- #1
nmsurobert
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- Homework Statement
- There is relatively little empty space between atoms in solids and liquids, so that the average density of an atom is about the same as matter on a macroscopic scale—approximately 10^3 kg/m^3 . The nucleus of an atom has a radius about 10^-5 that of the atom and contains nearly all the mass of the entire atom. (a) What is the approximate density of a nucleus?
- Relevant Equations
- ρ = m/V
I have the solution to the problem but I think I found either a typo or steps that were not included. I think I have made sense of the problem but I would like to double check that I'm doing this right. Thanks.
ρn = mn/Vn
mn ≈ ma
Rn = 10-5Ra
This is were I am having issue. The solution says:
ρn = ma/10-5Va = ρa/10-5
ρn = 103 kg/m3/10-5
It says 10-5Va. Shouldn't it be (10-5)3Va because of the radius is a cubed value? Same for the ρn/10-5. Shouldn't it also be ρa/(10-5)3?
If those values are not cubed then I end up with
ρn = 103 kg/m3/10-5 = 108... which is not right.
But if I cube that value then I get the correct answer. I wrote up a similar problem using simpler numbers and my logic checked out. However, like I said, I just wanted to double check.
ρn = mn/Vn
mn ≈ ma
Rn = 10-5Ra
This is were I am having issue. The solution says:
ρn = ma/10-5Va = ρa/10-5
ρn = 103 kg/m3/10-5
It says 10-5Va. Shouldn't it be (10-5)3Va because of the radius is a cubed value? Same for the ρn/10-5. Shouldn't it also be ρa/(10-5)3?
If those values are not cubed then I end up with
ρn = 103 kg/m3/10-5 = 108... which is not right.
But if I cube that value then I get the correct answer. I wrote up a similar problem using simpler numbers and my logic checked out. However, like I said, I just wanted to double check.
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