Calculating the Density of a Uranium Nucleus

[mawalker]

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.

 

[Max Eilerson]

The volume of a sphere is $$V= \frac{4}{3}\pi r^3$$ volume must have the units of length cubed. What you have calculated is the area of a circle (cross section of the sphere).

 

[mawalker]

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...

 

[quasar987]

When you cube something small it becomes even smaller.

 

[Max Eilerson]

You should 1.8E-42 (i presume m^3). You've forgotten that r = 0.5D.

 

[quasar987]

For instance (1/2)³ is of half of one half of one half. That's 1/8, which is smaller than 1/2.

 

[Astronuc]

i.e. $$V_{sphere}= \frac{\pi}{6} d^3$$

 

[mawalker]

thank you... i actually came up with 1.767 * 10^-42 as the volume... a really small number but it is correct.