Chapter 31.1 Notes

Notice the symbols for the positron and the neutrino. Since the sodium nuclide has gained a neutron but lost a proton, the mass number of the daughter nuclide (Neon in this case) remains the same, but the atomic number has decreased by one. How does this event compare to what goes on during beta decay?

There are other types of nuclear events whereby atoms absorb, rather than give off, particles.

For instance, Mercury-200 ( )can absorb a deuteron, the nucleus of Hydrogen-2 (). This produces a new daughter isotope, Thallium-202 which immediately alph-decays into another element.

You try: Write the nuclear equation for the above description, starting with the deuteron absorption by mercury, and continuing through the alpha decay of Thallium. What element is produced after the alpha particle is ejected from the nucleus of Thallium?

Fission.

When the nucleus of Uranium-235 absorbs a neutron, it temporarily becomes U-236, which almost immediately splits into two smaller nuclides.

• These nuclides may include various isotopes of borium,lanthanum, tellurium, cesium, krypton and numerous others.
• Each new nuclide will be roughly half the size of U-236, but since we don't know for sure what we'll get, we refer to these new daughter elements as "fission fragments".
• Also produced as a result of the fission are 2 or 3 fast neutrons and about 200 MeV of energy.

The equation for the fission of U-235 can be written as follows:

equation 1 :

We need to discuss this equation a little more thorougly to understand where the approximately 200 MeV of energy is coming from. We will take a look at the fuel, the "starter neutron" on the left, the fission fragments, the 2.5 (average) free neutrons on the right, and the 200 MeV of energy produced

The uranium fuel, U-235

First of all, if we add the masses of U-235's 92 protons to the masses of its 143 neutrons, we get a net mass that is DIFFERENT than the mass of the nuclues of U-235!

The mass of a proton = 1.007825 amu

• 92 protons x 1.007825 amu/proton = 92.7199 amu

The mass of a neutron = 1.008665 amu

• 143 neutrons x 1.008665 amu/neutron = 144.2391 amu

The mass of the U-235 nucleus should be : 144.2391 amu + 92.7199 amu = 236.9590 amu

but it isn't! The mass of U-235 is 235.0439

There is a difference in mass of -1.9151 amu. There seems to be some missing mass in the nucleus of U-235. This is known as the mass defect.

When the nucleons coalesce into a nucleus, some of their mass is converted to energy. In order to separate the nucleons back into free individuals would require the addition of this amount of energy back into the nucleus. This energy is called the binding energy of the nucleus, and we can calculate it's value according to Einstein's famous formula

E = mc²

To find the energy equivalent of the mass defect, we can find the value in Joules when 1 amu converts to energy if we use the kilogram equivalent of 1.0 amu for mass in the above formula, and light speed for c:

• E = (1.6605 x 10-²⁷ kg/amu x 1.0 amu)(2.9979 x 10⁸ m/s)²
• E = 14.923 x 10^-11 kg m²s² = 14.923 x 10^-11 Joule of energy result from each amu.
• converting this to electron volts, we see that
• E = 14.923 x 10^-11 Joule = 9.3149 x 10⁸ ev = 931.49 MeV/amu

Now, if we multiply our mass to energy conversion factor obtained above times the mass defect of U-235, we get the following:

• Binding energy = -1.9151 amu x 931.49 MeV/amu = -1783.90 MeV
• This is the binding energy of the nucleus of U-235. If we divide this value by the number of nucleons, we see that the binding energy per nucleon is:
• -1783.90 MeV / 235 nucleons = -7.6 MeV/nucleon
• The greater the binding energy/nucleon, the more "tightly bound" we say the nucleus is. That is to say, the more energy you would have to add to separate the nucleons from the nucleus.

You Try: do practice problems 1-4 in Ch 31.

• Note that as the nucleii get heavier, the binding energies/nucleon increase.
• This only continues until you get to Iron.
• After Iron, the binding energies/nucleon slowly taper off again. Iron has the most tightly bound nucleus.

The neutron on the left

This neutron is absorbed into the nucleus of U-235, making it U-236 and adding energy to the new nucleus.

• U-236 is very unstable and will split into 2 fission fragments.
• The neutron must be moving at an acceptable speed to enter a U-235 nucleus. If it is too fast, it is more likely to be absorbed by U-238, and there's lots of that in the fuel pellets used in a reactor.
• In fact, well over 90% of the fuel is U-238, and U-238 won't fission. It just becomes U-239.

The Fission Fragments

These fragments might be Krypton, Tellurium, etc. They will be elements found near Iron on the periodic table. The binding energies of the fission fragments in equation 1 above are closer to -8.5 MeV/nucleon.

• There is clearly a mass defect in these fission fragments that is greater than that of U-235.
• This additional mass defect is manifest as kinetic energy of the fragments and neutrons.
• This is the 200 MeV shown in the equation

The Neutrons:

The "2 or 3" neutrons that are released from the fission of U-235 are important players in the process.

• Not only is their kinetic energy an important resource, but they can be made to invade more U-235 nucleii and continue the reaction.
• When first liberated, these neutrons are too fast to successfully be absorbed by U-235, and are more likely to be taken up by U-238 which composes the majority of the Uranium fuel in a reactor.
• But, by adding water to the reactor vessel, when neutrons run into Hydrogen atoms, they slow down enough to the point that they can now cause fission of U-235!
• AND, the water has heated up to the point that it can produce steam to turn turbines connected to alternators!

The Energy:

The approximately 200 MeV produced from the fission of one atom of uranium doesn't sound like much, since 1 MeV is only 1.6 x 10^-13 Joule. But when we calculate the total energy released by a gram of pure U-235, the story looks a little different.

To make this calculation..

• Find the number of moles of U-235 you have in 1.0 g. The molar mass of U-235 is 235 g/mole.
• Thus 1.0 g of U-235 = = 4.25 x 10^-3 mole.
• At 6.02 x 10²³ atoms/mole, a gram of U-235 contains( )=2.56 x 10²¹ atoms.
• Producing 200 MeV/atom, we see that the fission of 1.0 gram of Uranium-235 can liberate 200MeV/atom x 2.56 x 10²¹ atoms, or 5.12 x 10²³ MeV.
• Converting this amount to Joules: 5.12 x 10²³ MeV x 1.6 x 10^-13 Joule/MeV, we arrive at 8.2 x 10¹⁰ Joule of energy liberated from the fission of a gram of pure U-235
• This value is approximately equal to 1 Megawatt-day of energy (1.0 J = 1.0 w-s, times 3600 seconds/hour x 24 hours/day = 8.6 x 10⁴ w-s, or 1.0 watt-day). A megawatt-day is a million (10⁶) watt-days, or 8.6 x 10¹⁰ watt-seconds.

The significance of this number is that while a gram of U-235 can produce a megawatt day of energy, it takes about 3.0 x 10⁶ grams of coal to produce the same amount. We see that fission of U-235 produces 3 million times the energy that the combustion of coal does. This is partly what makes the reaction so economically attractive, even in the face of lots of radio-active waste as by products. Even at that, the cost of nuclear power is now more on par with the cost of fossil fuel power production.

Currently, there are two operating nuclear power plants in California, each running two reactors. One plant is at Diablo Canyon near San Luis Obispo, and the other at San Onofre. These two plants generated 30,241 GWh of electricity in 2004. With an additional 6,729 GWh of nuclear-generated electricity imported from other states, about 13% of California's electricity was from nuclear sources in 2004.

Study the handout on Measuring radiation; HTML version PDF version(printable)

Internet: study this website about Nuclear Power Plants

• Learn the general layout of the power cycle of a PWR (pressurized water reactor),
• the major systems of such a plant,
• check out the reactor vessel and
• fuel systems.

click here for practice/supplemental problems