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If all electrical power were produced by nuclear fission of U-235, Earth’s known reserves of uranium would be depleted in less than a century. However, Earth’s supply of fissionable material can be expanded considerably using a breeder reactor    . A breeder reactor operates for the first time using the fission of U-235 as just described for the pressurized water reactor. But in addition to producing energy, some of the fast neutrons originating from the fission of U-235 are absorbed by U-238, resulting in the production of Pu-239 via the set of reactions

0 1 n + 92 238 U 92 239 U β 93 239 N p β 94 239 P u .

The Pu-239 is itself highly fissionable and can therefore be used as a nuclear fuel in place of U-235. Since 99.3 % of naturally occurring uranium is the U-238 isotope, the use of breeder reactors should increase our supply of nuclear fuel by roughly a factor of 100. Breeder reactors are now in operation in Great Britain, France, and Russia. Breeder reactors also have drawbacks. First, breeder reactors produce plutonium, which can, if leaked into the environment, produce serious public health problems. Second, plutonium can be used to build bombs, thus increasing significantly the risk of nuclear proliferation.

Calculating energy of fissionable fuel

Calculate the amount of energy produced by the fission of 1.00 kg of 235 U given that the average fission reaction of 235 U produces 200 MeV.

Strategy

The total energy produced is the number of 235 U atoms times the given energy per 235 U fission. We should therefore find the number of 235 U atoms in 1.00 kg.

Solution

The number of 235 U atoms in 1.00 kg is Avogadro’s number times the number of moles. One mole of 235 U has a mass of 235.04 g; thus, there are ( 1000 g ) / ( 235.04 g / mol ) = 4.25 mol . The number of 235 U atoms is therefore

( 4.25 mol ) ( 6.02 × 10 23 235 U / mol ) = 2.56 × 10 24 235 U .

Thus, the total energy released is

E = ( 2.56 × 10 24 235 U ) ( 200 MeV 235 U ) ( 1.60 × 10 −13 J MeV ) = 8.21 × 10 13 J .

Significance

This is another impressively large amount of energy, equivalent to about 14,000 barrels of crude oil or 600,000 gallons of gasoline. However, it is only one-fourth the energy produced by the fusion of a kilogram mixture of deuterium and tritium. Even though each fission reaction yields about 10 times the energy of a fusion reaction, the energy per kilogram of fission fuel is less, because there are far fewer moles per kilogram of the heavy nuclides. Fission fuel is also much scarcer than fusion fuel, and less than 1 % of uranium (the 235 U ) is readily usable.

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Check Your Understanding Which has a larger energy yield per fission reaction, a large or small sample of pure 235 U ?

the same

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Summary

  • Nuclear fission is a process in which the sum of the masses of the product nuclei are less than the masses of the reactants.
  • Energy changes in a nuclear fission reaction can be understood in terms of the binding energy per nucleon curve.
  • The production of new or different isotopes by nuclear transformation is called breeding, and reactors designed for this purpose are called breeder reactors.

Conceptual questions

Should an atomic bomb really be called nuclear bomb?

Yes. An atomic bomb is a fission bomb, and a fission bomb occurs by splitting the nucleus of atom.

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Why does a chain reaction occur during a fission reaction?

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In what way is an atomic nucleus like a liquid drop?

Short-range forces between nucleons in a nucleus are analogous to the forces between water molecules in a water droplet. In particular, the forces between nucleons at the surface of the nucleus produce a surface tension similar to that of a water droplet.

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Problems

A large power reactor that has been in operation for some months is turned off, but residual activity in the core still produces 150 MW of power. If the average energy per decay of the fission products is 1.00 MeV, what is the core activity?

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(a) Calculate the energy released in the neutron-induced fission n + 238 U 96 S r + 140 X e + 3 n , given m ( 96 S r ) = 95.921750 u and m ( 140 X e ) = 139.92164 .

(b) This result is about 6 MeV greater than the result for spontaneous fission. Why?

(c) Confirm that the total number of nucleons and total charge are conserved in this reaction.

a. 177.1 MeV; b. This value is approximately equal to the average BEN for heavy nuclei. c. n + 92 238 U 146 38 96 S r 58 + 54 140 X e 86 + 3 n , A i = 239 = A f , Z i = 92 = 38 + 54 = Z f

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(a) Calculate the energy released in the neutron-induced fission reaction n + 235 U 92 K r + 142 B a + 2 n , given m ( 92 K r ) = 91.926269 u and m ( 142 B a ) = 141.916361 u . (b) Confirm that the total number of nucleons and total charge are conserved in this reaction.

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The electrical power output of a large nuclear reactor facility is 900 MW. It has a 35.0 % efficiency in converting nuclear power to electrical power.

(a) What is the thermal nuclear power output in megawatts?

(b) How many 235 U nuclei fission each second, assuming the average fission produces 200 MeV?

(c) What mass of 235 U is fissioned in 1 year of full-power operation?

a. 2.57 × 10 3 MW ; b. 8.04 × 10 19 fissions/s ; c. 991 kg

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Find the total energy released if 1.00 kg of 92 235 U were to undergo fission.

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Practice Key Terms 5

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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