# How much faster does induced fission in a typical reactor occur than spontaneous fission in nature?

I know that the half-life of Uranium-235 is about 704 million years, but...

That includes alpha and beta radiation as well as spontaneous fission, though...

Also, I presume the 'spontaneous fission' rates usually given for specific isotopes actually include a great deal of naturally-induced fission, since they are measuring clumps of material, not individual atoms isolated from each other...

Elsewhere on S.E. physics I read that the natural, background rate for spontaneous fission of a clump of U-235 is about .00563 fissions per kg per second... Is that correct?

I also read that a 1000 MWe (or 1 GWe) nuclear reactor goes through about 3.14 kg of U-235 per day... Assuming pure U-235, I presume...

So, what is the rate of fissions per kg of U-235 per second in an average reactor?

P.S.: What is the rate for those 'natural' reactors in Gabon and Cameroon?

• Wikipedia's "isotopes of" pages list the percentage of decays which are spontaneous fission (SF) and cluster decay (CD). Eg, for U-235 en.wikipedia.org/wiki/Isotopes_of_uranium#List_of_isotopes gives $7×10^{-9}$% for SF. Apr 28, 2022 at 8:29
• 1. The natural reactor in Oklo, Gabon, happened ~1.7 billion years ago, when the percentage of U-235 in natural uranium was much higher than it is today. 2. "Assuming pure U-235, I presume" Even weapons grade U isn't pure U-235. Enriched uranium for reactors is typically 3% to 5% U-235. As I mentioned here, enrichment is difficult! Apr 28, 2022 at 8:46

The amount of fission will depend on the power being generated. However, if you know the power, then it is pretty easy to calculate the rate: just divide the total power by the energy per fission event. One $$^{235}_{92}\mathrm{U}$$ fission releases about 203 MeV of energy, or $$3.25 \times 10^{-17}\ \mathrm{MJ}$$. Thus if your reactor is producing a total power of 1 MW, say (a pretty darn small reactor, but you can scale up from here), then there will be about $$3.07 \times 10^{16}\ \mathrm{fission/s}$$, that's 30.7 quadrillion per second.

However, only some of this can be converted to electrical power. Conversion for a usual reactor is about 33%, so to get 1 MW of electrical power, we need about 3x that or 92 quadrillion fissions per second.

Now that's a big number, but it's still quite small on the scale of atoms and molecules: Avogadro's number is on the order of 10 million times larger still, and thus one can get some sense of the energy scales at work here and just how good nuclear fuel is when it comes to energy density.

As for how it compares to natural decay, depends on how much fuel, but given that the per-kg rate is so small, you can clearly tell the difference must be extremely large for any reasonable fuel amount.

"Spontaneous fission (SF)" and "fission" usually refer to two completely different processes. SF happens when an atom fissions all by itself. This is very rare. "Fission" occurs then an atom absorbs a neutron and then a fission event occurs. The probability of SF for uranium is much much smaller than the probability of alpha decay.

The number of fissions that occur in a reactor depends entirely on the power level of the reactor. In your example, you have a reactor that produces 1000 MWe. This translates to about 3000 MW thermal power. The amount of energy produced in a day is 3000 MW*day (need to convert this to Joules then MeV). A single fission produces about 200 MeV of energy, so you can find the total number of fissions that occur in a single day. You can then use Avogadro's number and the atomic mass of uranium to find the mass of uranium consumed every day.

This is a fairly common homework problem for nuclear engineering students. It works out to about 1 gram per day. Your value of 3.14 kg is very high.

Depends on fuel quantity. Fission is easy, put some uranium in a pile tight enough, you have fission. Spontaneous fission is a relatively slow process, so the energy amounts are similar in mass converted, but occurs at a much slower rate. Human induced fission compresses fuel to tight confine, when sufficient fuel abounds you get sufficient neutron economy, and fission

That includes alpha and beta radiation as well as spontaneous fission, though...

In fact U-235 is nearly a pure alpha emitter, with spontaneous fissions occurring in about 70 decays per trillion. Beta decay is energetically forbidden: Np-235 and Pa-235 can instead decay by $$\beta^\pm$$ to uranium.

Also, I presume the 'spontaneous fission' rates usually given for specific isotopes actually include a great deal of naturally-induced fission, since they are measuring clumps of material, not individual atoms isolated from each other...

Without looking at any source material today, I would expect any self-induced fission to have been corrected out of the data, so that the “spontaneous fission” rate is the rate for an isolated atom. For a given sample geometry (density, shape, nearby neutron moderators or absorbers, etc.), you can then compute the effective probability $$k_\text{eff}$$ that a neutron from any fission goes on to induce an additional fission. For $$k_\text{eff} < 1$$, any currently-active fission processes will exponentially die away; for $$k_\text{eff} > 1$$, the rate of fissions will exponentially increase. A stable reactor uses some negative feedback to maintain $$k_\text{eff} \approx 1$$.

the natural, background rate for spontaneous fission of a clump of U-235 is about .00563 fissions per kg per second... Is that correct?

This is a unit-conversion problem. If the number of atoms in the sample is given by

$$N(t) = N_0 2^{-t/t_{1/2}} = N_0 e^{-t/\tau}$$

then the decay rate per atom is

\begin{align} \frac{\dot N(t)}{N_0} &= -\frac{1}\tau = -\frac{\ln 2}{t_{1/2}} \\ &= -\frac{0.7}{0.7\times10^9\text{ year}} \color{lightgray}{ {}\times\left( \frac{1\text{ year}}{\pi\times 10^7\text s} \right) } \\ &\approx \frac1\pi \times 10^{-16} \text{ s}^{-1} \end{align} A mole of U-235, with $$2\pi\times10^{23}$$ atoms and mass 235 grams, would have total $$\approx 2\times10^7$$ decays per second, but spontaneous fissions only make up $$1.4\times10^{-3}\text{ s}^{-1}$$ of those. That’s roughly in line with your estimate per kilogram.

What is the rate of fissions per kg of U-235 per second in an average reactor?

If you know that each fission releases, on average, 200 MeV of kinetic energy, this is another unit conversion problem. We know that

\begin{align} \rm 1\,eV &= 1.6\times10^{-19}\rm\, J \\ \rm 1\,mol\cdot eV &= 6\times10^{23} \cdot 1.6\times 10^{-19} \rm\,J \approx 10\times 10^{4}\,J \\ \rm 1\,mol \cdot 200\,MeV &\approx \rm 200\times10^{11}\,J \end{align}

That is, fission at a rate of one mole per second would correspond to about twenty terawatts of thermal power. A two-gigawatt (thermal) power plant, a large commercial installation, apparently fissions a mole of U-235 every 10,000 seconds or so, or a little over a kilogram per day. Also consistent with your estimate, and consistent with values elsewhere of about one gram of fissioned fuel per megawatt-day.

What is the rate for those 'natural' reactors in Gabon and Cameroon?

Currently, zero. The natural reactors were active in the geological past, when natural uranium had more U-235. Apparently the uranium deposits would flood with water, which served as a neutron moderator and kickstarted the reaction, releasing enough heat to boil away the floodwater and turn the reaction off. This happened many times over many tens of thousands of years. I vaguely recall that the overall power at the Oklo reactor was never as much as a megawatt, and that the total fissioned uranium is appropriately measured in tons; Wikipedia confirms my recollections.