# Calculating released energy of a fission reaction

I have a question regarding the power of a fission reaction:

You have a Uranium U235 atom, in which you send a neutron at to cause it to split into two (or more?) new cores. This reaction also sends out 3 neutrons, which are to hit other U235 cores to form a chain reaction. You have a critical mass of U235 atoms, but how do you "know" how many cores the neutrons will hit? It seems coincidental to me which makes me think the power of a nuclear bomb could differ greatly as the number of cores that split can differ?

Is this correct or am I missing something?

• One has to know the energy dependent neutron cross sections and the composition of the fissile material as well as the geometry. It's a non-trivial numerical problem that, today, can be solved on pretty much any reasonably powerful computer (even your cell phone). In the early days of weapons development they had to use sophisticated theoretical models coupled to whatever amount of numerical math they could afford (which in the early days was done with "hand crank" mechanical calculators). – CuriousOne Feb 4 '16 at 19:27
• @CuriousOne I see, that you for the insight! Definitely seems logical, I'll give it some thought and do some research on how this works in depth – Erik Feb 4 '16 at 19:32

They usually don't hit and K for them is less then1. For the very reason we use moderaters like water to increase k greater than 1 for sustained reaction.

To make a bomb you need a nuclear fission chain reaction.
In essence what you need on average slightly more than one neutron produced by a fission to produce further fissions.

So of your 3 neutrons, if on average 1.1 neutrons initiated further fission then after 500 generations (one fission initiating others), which would not take very long, you would be initiating $5 \times 10^{20}$ fissions and a great deal of energy.

The problem with making a bomb is to get that multiplicative factor above 1 and doing it under your control.

Even in enriched Uranium your average 3 neutrons initiate very few fission in part because those neutrons choose to do other things, for example they can be absorbed by a Uranium nucleus and not initiate a fission, they can escape from the Uranium etc.

Now this is where the idea of critical mass comes in.
What you need is to reduce the proportion of neutrons escaping under controlled conditions.

The rate at which neutrons escape depends on the surface area of the Uranium the rate of production depends on the volume of the Uranium.
So first of all make the Uranium spherical s a sphere has the smallest surface area for a given volume.

The make the sphere bigger because the volume goes up faster ($\propto \text{radius}^3$) that the surface area ($\propto \text{radius}^2$).
So as the sphere gets bigger the rate of production of neutrons goes up faster than the rate of loss of neutrons.

The sums are not that easy to do and it is easier to look up a table of critical masses to see that the critical mass of U235 is 52 kg as a sphere of diameter 17 cm.
So anything bigger that this will go off with a bang.

The control of when the bomb is go off is very difficult but one way of doing it is to have two hemisphere of Uranium each with a less than critical mass and then join them together.

And, of course, don't try this at home.

Energy releases for one fission is approximatively the same . So, what changes ? The number of fissions during one second ( for instance ) giving thermal power.

Number of fissions = N * Sigma * Phi

N : number of fissile atoms

Sigma : probability , neutron cross section .

Phi : neutron flux

We suppose a bomb and a reactor with the same number N (fissile atoms ) and with the same probability Sigma . How they works ?

In the reactor , neutron flux is maintained on control at low level by control rods and energy is delivered during months .

In the bomb , neutron flux grows very quickly without control and energy appears in milliseconds .