If an atom were scaled up to be comparable to the extent of a mid-sized campus, how large would the nucleus be, and what sort of familiar object would be similar?

Short Answer

Expert verified
Answer: When the atom is scaled up to the size of a mid-sized campus, the nucleus would be about 10 centimeters in diameter. A familiar object with a similar size would be a grapefruit or a smaller soccer ball.

Step by step solution

01

Determine the approximate size ratio between the atom and its nucleus

To have an idea about the relative size, we will use the approximate sizes for an atom and its nucleus. Generally, an atom has a diameter of about 10^{-10} meters, or 1 angstrom, while its nucleus has a diameter of about 10^{-14} meters. The ratio between the nucleus and the atom's diameter is therefore: Ratio = (Diameter of nucleus) / (Diameter of atom) = (10^{-14} m) / (10^{-10} m) = 10^{-4} This means that the nucleus is about 10000 times smaller than the atom.
02

Assume a size for a mid-sized campus

Let's assume a mid-sized campus is approximately 1 kilometer (km) or 1000 meters (m) in diameter.
03

Calculate the size of the nucleus in the scaled-up version

Now, using our ratio, we can find the diameter of the nucleus in our scaled-up model. Since we found that the nucleus is about 10000 times smaller than the atom, and our atom is now represented by a diameter of 1000 meters, we can calculate: Diameter of scaled-up nucleus = (Diameter of campus) / 10000 = (1000 m) / 10000 = 0.1 m or 10 centimeters (cm) This means that when the atom is scaled up to the size of a mid-sized campus, the nucleus would be about 10 centimeters in diameter.
04

Find a familiar object with a similar size

Now we need to find a familiar object with a diameter of around 10 cm. A common object that has a diameter close to this size would be a grapefruit or a soccer ball (which has a diameter of around 22 cm). So, in the scaled-up version, the nucleus would be similar in size to a grapefruit or a smaller soccer ball.

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Most popular questions from this chapter

A particular fission of \({ }^{235} \mathrm{U}+\mathrm{n}(\) total \(A=236)\) breaks up. One fragment has \(Z=54\) and \(N=86\), making it \({ }^{140} \mathrm{Xe} .\) If no extra neutrons are produced in this event, what must the other fragment be, so all numbers add up? Refer to a periodic table (e.g., Fig. B.1; p. 375 ) to learn which element has the corresponding \(Z\) value, and express the result in the notation \({ }^{\mathrm{A}} \mathrm{X}\).

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