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A chemistry instructor makes the following claim: 鈥淐onsider that if the nucleus were the size of a grape, the electrons would be about 1 mile away on average.鈥 Is this claim reasonably accurate? Provide mathematical support.

Short Answer

Expert verified
The chemistry instructor's claim is not reasonably accurate. The actual size of a nucleus is approximately \(1.0 \times 10^{-14}\) meters, and the average distance of an electron from the nucleus is about \(5.3 \times 10^{-11}\) meters for a hydrogen atom. Scaling up the size of a nucleus to a grape (0.02 meters), we find that the scaled-up distance between the nucleus and electrons would be \(1.06 \times 10^5\) meters, which is approximately 65.85 times larger than the claimed distance (1 mile, or 1609.34 meters).

Step by step solution

01

Find the actual size of a nucleus and the distance between the nucleus and electrons

The size of a nucleus is approximately \(1.0 \times 10^{-14}\) meters, and the average distance of an electron from the nucleus is about \(5.3 \times 10^{-11}\) meters for a hydrogen atom.
02

Scale up the size of a nucleus to a grape and calculate the corresponding scale factor

Let's consider the size of a grape to be approximately 2 cm (0.02 meters). The scale factor is the ratio of the grape size to the actual nucleus size: \[\text{Scale factor} = \frac{\text{Size of grape}}{\text{Actual size of nucleus}}\] Now, let's calculate the scale factor: \[\text{Scale factor} = \frac{0.02}{1.0 \times 10^{-14}} = 2 \times 10^{15}\]
03

Determine the distance of the electrons using the scale factor

Now, let's multiply the actual distance between the nucleus and electrons by the scale factor to find the scaled-up distance: \[\text{Scaled-up distance} = \text{Actual distance} \times \text{Scale factor}\] Solving for the scaled-up distance: \[\text{Scaled-up distance} = (5.3 \times 10^{-11}) (2 \times 10^{15}) = 1.06 \times 10^5 \ \text{meters}\]
04

Compare the calculated distance with the claimed distance (1 mile)

To compare our calculated scaled distance with the claimed distance (1 mile), let's convert 1 mile to meters: 1 mile = 1609.34 meters Now let's see how close our calculated distance is to the claimed 1 mile away: \[\frac{\text{Calculated distance}}{\text{Claimed distance}} = \frac{1.06 \times 10^5}{1609.34} = 65.85\]
05

Analyze and conclude

The calculated scaled distance between the nucleus and the electrons is approximately 65.85 times larger than the claimed distance (1 mile). It shows that the claim "if the nucleus were the size of a grape, the electrons would be about 1 mile away on average" is not reasonably accurate.

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Key Concepts

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

Nucleus Size Comparison
In the realm of atomic physics, understanding the scale of a nucleus is crucial. The nucleus, located at the center of an atom, consists of protons and neutrons, and it is where most of the atom's mass is concentrated. However, its size is incredibly small, often described as being around \(1.0 \times 10^{-14}\) meters in diameter. To put this into perspective, the nucleus is like a tiny core hidden deep within a mostly empty space. This small size is what makes comparing the nucleus to everyday objects challenging. For instance, if we scale up the nucleus to the size of a grape (approximately 2 cm), the rest of the atom must also be considered in proportion. This scale helps illustrate the vast distances between the nucleus and electrons, providing insight into atomic structure beyond simple approximation.
Electron Distance
Electrons in an atom orbit the nucleus at varying distances, often portrayed as a cloud rather than fixed paths. For a hydrogen atom, the simplest example, the average distance from the nucleus is about \(5.3 \times 10^{-11}\) meters. This distance highlights how electrons are positioned relatively far from the nucleus compared to its size.When juxtaposed with the scaled version where the nucleus is the size of a grape, we need to multiply this electron distance by the previously calculated scale factor. This magnifying effect results in an electron orbit that would metaphorically place electrons approximately 65.85 miles away if the nucleus were indeed grape-sized.Understanding these vast electron distances underscores the atom鈥檚 largely empty space, reinforcing how this compares to familiar large-scale structures.
Scale Factor in Physics
A scale factor is a mathematical tool used to resize objects. In physics, it helps compare different sizes of physical quantities. Here it's used to translate atomic measurements into tangible, larger scales. The scale factor is determined by dividing the target size of the grape nucleus, \(0.02\) meters, by the actual size of the nucleus, \(1.0 \times 10^{-14}\) meters. This yields a staggering scale factor of \(2 \times 10^{15}\), illustrating the enormity required to visualize atomic structures in familiar terms. Applying this scale factor to electron distances transforms our understanding from the microscopic to a macroscopic intuition. By grasping this conceptual leap, one gains a better appreciation for the placement and distribution of components within an atom.
Hydrogen Atom Model
The hydrogen atom serves as the archetype in atomic physics, providing a fundamental model for atomic structure. Comprising a single proton nucleus and one electron, it simplifies complex interactions mostly due to its minimal structure, making it ideal for study.Using the Bohr model of the hydrogen atom, we see the electron orbiting at an average distance of \(5.3 \times 10^{-11}\) meters. This simplistic model enables calculations such as those explained previously to scale atomic sizes and understand structures representing all atoms.It is through such fundamental models that scientists can extrapolate atomic behaviors and interpret atomic scales, providing a better understanding of atoms in a manner that connects the microscopic world to the larger universe outside. The hydrogen atom model is not just significant for its simplicity, but also for demonstrating the basic principles of quantum mechanics that govern atomic interactions.

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Most popular questions from this chapter

Which of the following is(are) correct? a. \(^{40} \mathrm{Ca}^{2+}\) contains 20 protons and 18 electrons. b. Rutherford created the cathode-ray tube and was the founder of the charge- to-mass ratio of an electron. c. An electron is heavier than a proton. d. The nucleus contains protons, neutrons, and electrons

Each of the following statements is true, but Dalton might have had trouble explaining some of them with his atomic theory. Give explanations for the following statements. a. The space-filling models for ethyl alcohol and dimethyl ether are shown below. These two compounds have the same composition by mass \((52 \% \text { carbon, } 13 \% \text { hydrogen, and } 35 \% \text { oxygen }),\) yet the two have different melting points, boiling points, and solubilities in water. b. Burning wood leaves an ash that is only a small fraction of the mass of the original wood. c. Atoms can be broken down into smaller particles. d. One sample of lithium hydride is 87.4\(\%\) lithium by mass, while another sample of lithium hydride is 74.9\(\%\) lithium by mass. However, the two samples have the same chemical properties.

A binary ionic compound is known to contain a cation with 51 protons and 48 electrons. The anion contains one-third the number of protons as the cation. The number of electrons in the anion is equal to the number of protons plus 1. What is the formula of this compound? What is the name of this compound?

Why is the term 鈥渟odium chloride molecule鈥 incorrect whereas the term 鈥渃arbon dioxide molecule鈥 is correct?

Would you expect each of the following atoms to gain or lose electrons when forming ions? What ion is the most likely in each case? a. Ra b. In c. P d. Te e. Br f. Rb

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