Principal Quantum number by size?

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The discussion centers on calculating the principal quantum number (n) for a hydrogen atom to achieve a size of one micron, using the formula r=(5.3x10^-11)(n^2). The calculation yields n approximately equal to 97, raising questions about the practicality of such a high quantum number since typical values range from 1 to 6. It is noted that while n can theoretically extend to infinity, in practice, an atom with such a high n would likely lose its electron due to collisions. The conversation concludes with an acknowledgment of the complexities involved in such high quantum states. Understanding these quantum numbers is crucial for grasping atomic behavior at extreme scales.
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Homework Statement


What is the value of the principal quantum number for a hydrogen atom to have a size of one micron?

Homework Equations


r=(5.3x10-11)(n2)
r=1x10-6/2=5x10-7

The Attempt at a Solution



5x10-7=(5.3x10-11)(n2)

97.1 = n

How is this possible? lol Maybe it's more complicated then I'm thinking.
 
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Just use the nearest integer value.
 
So I'm right? The answer is just 97? I thought the principal quantum numbers (n) only go up to like 5 or 6 or something like that though.
 
Yes, you're right. n can go up to infinity in theory. In practice, an atom with such a high n would eventually collide with another atom or perhaps a container wall, and completely lose the electron.
 
Alrighty, thank you very much :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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