Chapter 31.1 Notes
Other Nuclear Activities
|Sometimes nuclides can undergo positronic decay. In this case, a proton re-organizes into a neutron and a positron is ejected from the nucleus, along with a neutrino. The positron is the anti-matter counterpart of an electron, and carries a charge of plus 1. The nuclear equation for such a decay can be written as follows:
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?
When the nucleus of Uranium-235 absorbs a neutron, it temporarily becomes U-236, which almost immediately splits into two smaller nuclides.
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
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
The mass of a neutron = 1.008665 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:
Now, if we multiply our mass to energy conversion factor obtained above times the mass defect of U-235, we get the following:
You Try: do practice problems 1-4 in Ch 31.
This neutron is absorbed into the nucleus of U-235, making it U-236 and adding energy to the new nucleus.
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.
The "2 or 3" neutrons that are released from the fission of U-235 are important players in the process.
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..
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.
homework: Set 1: Ch31 questions 1-8 , problems 21-25
Internet: study this website about Nuclear Power Plants