Volume of critical mass?

Question

I'm having problem with this question... Hopefully someone can help? It goes like this:

A particular uranium alloy has a density of 18.75 g/cm^3. What volume if occupied by a critical mass of 49 kg of this alloy? [I calculated 2613.3 cm^3.] The critical mass can be decreased to 16kg if the alloy is surrounded by a layer of natural uranium (which acts as a neutron reflector). What is the volume of the smaller mass?

Actually, I'm a little confused here as to whether the density of the alloy changes when it is surrounded by a neutron reflector. Can anyone help me? Thanks.

Answer 1

The density of the alloy does not change, the natural U shielding just means that less mass will be required to achieve a critical nuclear reaction.

So, using the density provide, calcualte a new volume (should be slightly more than 1 ml (cc)

Ken

Answer 2

For 49 kg you need a volume of
49'000 g / 18.75 g/cm^3 = 2'613 cm^3 = .ca. 2,6 litres.

The density does not change when it is surrounded by a neutron reflector. All that changes is the fact that neutrons get reflected back into the uranium thus increasing the probability of a reaction. Hence, you need less mass to start a chain reaction. For that, you will then need:
16'000 g / 18.75 g/cm^3 = 853 cm^3 = .ca. 0,8 litres.

Before the critical mass is reached everything seems normal, but after that you have a big mess - so fast that you can not even blink an eyelid. The critical mass also depends on the shape of the volume - is it spherical, bars, rods or blocks, powder etc...

Answer 3

Density = Mass/Volume.