Calculating the Density of a Uranium Nucleus

  • #1
this seems like a fairly simple problem, however I'm not sure if my calculation seems right...

the question:
The nucleus of a uranium atom has a diameter of 1.5 * 10 to the -14 and a mass of 4.0 * 10 to the -25.

It then asks what the density of the nucleus is.

I know that nucleus density is Mass/Volume, and I calculated the volume to be the radius (.5(1.5 * 10 to the -14)) squared times pi. this gave me 1.76 * 10 to the -28. So am i correct with an answer of 4.0 * 10 to the -25 / 1.76 * 10 to the -28? or am i going about this problem wrong? please help.

Answers and Replies

  • #2
The volume of a sphere is [tex]V= \frac{4}{3}\pi r^3 [/tex] volume must have the units of length cubed. What you have calculated is the area of a circle (cross section of the sphere).
  • #3
ahh, that makes sense... but that still doesn't seem right for some reason. that yielded me 1.4137 * 10 to the -41 for the volume...
  • #4
When you cube something small it becomes even smaller.
  • #5
You should 1.8E-42 (i presume m^3). You've forgotten that r = 0.5D.
  • #6
For instance (1/2)³ is of half of one half of one half. That's 1/8, which is smaller than 1/2.
  • #7
i.e. [tex]V_{sphere}= \frac{\pi}{6} d^3 [/tex]
  • #8
thank you... i actually came up with 1.767 * 10^-42 as the volume... a really small number but it is correct.

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