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Calculating energy from a kilogram of fissionable fuel

Calculate the amount of energy produced by the fission of 1.00 kg of 235 U size 12{ {} rSup { size 8{"235"} } U} {} , given the average fission reaction of 235 U size 12{ {} rSup { size 8{"235"} } U} {} produces 200 MeV.

Strategy

The total energy produced is the number of 235 U size 12{ {} rSup { size 8{"235"} } U} {} atoms times the given energy per 235 U size 12{ {} rSup { size 8{"235"} } U} {} fission. We should therefore find the number of 235 U size 12{ {} rSup { size 8{"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 size 12{ {} rSup { size 8{"235"} } U} {} atoms is therefore,

4.25 mol 6.02 × 10 23 235 U/mol = 2 . 56 × 10 24 235 U . size 12{ left (4 "." "25"`"mol" right ) left (6 "." "02" times "10" rSup { size 8{"23"} } `"" lSup { size 8{"235"} } "U/mol" right )=2 "." "56" times "10" rSup { size 8{"24"} } `"" lSup { size 8{"235"} } U} {}

So 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 . alignl { stack { size 12{E= left (2 "." "56" times "10" rSup { size 8{"24"} } `"" lSup { size 8{"235"} } U right ) left ( { {"200"`"MeV"} over {"" lSup { size 8{"235"} } U} } right ) left ( { {1 "." "60" times "10" rSup { size 8{ - "13"} } `J} over {"MeV"} } right )} {} #" "=" 8" "." "20" times "10" rSup { size 8{"13"} } `J "." {} } } {}

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 (the 235 U ) size 12{ {} rSup { size 8{"235"} } U} {} is readily usable.

One nuclide already mentioned is 239 Pu size 12{ {} rSup { size 8{"239"} } ital "Pu"} {} , which has a 24,120-y half-life and does not exist in nature. Plutonium-239 is manufactured from 238 U size 12{ {} rSup { size 8{"238"} } 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 239 Pu size 12{ {} rSup { size 8{"239"} } ital "Pu"} {} . Breeding begins with neutron capture by 238 U size 12{ {} rSup { size 8{"238"} } U} {} :

238 U + n 239 U + γ .

Uranium-239 then β decays:

239 U 239 Np + β + v e ( t 1/2 = 23 min) .

Neptunium-239 also β decays:

239 Np 239 Pu + β + v e ( t 1/2 = 2 . 4 d ).

Plutonium-239 builds up in reactor fuel at a rate that depends on the probability of neutron capture by 238 U size 12{ {} rSup { size 8{"238"} } U} {} (all reactor fuel contains more 238 U size 12{ {} rSup { size 8{"238"} } U} {} than 235 U size 12{ {} rSup { size 8{"235"} } 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 235 U size 12{ {} rSup { size 8{"235"} } 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 239 Pu size 12{ {} rSup { size 8{"239"} } ital "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 FF 1 + FF 2 + xn , where FF 1 size 12{"FF" rSub { size 8{1} } } {} and FF 2 size 12{"FF" rSub { size 8{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 239 Pu ) by nuclear transformation is called breeding, and reactors designed for this purpose are called breeder reactors.

Questions & Answers

How we are making nano material?
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scanning tunneling microscope
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The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
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yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
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biomolecules are e building blocks of every organics and inorganic materials.
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research.net
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sciencedirect big data base
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Introduction about quantum dots in nanotechnology
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what does nano mean?
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nano basically means 10^(-9). nanometer is a unit to measure length.
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absolutely yes
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there is no specific books for beginners but there is book called principle of nanotechnology
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Devang Reply
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s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
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Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
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Mostly, they use nano carbon for electronics and for materials to be strengthened.
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carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
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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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