# 1.6 Fission  (Page 4/14)

 Page 4 / 14

## Calculating energy from a kilogram of fissionable fuel

Calculate the amount of energy produced by the fission of 1.00 kg of ${}^{\text{235}}\text{U}$ , given the average fission reaction of ${}^{\text{235}}\text{U}$ produces 200 MeV.

Strategy

The total energy produced is the number of ${}^{\text{235}}\text{U}$ atoms times the given energy per ${}^{\text{235}}\text{U}$ fission. We should therefore find the number of ${}^{\text{235}}\text{U}$ atoms in 1.00 kg.

Solution

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

$\left(4.25 mol\right)\left(6.02×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}{}^{\text{235}}\text{U/mol}\right)=2\text{.}\text{56}×{\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}{}^{\text{235}}\text{U}.$

So the total energy released is

$\begin{array}{lll}E& =& \left(2\text{.}\text{56}×{\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}{}^{\text{235}}\text{U}\right)\left(\frac{\text{200 MeV}}{{}^{\text{235}}\text{U}}\right)\left(\frac{1.60×{\text{10}}^{-\text{13}}\phantom{\rule{0.25em}{0ex}}\text{J}}{\text{MeV}}\right)\\ & =& \text{}\text{8.21}×{\text{10}}^{\text{13}}\phantom{\rule{0.25em}{0ex}}\text{J}\text{.}\end{array}$

Discussion

This is another impressively large amount of energy, equivalent to about 14,000 barrels of crude oil or 600,000 gallons of gasoline. But, it is only one-fourth the energy produced by the fusion of a kilogram mixture of deuterium and tritium as seen in [link] . Even though each fission reaction yields about ten 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 more scarce than fusion fuel, and less than 1% of uranium $\text{(the}\phantom{\rule{0.25em}{0ex}}{}^{\text{235}}\text{U}\text{)}$ is readily usable.

One nuclide already mentioned is ${}^{\text{239}}\text{Pu}$ , which has a 24,120-y half-life and does not exist in nature. Plutonium-239 is manufactured from ${}^{\text{238}}\text{U}$ in reactors, and it provides an opportunity to utilize the other 99% of natural uranium as an energy source. The following reaction sequence, called breeding    , produces ${}^{\text{239}}\text{Pu}$ . Breeding begins with neutron capture by ${}^{\text{238}}\text{U}$ :

${}^{\text{238}}\text{U}+n\to {}^{\text{239}}\text{U}+\gamma .$

Uranium-239 then ${\beta }^{–}$ decays:

${}^{\text{239}}\text{U}\to {}^{\text{239}}\text{Np}+{\beta }^{-}+{v}_{e}\text{(}{\text{t}}_{\text{1/2}}=\text{23}\phantom{\rule{0.25em}{0ex}}\text{min)}.$

Neptunium-239 also ${\beta }^{–}$ decays:

${}^{\text{239}}\text{Np}\to {}^{\text{239}}\text{Pu}+{\beta }^{-}+{v}_{e}\text{(}{\text{t}}_{\text{1/2}}=2\text{.}4\phantom{\rule{0.25em}{0ex}}\text{d}\text{).}$

Plutonium-239 builds up in reactor fuel at a rate that depends on the probability of neutron capture by ${}^{\text{238}}\text{U}$ (all reactor fuel contains more ${}^{\text{238}}\text{U}$ than ${}^{\text{235}}\text{U}$ ). Reactors designed specifically to make plutonium are called breeder reactors    . They seem to be inherently more hazardous than conventional reactors, but it remains unknown whether their hazards can be made economically acceptable. The four reactors at Chernobyl, including the one that was destroyed, were built to breed plutonium and produce electricity. These reactors had a design that was significantly different from the pressurized water reactor illustrated above.

Plutonium-239 has advantages over ${}^{\text{235}}\text{U}$ as a reactor fuel — it produces more neutrons per fission on average, and it is easier for a thermal neutron to cause it to fission. It is also chemically different from uranium, so it is inherently easier to separate from uranium ore. This means ${}^{\text{239}}\text{Pu}$ has a particularly small critical mass, an advantage for nuclear weapons.

## Phet explorations: nuclear fission

Start a chain reaction, or introduce non-radioactive isotopes to prevent one. Control energy production in a nuclear reactor! Nuclear Fission

## Section summary

• Nuclear fission is a reaction in which a nucleus is split.
• Fission releases energy when heavy nuclei are split into medium-mass nuclei.
• Self-sustained fission is possible, because neutron-induced fission also produces neutrons that can induce other fissions, $n+{}^{A}X\to {\text{FF}}_{1}+{\text{FF}}_{2}+\text{xn}$ , where ${\text{FF}}_{1}$ and ${\text{FF}}_{2}$ are the two daughter nuclei, or fission fragments, and x is the number of neutrons produced.
• A minimum mass, called the critical mass, should be present to achieve criticality.
• More than a critical mass can produce supercriticality.
• The production of new or different isotopes (especially ${}^{\text{239}}\text{Pu}$ ) by nuclear transformation is called breeding, and reactors designed for this purpose are called breeder reactors.

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Source:  OpenStax, Physics for the modern world. OpenStax CNX. Sep 16, 2015 Download for free at http://legacy.cnx.org/content/col11865/1.3
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