The claim: "If the nucleus were the size of a grape, the electrons would be one mile away on average" is reasonably accurate because the ratios between the nucleus's sizes and the distances (between electrons and nucleus) for the two given examples are in the same order of magnitude.
To know if the claim is accurate we need to calculate the ratio of the size of the nucleus (the same as a grape) and the distance between the electrons and the nucleus for example 1 (r₁):
[tex] r_{1} = \frac{s_{1}}{d_{1}} [/tex] (1)
and to compare it with the ratio of the size and the distance given in example 2 (r₂):
[tex] r_{2} = \frac{s_{2}}{d_{2}} [/tex] (2)
Where:
s₁: is the size of the nucleus (like the size of a grape)
d₁: is the distance between electrons and nucleus of example 1 = 1 mile
s₂: is the average diameter of the nucleus = 10⁻¹³ cm
d₂: is the average distance between electrons and nucleus of example 2 = 10⁻⁸ cm
Assuming that the diameter of a grape is 3 cm (in a spherical way), the ratio of the first example is (eq 1):
[tex] r_{1} = \frac{3 cm}{1 mi*\frac{160934 cm}{1 mi}} = 1.86 \cdot 10^{-5} [/tex]
Now, the ratio of the second example is (eq 2):
[tex] r_{2} = \frac{10^{-13} cm}{10^{-8} cm} = 1.00 \cdot 10^{-5} [/tex]
Since r₁ and r₂ are in the same order of magnitude (10⁻⁵), we can conclude that the given claim is reasonably accurate.
You can learn more about the nucleus of an atom here: https://brainly.com/question/10658589?referrer=searchResults
I hope it helps you!
