Hydrogen atom vs hydrogen nucleus.

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Homework Help Overview

The discussion revolves around the comparison of the sizes of a hydrogen atom and its nucleus, specifically focusing on scaling these dimensions to a more relatable size, such as an American football field. The problem involves calculating the scaled diameter of the hydrogen nucleus and determining the volume ratio between the atom and the nucleus.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to convert measurements appropriately and scale the sizes correctly. There is a focus on understanding how to relate the size of the hydrogen atom to that of a football field and the implications for the nucleus's size. Questions arise about the correct method for scaling and whether the calculations align with the problem's requirements.

Discussion Status

Some participants have offered pointers regarding unit conversions and scaling methods, while others express uncertainty about the approach to take. There is an ongoing exploration of how to correctly interpret the problem and apply the relevant equations for volume comparison.

Contextual Notes

Participants note the importance of maintaining proportionality when scaling dimensions and highlight the potential for confusion in interpreting the problem's requirements. There is also mention of the need for clarity in showing work to facilitate better responses.

albinoboy
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Homework Statement


A hydrogen atom has a diameter of approximately 1.06x10^-10 m, as defined by the diameter of the spherical electron cloud around the nucleus. The hydrogen nucleus has a diameter of approximately 2.40x10^-15 m.

(a) For a scale model, represent the diameter of the hydrogen atom by the length of an American football field (100 yards = 300 ft) and determine the diameter of the nucleus in millimeters.

(b) The atom is how many times larger in volume than its nucleus?

Homework Equations


2.40x10^-15 (1000mm/1m) = 2.40e-12

V=(4/3)*pi*r^3

The Attempt at a Solution


For (a) I got 2.40e-12
It says "Your response differs from the correct answer by orders of magnitude."

For (b) I got 4.42e4
It says "Your response differs from the correct answer by orders of magnitude."

Any help would be greatly appreciated!
 
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Hello albinoboy,

Just a couple of general pointers:

Don't forget to convert the length of a football field to meters.

How much bigger is the length of a football field than the diameter of a hydrogen atom? So scaling the diameter of the nucleus by the same amount gives us...

You used "2.40x10^-15 (1000mm/1m) = 2.40e-12", but that's just converting the actual diameter of the nucleus from meters to millimeters. I believe the question is asking you to scale up the whole atom to the size of a football field, and then express the scaled-up diameter of the nucleus in millimeters. In other words, "if hydrogen atoms were the size of football fields, how big would hydrogen nuclei be?"

You have the correct relevant equation given for volume. But your answer isn't correct. Showing your work would help get better responses.
 
collinsmark said:
Hello albinoboy,

Just a couple of general pointers:

Don't forget to convert the length of a football field to meters.

How much bigger is the length of a football field than the diameter of a hydrogen atom? So scaling the diameter of the nucleus by the same amount gives us...

You used "2.40x10^-15 (1000mm/1m) = 2.40e-12", but that's just converting the actual diameter of the nucleus from meters to millimeters. I believe the question is asking you to scale up the whole atom to the size of a football field, and then express the scaled-up diameter of the nucleus in millimeters. In other words, "if hydrogen atoms were the size of football fields, how big would hydrogen nuclei be?"

You have the correct relevant equation given for volume. But your answer isn't correct. Showing your work would help get better responses.

I am unsure if I am following correctly. Do I take 2.40x10^-15 and multiply it by how many meters are in a football field?
 
albinoboy said:
I am unsure if I am following correctly. Do I take 2.40x10^-15 and multiply it by how many meters are in a football field?

Not quite. What I'm saying is find,

x = \frac{ \mbox{length of a football field, in meters}}{\mbox{diameter of a hydrogen atom, in meters}}

Now you can say, Football fields are x times bigger than hydrogen atoms."

Now scale the nucleus diameter by the same amount.

If you're having trouble, think of it this way. You've multiplied the diameter of a hydrogen atom by x to make it as big as a football field. Now multiply the diameter of the nucleus by x to keep it proportional to the size of football field sized atom.

Moving on to part (b). Here you need to find the volume of a hydrogen atom and the volume of a hydrogen nucleus, then divide the two. You can do this directly using your relevant equation. (But you might find it easier to keep things in terms of equations before you divide, because a lot of terms will cancel out and you'll have fewer calculations; although this is not absolutely necessary. You should get the same result either way.)
 

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