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Chapter 20: Problem 73

(a) Assuming nuclei are spherical in shape, show that the radius \((r)\) of a nucleus is proportional to the cube root of mass number \((A)\). (b) In general, the radius of a nucleus is given by \(r=r_{0} A^{1 / 3},\) where \(r_{0},\) the proportionality constant, is given by \(1.2 \times 10^{-15} \mathrm{~m}\). Calculate the volume of the \({ }^{238} \mathrm{U}\) nucleus.

Short Answer

Expert verified
The volume of the \(^{238}U\) nucleus is approximately \(1.71 \times 10^{-45} \text{ m}^3\).

Step by step solution

01

Understanding the Nuclear Radius

The radius of a nucleus is assumed to be proportional to the cube root of its mass number, i.e., \( r \propto A^{1/3} \). This means the radius should increase as the cube root of the number of nucleons (mass number \( A \)) increases. This idea stems from the concept that nucleons are distributed evenly inside the nucleus, giving it a spherical shape.
02

Expressing the Proportionality

We express the proportionality mathematically as \( r = r_0 A^{1/3} \), where \( r_0 \) is a constant of proportionality known as the 'nuclear proportionality constant'. This formula is essential in calculating specific nuclear properties, such as the radius or volume of the nucleus.
03

Identify the Given Constant

For this exercise, we are provided with the constant \( r_0 = 1.2 \times 10^{-15} \text{ m} \). This constant is an empirical value obtained from experimental data and helps us find the actual radius of any nucleus with mass number \( A \).
04

Determine the Radius of the \(^{238}U\) Nucleus

Substitute \( A = 238 \) and \( r_0 = 1.2 \times 10^{-15} \text{ m} \) into the equation. Calculate:\[ r = 1.2 \times 10^{-15} \times 238^{1/3}\]
05

Calculate the Cube Root of 238

Find the cube root of 238. Use a calculator to get:\[ 238^{1/3} \approx 6.193\]
06

Compute the Radius

Substitute the calculated cube root into the equation:\[ r = 1.2 \times 10^{-15} \times 6.193 \approx 7.43 \times 10^{-15} \text{ m}\].
07

Understanding Volume of a Sphere

The formula for the volume \( V \) of a sphere is given by \( V = \frac{4}{3}\pi r^3 \). Apply this formula to find the volume of the \(^{238}U\) nucleus.
08

Calculate the Volume of the \(^{238}U\) Nucleus

Use the radius \( r = 7.43 \times 10^{-15} \text{ m} \) to find the volume:\[ V = \frac{4}{3}\pi \times (7.43 \times 10^{-15})^3\]Calculate to get:\[ V \approx 1.71 \times 10^{-45} \text{ m}^3\]

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Key Concepts

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

Nuclear Radius Calculation
In nuclear physics, understanding the size of a nucleus is vital. The nuclear radius is not directly proportional to the mass number; rather, it follows a cube root relationship. This means that if you have a nucleus with a mass number \( A \), the radius \( r \) can be represented as \( r = r_0 A^{1/3} \). Here, \( r_0 \) is a constant known as the nuclear proportionality constant. This particular relationship arises because nuclei are generally spherical, and when nucleons (protons and neutrons) come together, they form a dense, compact structure.
Mass Number
The mass number is the total count of protons and neutrons in an atomic nucleus, represented by the symbol \( A \). For example, in \({^{238}U}\), the mass number is 238. Mass number is crucial because it dictates the nuclear radius in our calculations. Since the relationship is cubic, a small change in mass number can still have a significant impact on the nuclear radius and volume calculations.
Nuclear Volume
The volume of a nucleus is calculated assuming it is a sphere, using the formula \( V = \frac{4}{3}\pi r^3 \). This approach makes it easier to estimate the space within a nucleus. Once the radius \( r \) is known, which can be found through \( r = r_0 A^{1/3} \), the volume can be instantly calculated. For the \({^{238}U}\) nucleus with a radius of approximately \( 7.43 \times 10^{-15} \text{ m} \), its volume becomes \( V \approx 1.71 \times 10^{-45} \text{ m}^3 \). This tells us about how densely nucleons are packed inside the nucleus.
Spherical Nuclei
Nuclei are often modeled as spheres because this shape simplifies calculations and accurately reflects the uniform distribution of nucleons. In a spherical model, the radius formula \( r = r_0 A^{1/3} \) becomes especially useful. By treating the nucleus as spherical, we can assume symmetry in all directions, making it straightforward to apply mathematical formulas for volume and radius. This approach is commonly used, despite some nuclei being slightly deformed or ellipsoidal in reality.
Nuclear Constants
Key constants in nuclear physics, such as the nuclear proportionality constant \( r_0 \), are crucial for calculations relating to nuclear structure. The value \( r_0 = 1.2 \times 10^{-15} \text{ m} \) is derived from empirical data, providing a standard measure. This constant helps us uniformly describe nuclei sizes across different elements. Consistency in measurements is essential for scientific accuracy, allowing comparisons to be made between different nuclei and supporting the theoretical models of nuclear structure.

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Most popular questions from this chapter

Consider the decay series \(\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C} \longrightarrow \mathrm{D}\) where \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are radioactive isotopes with halflives of \(4.50 \mathrm{~s}, 15.0\) days, and \(1.00 \mathrm{~s},\) respectively, and \(\mathrm{D}\) is nonradioactive. Starting with 1.00 mole of \(\mathrm{A},\) and none of \(\mathrm{B}, \mathrm{C},\) or \(\mathrm{D},\) calculate the number of moles of \(\mathrm{A}\), \(\mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) left after 30 days.

Write the abbreviated forms for the following reactions: (a) \({ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \alpha \longrightarrow{ }_{8}^{17} \mathrm{O}+{ }_{1}^{1} \mathrm{p}\) (b) \({ }_{4}^{9} \mathrm{Be}+{ }_{2}^{4} \alpha \longrightarrow{ }_{6}^{12} \mathrm{C}+{ }_{0}^{1} \mathrm{n}\) (c) \({ }_{92}^{238} \mathrm{U}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{93}^{238} \mathrm{~Np}+2{ }_{0}^{1} \mathrm{n}\)

An electron and a positron are accelerated to nearly the speed of light before colliding in a particle accelerator. The resulting collision produces an exotic particle having a mass many times that of a proton. Does this result violate the law of conservation of mass? Explain.

Given that the half-life of \({ }^{238} \mathrm{U}\) is \(4.51 \times 10^{9}\) years, determine the age of a rock found to contain \(1.09 \mathrm{mg}\) \({ }^{238} \mathrm{U}\) and \(0.08 \mathrm{mg}{ }^{206} \mathrm{~Pb}\).

Given that: $$ \mathrm{H}(g)+\mathrm{H}(g) \longrightarrow \mathrm{H}_{2}(g) \quad \Delta H^{\circ}=-436.4 \mathrm{~kJ} / \mathrm{mol} $$ calculate the change in mass (in \(\mathrm{kg}\) ) per mole of \(\mathrm{H}_{2}\) formed.
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