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Chapter 6: Problem 17

Show that the nuclear density is the same for all nuclei. (Take the masses of the proton and neutron to be the same.)

Short Answer

Expert verified
The nuclear density is \( \frac{3m}{4\pi r_0^3} \) and it is independent of the mass number.

Step by step solution

01

- Understand Nuclear Density

Nuclear density is defined as the mass of the nucleus divided by its volume. We represent the mass of a nucleus with mass number A, which is the sum of protons and neutrons.
02

- Compute the Mass of the Nucleus

The mass of the nucleus (M) can be approximated as the mass of one nucleon (either proton or neutron) multiplied by the mass number (A). Given the masses of the proton and neutron are the same, we can denote the nucleon mass as m. Hence, the mass of the nucleus is approximately: \[ M = A \times m \]
03

- Determine the Volume of the Nucleus

The radius (r) of a nucleus is given by the empirical formula: \[ r = r_0 A^{1/3} \] where \( r_0 \) is a constant approximately equal to 1.2 femtometers (fm). The volume (V) of the nucleus can be approximated as a sphere: \[ V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (r_0 A^{1/3})^3 \]
04

- Simplify the Volume Expression

Simplify the expression for the volume: \[ V = \frac{4}{3} \pi (r_0^3) A \]
05

- Calculate the Nuclear Density

Nuclear density (\( \rho \)) is defined as mass divided by volume. Hence, \[ \rho = \frac{M}{V} = \frac{A \times m}{(\frac{4}{3} \pi r_0^3) A} \] The A terms cancel each other out, leaving us with: \[ \rho = \frac{3m}{4\pi r_0^3} \]
06

- Conclude the Density is Constant

The final expression for nuclear density \( \rho \) does not depend on the mass number A but only on the nucleon mass m and the constant \( r_0 \): \[ \rho = \frac{3m}{4\pi r_0^3} \] Thus, the nuclear density is the same for all nuclei.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Number
The mass number, often symbolized as A, is a crucial property of atomic nuclei. It represents the total number of nucleons in a nucleus, which includes both protons and neutrons. This number is vital in determining the overall mass of a nucleus. For instance, if a nucleus contains 6 protons and 6 neutrons, its mass number would be 12.
Mass number is not the same as atomic mass, although they are related. Atomic mass accounts for the entire atom, including electrons and binding energy differences.
Understanding the mass number helps us compute other essential properties like nuclear density since it directly influences the mass of the nucleus.
Nucleon Mass
Nucleons are the particles that make up the nucleus, specifically protons and neutrons. For simplicity in many physics problems, including this one, we approximate the masses of protons and neutrons to be nearly identical. This approximated mass is referred to as the nucleon mass, symbolized as m. It simplifies calculations involving nuclear properties.
Typically, the nucleon mass is about 1.67 x 10^-27 kilograms. Multiplying the nucleon mass by the mass number (A) gives us the approximate mass of the entire nucleus:
\[ M = A \times m \] Acquaintance with the nucleon mass is essential for understanding how mass number and mass relate directly in nuclear physics.
Nuclear Radius
The nuclear radius is a measure of the size of the nucleus. It doesn't increase linearly with the number of nucleons but rather approximates as the cube root of the mass number. The empirical formula used to calculate the nuclear radius is:
\[ r = r_0 A^{1/3} \]
Here, \( r_0 \) is a constant around 1.2 femtometers (fm). Knowing the radius is crucial for computing other properties like volume. Since volume is determined based on radius, this non-linear increase ensures that we can accurately estimate the volume and, hence, density for nuclei of varying sizes.
Volume of Nucleus
The volume of a nucleus can be derived from its radius, assuming the nuclear shape is roughly spherical. Using the sphere volume formula, we have:
\[ V = \frac{4}{3} \pi r^3 \]
Incorporating our formula for the nuclear radius, it expands to:
\[ V = \frac{4}{3} \pi (r_0 A^{1/3})^3 \]
Simplifying this gives us:
\[ V = \frac{4}{3} \pi r_0^3 A \]
This formula shows that the volume scales linearly with the mass number A. Thus, even though the radius increases slowly with A, the volume reflects a more significant growth. This understanding is especially helpful when calculating nuclear density, as it factors into the mass-to-volume ratio straightforwardly.

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Most popular questions from this chapter

singly ionized atoms of neon-20 of velocity \(2.0 \times 10^{5} \mathrm{m} \mathrm{s}^{-1}\) enter the velocity selector of a mass spectrometer, where the magnetic field has the value \(0.15 \mathrm{T}\) (a) What electric field is established in the velocity selector? (b) The ions then enter the region where a second magnetic field of value \(0.50 \mathrm{T}\) deflects them into circular paths. What is the radius of the circular path? (c) If the beam of ions also contains traces of neon-22, what would the detection radius for this isotope be? (Molar masses: neon- \(20=19.992 \mathrm{g} \mathrm{mol}^{-1}\) neon- \(22=21.99 \mathrm{g} \mathrm{mol}^{-1}\).)

The radius of an atomic nucleus is given by the expression $$R=1.2 \times A^{1 / 3} \times 10^{-15} \mathrm{m}$$ (a) Use this expression to find the density of a $$\text { nucleus of iron }\left(_{26}^{56} \mathrm{Fe}\right) \text { in } \mathrm{kg} \mathrm{m}^{-3}$$ (b) How does this density compare with the normal density of iron? (c) If a star with a mass equal to 1.4 times the mass of our sun (solar mass \(=2.0 \times 10^{30}\) kg were to have this density, what should its radius be? (Such stars are formed in the end stage of the evolution of normal stars and are called neutron stars.)

How much energy is required to remove one proton from the nucleus of \(^{16}\) O A rough answer to this question is obtained by giving the binding energy per nucleon. A better answer is obtained when we write a reaction that removes a proton from the nucleus. In this case \(^{16} \mathrm{O} \rightarrow_{1}^{1} \mathrm{p}+^{15} \mathrm{N}\) Find the energy required for this reaction to take place. This is the proton separation energy. Get both values and compare them. (The atomic mass of oxygen is \(15.994 \mathrm{u} ;\) that of nitrogen is 15.000 u).

What is the dominant force between two protons separated by a distance of: (a) \(1.0 \times 10^{-15} \mathrm{m}\) (b) \(1.0 \times 10^{-14} \mathrm{m} ?\)

A radioactive source has a half-life of \(3.00 \mathrm{min}\). At the start of an experiment there was \(32.0 \mathrm{mg}\) of the radioactive material present. How much will there be after \(18.0 \mathrm{min}\) ?
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