SCI 270: On Nuclear Technology Practice Problems
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# 9. Nuclear Fission and Nuclear Reactors

1. Give the nuclear reactions for the syntheses of neptunium and plutonium used in the fast breeder reactor for their productions.

 Nuclear reaction:____238U + n ® 239U + g ® Np + b ® Pu + b ____ Half lives: 239Np ___ 2.35 d _____ 239Pu __ 244e4 y __

2. Consult a handbook or a nuclear data source and give the range of known (or synthesized) plutonium isotopes. Which one has the longest half life and what is its decay mode?

 Source:_____________ Varies __________ Range of mass numbersof Pu isotopes: From _228__ to _ 247 _ Isotope with thelongest half life _244, 8.2e7 y __ Decay (modes): __alpha, beta, EC, SF,____

3. Assume the thermal-neutron induced fission of 235U (mass = 235.0439) gives two fragments of mass 140 and 93, and the some neutrons. After having consulted the properties of nuclides of mass numbers 140 and 93, we know that 140Ce (mass = 139.90539) and 93Nb (mass = 92.90638) are stable nuclides. Write the nuclear reaction equation for this fission process. Estimate the total energy (including energies of beta-decay of the fission products) in this fission process.

 Fission reaction:__ 235U + n ® 140Ce + 93Nb + 3 n ___ Energy released: _ Q = __ 200.08 ___ MeV/fission Energy released: _ Q = __ 1.93e13 __ J/mole

4. How much natural uranium must be processed in order to produce 1.0 kg of 235U, if only 50% of this isotope is extracted from natural uranium? What methods are used to separate 235U from natural uranium?

 Abundancesof 235U __0.72 %__;   238U __99.3 % ___ Amount of natural uranium required: _ 285.7 kg _ Methods used for extracting 235U: Gas diffusion of UF6 based on mass difference. _ __ Centrifuging of UF6 gas_____ __ Electromagnetic method to separate UF6+ ions ___ ___Thermal diffusion __

The following question are provided to help your study, and they will not be marked.

• What is a moderator? What are the desirable properties of a compound to make it a good moderator? Why are heavy water and graphite used as moderator for nuclear reactors using natural uranium as fuel, but normal distilled water not a suitable moderator?

See the end

• This problem may be a challenge to some of you, but hint will be given on the course website later. Assume the abundance of 235U to be 5.0% when the earth was created. Calculate the age of the earth from the distribution of 235U and 238U (now 0.72 % and 99.28 %). Note that the assumption may not be valid.

See below

• What is the percentage of deuterium in natural hydrogen? Assume that 70% of deuterium is extracted from natural water, how much water must be processed in order to produce 1.0 L of heavy water?

Is it possible to estimate the energy required to produce 1.0 L of heavy water? If it is, how much is required? If not, what information is required or why not?

Hint
Deuterium consists of 0.015% of natural hydrogen. To extract 1.0 L heavy water, we carry out the conversion in the following way:

```           100 L water   100 %
1.0 L D2O ------------ ------- = 9524 L water
0.015 L D2O   70 %
```
A large amount of water must be processes.

Many variables are present in the second half of the question, and assumptions have to be made in order to provide answers.

• Assume the abundance of 235U to be 5.0% when the earth was created. Calculate the age of the earth from the distribution of 235U and 238U (now 0.72 % and 99.28 %). Note that the assumption may not be valid.

• What is a moderator? What are the desirable properties of a compound to make it a good moderator? Why are heavy water and graphite used as moderator for nuclear reactors using natural uranium as fuel, but normal distilled water not a suitable moderator?

Hint
Assume that at time = 0, the atomic ratio of 235U to 238U was 5 to 95. After a period of time t, the ratio became 0.72/99.28.

l235 = 0.693/7.038e-8 y = 9.85e-10 y
l238 = 0.693/4.468e-9 y = 1.55e-10 y

235N = 235No exp (-l235 t )
-----     ----   ------------- dividing the two equations
238N = 238No exp (-l238 t )
Thus,

```0.72    5    exp (-l235 t )
---- = ---- --------------- = 0.55 exp -(l235 - l238) t
99.28  95.0  exp (-l238 t )
```
0.00725 = 0.55 exp -(l235 - l238) t
Solving for t gives t = 2.4e9 years. (Almost two and half billions years)

Discussion:
There are several methods for solving this problem, and all are acceptable. For a period this long, the estimates may be different due to difference in assumptions and methods. The assumption of 5 percent may not be correct, but making it allows us to look at the problem with some values.