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Chapter 2: Problem 102

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 chemistry instructor's claim can be considered reasonably accurate. After calculating and comparing the scaling factors for the size of a grape and 1 mile, we found that they are relatively close. \(2.54 * 10^{13}\) is close to \(3.04 * 10^{13}\). The electrons are far away from the nucleus when compared to the nucleus's size, but the exact distance can vary depending on the atom and the electron's orbital.

Step by step solution

01

Size of a nucleus

A typical size of a nucleus can be approximated by its radius, which is approximately in the order of \(10^{-15}\) meters (1 femtometer).
02

Size of an atom

The size of an atom can be approximated by the radius of the outermost electron orbital. For hydrogen, the smallest atom, it is about \(5.3 * 10^{-11}\) meters (53 picometers). We will use this value for our calculation. #Step 2: Calculate the ratio between nucleus size and atom size#
03

Ratio between the sizes

We need to find the ratio between the size of the nucleus and the size of the atom. This can be calculated as follows: Ratio = \(\frac{\text{Size of nucleus}}{\text{Size of atom}}\) Ratio = \(\frac{10^{-15}}{5.3 * 10^{-11}}\) #Step 3: Calculate the scaling factor for "Size of a grape" and "1 mile away" equivalence#
04

Scaling factor for size of a grape

Now, we need to find out how many times we have to scale up the size of the nucleus (1 femtometer) to reach the size of a grape. Let's assume the size of a grape to be about 1 inch in diameter (2.54 cm). Scaling factor for grape = \(\frac{2.54 * 10^{-2}\,\text{meters}}{10^{-15}\,\text{meters}}\)
05

Scaling factor for 1 mile

We also need to find out the scaling factor for distance of 1 mile. 1 mile is approximately equal to 1609.34 meters. Scaling factor for 1 mile = \(\frac{1609.34\,\text{meters}}{5.3 * 10^{-11}\,\text{meters}}\) #Step 4: Compare the scaling factors and comment on the claim's accuracy# After calculating the scaling factors in Step 3, we can compare them to evaluate the claim. If the calculated scaling factors are close to each other, the claim can be considered reasonably accurate. If there is a significant difference, the claim might not be accurate. Solution: Let's calculate the actual scaling factors:
06

Calculate the scaling factors

Scaling factor for grape = \(\frac{2.54 * 10^{-2}\,\text{meters}}{10^{-15}\,\text{meters}}\) 鈮� \(2.54 * 10^{13}\) Scaling factor for 1 mile = \(\frac{1609.34\,\text{meters}}{5.3 * 10^{-11}\,\text{meters}}\) 鈮� \(3.04 * 10^{13}\) Now, let's compare them: \(2.54 * 10^{13}\) is relatively close to \(3.04 * 10^{13}\). Considering the approximations and assumptions made during the calculations, the chemistry instructor's claim can be considered reasonably accurate. The electrons are indeed far away from the nucleus when compared to the nucleus's size, but the exact distance can vary depending on the atom and the electron's orbital.

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