Why can't you just turn a nuclear reactor off?

Printer-friendly versionPrinter-friendly version Share this

When a nuclear reactor gets into trouble, what's wrong with hitting the off switch?

Big Ideas: 
  • The difference between nuclear fission and nuclear decay: fission is neutron-induced and you can stop it with neutron-absorbing material; nuclear decay just happens.

Nuclear reactors generate heat by the splitting into two of uranium (sometimes plutonium) nuclei. This process is called fission. The reaction is self-sustaining because it is initiated by a neutron striking and being absorbed onto a uranium nucleus, and the fission process produces more neutrons, which go on to cause more fission. This is called a chain reaction. The chain reaction is easy to stop. Neutrons can be absorbed harmlessly on many different nuclei ("neutron poisons", e.g. cadmium), and the introduction of these poisons, in the form of control rods pushed into the reactor core, turn off the chain reaction very quickly.

Now let us consider the fission products. Uranium-235 abs:orbs a neutron, briefly becomes uranium-236, and splits into two "daughters". Uranium-236 has 92 protons and 144 neutrons. Note there are more than 1.5 times as many neutrons as protons. This is typical of heavy nuclei but not of light and medium-mass nuclei (where the number of neutrons equals that of protons for elements up to around 32S and 40Ca). The reason for the extra neutrons is that there is the enormous electrostatic repulsion between the 92 protons that needs to be balanced by the attractive nuclear force provided by the uncharged neutrons. When the 236U splits, a couple of these excess neutrons escape immediately (and are used to propagate the chain reaction):

n + 235U → 236U → 117Pd + 117Pd + 2n

It doesn't often happen that the uranium splits into two identical daughters (palladium, Pd,  is element 92/2 = 46), but we use this example for simplicity. Palladium has several stable isotopes, mass 102 to 110, but mass 117 is not one of them! There are at least 7 too many neutrons. Palladium, with only 46 protons, doesn't need the same excess of neutrons that uranium does to be stable. As a result, the 117Pa has to undergo at several beta-decays (which convert neutrons into protons in the nucleus with the emission of an electron and a neutrino) to get to stability (tin-117), and in doing so, releases a lot of energy.

117Pd → 117Ag +e-+ν ... etc., then  117Ag → 117Cd → 117In → 117Sn

Have a look at these three plots of nuclei, modified from the Interactive Chart of Nuclides1 available on the web from  the National Nuclear Data Center in Brookhaven, New York.


 Fig.1 All nuclides plotted according to decay mode. Stable = black, electron emission (β- decay) = pink, positron emission (β+ decay) or electron capture = blue, alpha-decay = yellow.

The red line connecting 236U to the origin shows roughly where the fission products can lie. The red fission product line never intersects with the black line of stability, because the black line is curved. Heavy nuclei need more neutrons for stability, proportionally, than do lighter ones.

 Fig.2 Products of neutron-induced 235U fission.  

Real fission is usually asymmetric. This gets the daughter nuclei slightly closer to the line of stability, which releases slightly more energy in the original fission, and is therefore more probable than symmetric fission. But this is a detail that doesn't affect the basic question here. The fission products are nearly all unstable, and are going to decay their way upwards and to the left, toward the line of stability, by β-decay. How long will this take? See the next figure:


 Fig. 3 All nuclides colour-coded by half-life, from light pink (<1μs) to green (~1s) to blue (>1y) and black (stable).

The half-life gets longer and longer the closer one gets to stability. Some of the last steps before reaching a stable nucleus take tens of years or more. This is why reactor cores continue to generate heat long after the chain-reaction is shut down.

At the time of writing (March 29th, 2011) each reactor at Fukushima was still generating several MW of heat2. To set this in perspective, if one tries to cool this with water, we can easily calculate how much. Raising water from 10C to steam at 100C absorbs (90 K)(4.180 kJ/kg/K)+2260 kJ/kg = 2.6 MJ/kg. That is to say, 2.6 MW will boil off 1 kg of water per second, a tonne in 17 minutes, etc.

Currently Fukushima reactor 1-3 are generating about 14 MW2 of heat and the prediction is that this rate will fall very slowly. A year from now the total power will have dropped to 8 MW.

Fig.4 Power generated by Fukushima reactors 1-3 as function of time after the tsunami it. The data data for the first five days are measured, the rest are a prediction2.

 Chart of Nuclides Introduction:

Nuclear Fission and Decay Heat:

Rate of Boiling the Cooling Water at Fukushima:


Yes that's right this will

Yes that's right this will gets the daughter nuclei slightly closer which releases slightly more energy and is therefore more probable than symmetric fission and doesn't affect the basic question here.

Post new comment

Please note that these comments are moderated and reviewed before publishing.

The content of this field is kept private and will not be shown publicly.
By submitting this form, you accept the Mollom privacy policy.

a place of mind, The University of British Columbia

C21: Physics Teaching for the 21st Century
UBC Department of Physics & Astronomy
6224 Agricultural Road
Vancouver, BC V6T 1Z1
Tel 604.822.3675
Fax 604.822.5324

Emergency Procedures | Accessibility | Contact UBC | © Copyright The University of British Columbia