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Chapter 12: Problem 30

What is the density of the material in the nucleus of the hydrogen atom? The nucleus can be considered to be a sphere of radius \(1.2\) \(\times 10^{-15} \mathrm{~m}\) and its mass is \(1.67 \times 10^{-27} \mathrm{~kg} .\) The volume of a sphere is \((4 / 3) \pi r^{3}\).

Short Answer

Expert verified
The density of the nucleus is approximately \(2.31 \times 10^{17} \text{ kg/m}^3\).

Step by step solution

01

Understand the Formula for Density

Density is given by the formula \( \text{density} = \frac{\text{mass}}{\text{volume}} \). We need both the mass of the nucleus and its volume to calculate its density.
02

Identify the Given Mass

The mass of the hydrogen atom nucleus is given as \( 1.67 \times 10^{-27} \text{ kg} \). This value will be used directly in the density formula.
03

Calculate the Volume of the Nucleus

We use the formula for the volume of a sphere: \[V = \frac{4}{3} \pi r^3\] where \( r = 1.2 \times 10^{-15} \text{ m} \). Plugging in the radius, we calculate the volume:\[V = \frac{4}{3} \pi (1.2 \times 10^{-15})^3\]\[V = \frac{4}{3} \pi (1.728 \times 10^{-45}) \text{ m}^3\]\[V \approx 7.238 \times 10^{-45} \text{ m}^3\]Thus, the volume of the nucleus is approximately \( 7.238 \times 10^{-45} \text{ m}^3 \).
04

Calculate the Density of the Nucleus

Now that we have both the mass and volume, we can find the density:\[\text{density} = \frac{1.67 \times 10^{-27}}{7.238 \times 10^{-45}}\]Calculating this gives:\[\text{density} \approx 2.31 \times 10^{17} \text{ kg/m}^3\]
05

Conclusion

Therefore, the density of the material in the nucleus of the hydrogen atom is approximately \(2.31 \times 10^{17} \text{ kg/m}^3\).

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Key Concepts

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

Density Calculation
Density is a fundamental concept in physics that describes how much mass is contained in a given volume. This is expressed with the formula \( \text{density} = \frac{\text{mass}}{\text{volume}} \). In the case of an atomic nucleus, understanding its density can reveal insights into its composition and structure.

To calculate the density of a nucleus, you need two key pieces of information:
  • The mass of the nucleus, which could be provided in various units, but is commonly measured in kilograms (kg).
  • The volume of the nucleus, a value derived from its geometric shape, usually a sphere.
Once these are known, plug them into the density formula to find the density. The result tells us how much mass per unit volume the nucleus has.
Understanding nuclear density is vital as it helps in distinguishing between various elements and isotopes based on their atomic structure.
Volume of a Sphere
The volume of a sphere is a well-known geometrical formula that applies to three-dimensional spherical objects. The formula is \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere. This formula tells us how much space is enclosed within a sphere's boundary.

Calculating the volume involves the following:
  • First, identify the radius of the sphere. This measure extends from the center to any point on the surface.
  • Next, cube the radius: multiply the radius by itself three times. This step accounts for the three-dimensional nature of volume.
  • Multiply the result by \( \pi \) (approximately 3.14159) and finally by \( \frac{4}{3} \).
Knowing how to calculate the volume of a sphere is essential for diverse fields, from physics to engineering, where spherical shapes often occur. In our exercise, this formula allows us to compute the nucleus's volume, a crucial step in determining its density.
Atomic Nucleus
The atomic nucleus is a small yet incredibly dense region at the center of an atom. It contains protons and neutrons, collectively known as nucleons. Although the nucleus is minuscule compared to the entire atom, it contains nearly all the atom's mass.

Several characteristics define the atomic nucleus:
  • Size: Typically measured in femtometers. The radius of a nucleus is extraordinarily tiny, around 1.2 x 10\(^{-15}\) meters.
  • Mass: Composed mostly of nucleons, with significant contributions from both protons and neutrons.
  • Shape: Often modeled as a sphere to simplify calculations, although real nuclei may deviate slightly from this shape.
Understanding the atomic nucleus is crucial, as it is the focus of nuclear physics鈥攊ncluding phenomena such as radioactivity and nuclear reactions. The density of a nucleus, as illustrated in the exercise, shows us how densely packed these subatomic particles are within the extremely small volume.

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Most popular questions from this chapter

What volume does 300 g of mercury occupy? The density of mercury is 13600 . From \(\rho=m / v\) $$ V=\frac{m}{\rho}=\frac{0.300 \mathrm{~kg}}{13600 \mathrm{~kg} / \mathrm{m}^{3}}=2.21 \times 10^{-5} \mathrm{~m}^{3}=22.1 \mathrm{~cm}^{3} $$

An electrolytic tin-plating process gives a tin coating that is \(7.50 \times\) \(10^{-5}\) thick. How large an area can be coated with \(0.500 \mathrm{~kg}\) of tin? The density of tin is \(7300 \mathrm{~kg} / \mathrm{m}^{3}\). The volume of \(0.500\) kg of tin is given by \(\rho=m / V\) to be $$ V=\frac{m}{\rho}=\frac{0.500 \mathrm{~kg}}{7300 \mathrm{~kg} / \mathrm{m}^{3}}=6.85 \times 10^{-5} \mathrm{~m}^{3} $$ The volume of a film with area \(A\) and thickness \(d\) is \(V=A d\). Solving for \(A\), we find $$ A=\frac{V}{d}=\frac{6.85 \times 10^{-5} \mathrm{~m}^{3}}{7.50 \times 10^{-7} \mathrm{~m}}=91.3 \mathrm{~m}^{2} $$ as the area that can be covered.

A metal wire \(75.0 \mathrm{~cm}\) long and \(0.130 \mathrm{~cm}\) in diameter stretches \(0.0350 \mathrm{~cm}\) when a load of \(8.00 \mathrm{~kg}\) is hung on its end. Find the stress, the strain, and the Young's modulus for the material of the wire. $$ \begin{array}{l} \sigma=\frac{F}{A}=\frac{(8.00 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}{\pi\left(6.50 \times 10^{-4} \mathrm{~m}\right)^{2}}=5.91 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}=5.91 \times 10^{7} \mathrm{~Pa} \\ \varepsilon=\frac{\Delta L}{L_{0}}=\frac{0.0350 \mathrm{~cm}}{75.0 \mathrm{~cm}}=4.67 \times 10^{-4} \\ Y=\frac{\sigma}{\varepsilon}=\frac{5.91 \times 10^{7} \mathrm{~Pa}}{4.67 \times 10^{-4}}=1.27 \times 10^{11} \mathrm{~Pa}=127 \mathrm{GPa} \end{array} $$

A 60 -kg woman stands on a light, cubical box that is \(5.0 \mathrm{~cm}\) on each edge. The box sits on the floor. What pressure does the box exert on the floor? $$ P=\frac{F}{A}=\frac{(60)(9.81) \mathrm{N}}{\left(5.0 \times 10^{-2} \mathrm{~m}\right)^{2}}=2.4 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} $$

Determine the fractional change in volume as the pressure of the atmosphere \(\left(1 \times 10^{5} \mathrm{~Pa}\right)\) around a metal block is reduced to zero by placing the block in vacuum. The bulk modulus for the metal is \(125 \mathrm{GPa}\)
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