# [mechanics] radius of a sphere with critical mass

1. Sep 2, 2015

1. The problem statement, all variables and given/known data
Here is the question:

"In the fall of 2002, a group of scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a chain reaction. This element has a density of 19.5 g/cm3 .

What would be the radius of a sphere of this material that has a critical mass?

r = ______ cm ? "

So my answer should be in CM, and I should express my answer using 2 significant figures.

2. Relevant equations
Here, equations I used were

V = m * d

3. The attempt at a solution

What I've done so far:

Ok you have

Mass = 60 kg

D = 19.5g/cm3

I converted 60 kg to 60,000 g because D is in terms of g/cm3 and my final answer is suppose to be cm, so I think this helps simplify it.

anyways, I have the equation:

V = m * d

so

(4/3)pi*r3 = 60,000 g * 19.5g/cm3

Okay, I'm unsure what happens when I multiply 60,000 g by 19.5g/cm3

Does my result become 1170000 g2/cm3? or does it remain 1170000 g/cm3 ???

So from what I understand after this, I'm going to get

r3 = (3 (result of the above multiplication) ) / 4pi

and then take that result and 1/3 it to get my final answer.

I'm mostly stuck here on the multiplying 60,000 g by 19.5g/cm3 part.

I have the feeling that I understand how to do it but I actually can't do it :/!

2. Sep 2, 2015

### SteamKing

Staff Emeritus
Really? It should be very clear that multiplying grams × grams / cm3 ≠ cm3

Figure out how to get only cm3 [Hint: multiplication isn't the only thing you can do here.]

3. Sep 2, 2015

Sorry, what I meant was, when multiplying grams by grams/cm^3, is my answer going to be grams^2/cm^3 or is it going to stay grams/cm^3?

Also, do you mean dividing?

If so, I tried the following:

60,000 g x cm^3 / 19.5 g to cancel out the g's, resulting with 3076.923077 cm^3.

After that I got ( (3076.923077 cm^3.) x 3 ) / 4pi = 7249.829201 cm^3

so now I take (7249.829201 cm^3)^1/3 = 19.35423112 .

see, now I'm not sure what happened to the cm^3. when I did ^1/3, did that get rid of the ^3 and only make it cm? Or does it stay cm^3?

4. Sep 2, 2015

### SteamKing

Staff Emeritus

Density is defined as $ρ = \frac{M}{V}$. Obviously, if you are looking for V, then you must manipulate this equation algebraically to obtain $V = \frac{M}{ρ}$.
The units are manipulated in a similar fashion: $\frac{g}{cm^3}=g\div cm^3$, so doing a similar manipulation to the one above, $cm^3=\frac{g}{\frac{g}{cm^3}}$, which can be simplified by inverting the units in the denominator and multiplying by the numerator, $\frac{g}{\frac{g}{cm^3}} = g × \frac{cm^3}{g}=cm^3$.

You get a volume of $V = 3076.9\: cm^3 = \frac{4πr^3}{3}$.

When you solve for r3, the units are still cm3. By taking the cube root of r3 in cm3, you are left with the radius in cm.

5. Sep 2, 2015