \(m_{\mathrm{p}}\) is the mass of a proton. (c) Use the results of parts (a) and (b) to show that all nuclei should have about the same density. Then calculate this density in \(\mathrm{kg} / \mathrm{m}^{3},\) and compare it with the density of lead (which is 11.4 \(\mathrm{g} / \mathrm{cm}^{3} )\) and a neutron star (about \(10^{17} \mathrm{kg} / \mathrm{m}^{3} ) .\)

Step-by-Step Solution

Answer

The nuclear volume is proportional to the nucleon number, mass is approx. \( m = m_p A \), and density is constant, \( \approx 2.3 \times 10^{17} \text{ kg/m}^3 \).

Step 1: Understanding the empirical formula for the nuclear radius

The empirical formula for the radius of a nucleus is given by \( R = R_0 A^{1/3} \) where \( R_0 \) is a constant and \( A \) is the nucleon number. This means that the radius of a nucleus grows proportionally to \( A^{1/3} \).

Step 2: Determining the volume of a nucleus

The volume \( V \) of a sphere (which the nucleus is approximated as) is given by \( V = \frac{4}{3} \pi R^3 \). Substituting the expression for the radius from Step 1, we have \( V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A \). Hence, the volume is directly proportional to \( A \).

Step 3: Explaining the mass approximation for the nucleus

Since each nucleon (either a proton or neutron) has roughly the same mass, the total mass \( m \) of the nucleus with nucleon number \( A \) can be approximated as \( m = m_p A \), where \( m_p \) is the mass of a single proton. This is because the contributions to mass by electrons and binding energy are relatively small.

Step 4: Showing the density of a nucleus is constant

Density \( \rho \) is defined as mass divided by volume, \( \rho = \frac{m}{V} \). Substituting from Steps 2 and 3, we get \( \rho = \frac{m_p A}{\frac{4}{3} \pi R_0^3 A} = \frac{3m_p}{4\pi R_0^3} \). Thus, the density is independent of \( A \) and is a constant for all nuclei.

Step 5: Calculating the density of a nucleus

Given \( R_0 \approx 1.2 \times 10^{-15} \text{ m} \) and \( m_p \approx 1.67 \times 10^{-27} \text{ kg} \), the density \( \rho \) is calculated as \[ \rho = \frac{3 \times 1.67 \times 10^{-27}}{4 \pi (1.2 \times 10^{-15})^3} \approx 2.3 \times 10^{17} \text{ kg/m}^3. \]

Step 6: Comparing nuclear density to other objects

The density of a nucleus, \( 2.3 \times 10^{17} \text{ kg/m}^3 \), is vastly greater than both the density of lead (\( 11.4 \text{ g/cm}^3 = 11,400 \text{ kg/m}^3 \)) and comparable to that of a neutron star (\( 10^{17} \text{ kg/m}^3 \)). This illustrates the extreme density of nuclear matter.

Key Concepts

Nucleus Volume

When discussing the size of an atomic nucleus, we often approximate its shape as a sphere. This simplification allows us to calculate its volume using the formula for the volume of a sphere: \( V = \frac{4}{3} \pi R^3 \). Here, \( R \) represents the radius of the nucleus. To be more precise in our calculations of nuclear volume, we use an empirical formula for the radius: \( R = R_0 A^{1/3} \).
This formula implies that the radius is dependent on the nucleon number \( A \), which makes sense as larger nuclei with more nucleons should have larger radii.
Substituting \( R \) in the sphere volume formula, we get \( V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A \). This indicates that the volume of the nucleus is directly proportionate to \( A \), the nucleon count.
In simple terms, the more nucleons a nucleus contains, the more volume it occupies, following this cubic root pattern within the radius formula.

Nucleon Number

The nucleon number \( A \) of an atomic nucleus plays a significant role in defining its properties. It represents the total number of protons and neutrons within the nucleus. Each of these particles is referred to as a nucleon.
This number not only determines the element of the atom but also its stability and how it behaves in nuclear reactions.
When calculating other nucleus properties like volume or mass, the nucleon number is crucial.

• For instance, the radius of an atomic nucleus scales as \( A^{1/3} \), highlighting the dependence on nucleon number.
• The mass, similarly, is approximated by considering the sum of the individual masses of \( A \) nucleons.

In essence, knowing the nucleon number allows scientists to predict and compute fundamental nuclear characteristics, correlating directly to real-world nuclear applications and phenomena.

Mass Approximation

When estimating the mass of a nucleus, a handy concept is the mass approximation. Imagine a nucleus as a compact collection of nucleons. Since these nucleons (protons and neutrons) have nearly equal mass, a reasonable approximation is to use the mass of one nucleon and multiply by the nucleon number \( A \). Thus, the total mass \( m \) of the nucleus is given by \( m = m_{p} A \), where \( m_{p} \) is the mass of a proton.
This approach simplifies calculations significantly because we neglect small contributions from electrons and binding energy differences.
While the binding energy—which holds nucleons together—does technically slightly reduce the nuclear mass, for many practical purposes, this approximation suffices.
Through this approximation, we can understand the consistent density across different nuclei, leading to deeper insights into nuclear physics.

Nuclear Radius Formula

A compelling insight into nuclear structure comes from the nuclear radius formula. This empirical formula \( R = R_0 A^{1/3} \) neatly encapsulates how nuclear size varies with nucleon number \( A \).

• \( R_0 \) is a constant that reflects a typical nucleon radius and is approximately \( 1.2 \times 10^{-15} \text{ meters} \).
• The \( A^{1/3} \) term indicates that for each nucleus with a higher nucleon count, the radius grows at a rate proportional to the cubic root of the nucleon number.

The utility of this formula lies in its predictive power; it helps nuclear physicists estimate nuclear sizes without direct measurement, smoothing many theoretical and practical pathways.
With this formula, calculations pertaining to volume, like in the step-by-step solution, become straightforward, establishing a consistent guideline across diverse nuclear entities.