Energy released in fission reaction

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Can someone help me on this problem? Here's my work :)

Calculate the energy released in the fission reaction:
n + 235/92U-> Sr-88 + Xe-136 + 12n
Use appendix D, assume the initial KE of the neutron is very small

It can be seen that when the compound nucleus splits, it breaks into fission fragments, Strontium-88, Xenon-136, and some neutrons. Both fission products then decay by multiple emissions as a result of the high neutron-to-proton ratio possessed by these nuclides.

1/0 n + 235/92U ->236/92 U -> 88 / 38 Sr + 136/54Xe + 12/0n

ΔBE = BEproducts – BEreactamts

ΔBE = BEproducts – BEreactamts

ΔBE = (BESr-88 + BEXe-136 ) – (BEU-235)

ΔBE = (BESr-88 + BEXe-136 ) – (BEU-235)

ΔBE = (783.2MeV + 1156MeV ) – (1786 MeV)

= 153.2 MeV
 

Answers and Replies

  • #2
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I think this post would be more appropriate for the nuclear physics forum.
 
  • #3
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Could you restate your question, airkapp? I am not sure what you need help with. Do you just want your work so far checked or do you need to know about the decay of Xenon?
 
  • #4
Clausius2
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airkapp said:
Can someone help me on this problem? Here's my work :)

Calculate the energy released in the fission reaction:
n + 235/92U-> Sr-88 + Xe-136 + 12n
Use appendix D, assume the initial KE of the neutron is very small

It can be seen that when the compound nucleus splits, it breaks into fission fragments, Strontium-88, Xenon-136, and some neutrons. Both fission products then decay by multiple emissions as a result of the high neutron-to-proton ratio possessed by these nuclides.

1/0 n + 235/92U ->236/92 U -> 88 / 38 Sr + 136/54Xe + 12/0n

ΔBE = BEproducts – BEreactamts

ΔBE = BEproducts – BEreactamts

ΔBE = (BESr-88 + BEXe-136 ) – (BEU-235)

ΔBE = (BESr-88 + BEXe-136 ) – (BEU-235)

ΔBE = (783.2MeV + 1156MeV ) – (1786 MeV)

= 153.2 MeV
It only consists in looking at the tables you have just in front of you. Just to say the number 153.2 MeV sounds good, a typical fission reaction delivers approximately 200 MeV.
 
  • #5
Astronuc
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airkapp said:
Can someone help me on this problem? Here's my work :)

Calculate the energy released in the fission reaction:
n + 235/92U-> Sr-88 + Xe-136 + 12n
Use appendix D, assume the initial KE of the neutron is very small

It can be seen that when the compound nucleus splits, it breaks into fission fragments, Strontium-88, Xenon-136, and some neutrons. Both fission products then decay by multiple emissions as a result of the high neutron-to-proton ratio possessed by these nuclides.

1/0 n + 235/92U ->236/92 U -> 88 / 38 Sr + 136/54Xe + 12/0n

ΔBE = BEproducts – BEreactamts

ΔBE = BEproducts – BEreactamts

ΔBE = (BESr-88 + BEXe-136 ) – (BEU-235)

ΔBE = (BESr-88 + BEXe-136 ) – (BEU-235)

ΔBE = (783.2MeV + 1156MeV ) – (1786 MeV)

= 153.2 MeV
The 12 n produced and the incident neutron are ignored in the energy equation. The comment "assume the initial KE of the neutron is very small" would apply to the incident neutron, which maybe assumed to have a thermal energy, or about 0.025 eV, as opposed to a fast neutron, which would have an energy on the order of 1 MeV.

The release of 12 neutrons would be extremely unlikely. The typical neutron production is between 2 or 3 neutrons per fission.

See - http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/fisfrag.html#c1

More likely, the fission reaction will produce fission product isotopes with masses of 97 and 137. Typically one observes Xe-133, 135 and 138, with the corresponding I isotopes as presursors. Xe-136 is a stable isotope of Xe.
 
  • #6
Clausius2
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Astronuc said:
The 12 n produced and the incident neutron are ignored in the energy equation. The comment "assume the initial KE of the neutron is very small" would apply to the incident neutron, which maybe assumed to have a thermal energy, or about 0.025 eV, as opposed to a fast neutron, which would have an energy on the order of 1 MeV.

The release of 12 neutrons would be extremely unlikely. The typical neutron production is between 2 or 3 neutrons per fission.

See - http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/fisfrag.html#c1

More likely, the fission reaction will produce fission product isotopes with masses of 97 and 137. Typically one observes Xe-133, 135 and 138, with the corresponding I isotopes as presursors. Xe-136 is a stable isotope of Xe.
Some stupid question, Astronuc:

would it be possible to generate Aurum from a fission reaction?
 
  • #7
Astronuc
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Clausius2 said:
Some stupid question, Astronuc:

would it be possible to generate Aurum from a fission reaction?
:rofl:

I think lots of folks would like to do that, I am sure.

However, [itex]_{79}^{197}Au[/itex] is the only stable isotope, and with such a heavy mass, the amount produced in a fission reaction is neglible - probably something like pico-grams per kgU or MgU.

One could employ an (n,[itex]\gamma[/itex]) reaction with [itex]_{80}^{196}Hg[/itex], whereby one must induce neutron ([itex]_{0}^{1}n[/itex]) capture in [itex]_{80}^{196}Hg[/itex] (which has low abundance (0.15%) of natural Hg). The resulting radionuclide [itex]_{80}^{197}Hg[/itex] transmutes (decays) by electron capture (EC) and becomes [itex]_{79}^{197}Au[/itex]. :biggrin: But then one needs a source of neutrons. Not only that, the gold atoms would also absorb neutrons becoming [itex]_{79}^{198}Au[/itex] which then ß-decay to [itex]_{80}^{198}Hg[/itex].

Another way to get Au-197 would be to induce neutron capture by [itex]_{78}^{196}Pt[/itex] and have it decay by ß-decay to [itex]_{79}^{197}Au[/itex]. Pt is more valuable than Au, so this would be counter-productive.

Nature has a way of teasing people. :biggrin:
 
  • #8
Clausius2
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Astronuc said:
:rofl:

I think lots of folks would like to do that, I am sure.

However, [itex]_{79}^{197}Au[/itex] is the only stable isotope, and with such a heavy mass, the amount produced in a fission reaction is neglible - probably something like pico-grams per kgU or MgU.

One could employ an (n,[itex]\gamma[/itex]) reaction with [itex]_{80}^{196}Hg[/itex], whereby one must induce neutron ([itex]_{0}^{1}n[/itex]) capture in [itex]_{80}^{196}Hg[/itex] (which has low abundance (0.15%) of natural Hg). The resulting radionuclide [itex]_{80}^{197}Hg[/itex] transmutes (decays) by electron capture (EC) and becomes [itex]_{79}^{197}Au[/itex]. :biggrin: But then one needs a source of neutrons. Not only that, the gold atoms would also absorb neutrons becoming [itex]_{79}^{198}Au[/itex] which then ß-decay to [itex]_{80}^{198}Hg[/itex].

Another way to get Au-197 would be to induce neutron capture by [itex]_{78}^{196}Pt[/itex] and have it decay by ß-decay to [itex]_{79}^{197}Au[/itex]. Pt is more valuable than Au, so this would be counter-productive.

Nature has a way of teasing people. :biggrin:
What a pity!... :cry:
I was hoping to buy a car with this invent.... :tongue2:
 

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