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

Short Answer

Expert verified
The instructor's claim is not very accurate, as the actual ratio between the nucleus size and electron's orbital distance in a hydrogen atom is about 1:31,200, while the claim's ratio is 1:80,467. The claim is about 58% larger than the actual ratio. While it does illustrate the vast space between the nucleus and electrons, the specific numbers used are not an accurate representation of the scale.

Step by step solution

01

Scale for the instructor's claim

First, determine the scale factor for the chemistry instructor's claim. Assume that the grape has a diameter of 2 cm. Convert 1 mile to centimeters. 1 mile = 1.609 km = 160,934 cm Now, find the scale factor. Since 1 mile is 160,934 cm, the scale factor is: Scale factor = 160,934 / 2 = 80,467 According to the instructor's claim, the actual nucleus-electron distance ratio in an atom would be similar to 1:80,467.
02

Actual size of nucleus and electron orbits

Now let's look at the actual size of the nucleus and electron orbits in a hydrogen atom, the simplest atom. The diameter of a hydrogen nucleus (a single proton) is about 1.7 * 10^{-15} meters, while the electron orbits around the nucleus at an average distance of approximately 5.3 * 10^{-11} meters (Bohr radius).
03

Calculate the actual ratio

Now we will calculate the actual ratio between the nucleus size and the electron's orbital distance. Divide the electron's orbital distance (5.3 * 10^{-11} m) by the nucleus diameter (1.7 * 10^{-15} m): Actual ratio = (5.3 * 10^{-11}) / (1.7 * 10^{-15}) = 3.12 * 10^4
04

Compare the instructor's claim with the actual ratio

Now we can compare the instructor's claim with the actual ratio. The instructor's claim had a ratio of 1:80,467, while the actual ratio is 1:31,200. To see how accurate the claim is, we can calculate the relative error between the claim and the actual ratio: Relative error = (80,467 - 31,200) / 31,200 = 49,267 / 31,200 ≈ 1.58 The relative error is approximately 1.58, which means the instructor's claim is about 58% larger than the actual ratio. #Conclusion# The instructor's claim is not very accurate, as the actual ratio is about 58% lower than the claim's ratio. While the claim helps to illustrate the vast space between the nucleus and electrons, the specific numbers used are not an accurate representation of the scale.

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