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Antimatter bomb vs tsar bomb

  1. Jan 7, 2009 #1
    This is hypothetical because I know that we do not have the ability to create or store antimatter. The tsar Bomb had a yeild of 50 megatons is it true that a bomb made of 1 kg of antimatter and 1kg of matter could yeild around the same (42 Mt)? If this could be achieved would this be considered a "clean bomb"?
  2. jcsd
  3. Jan 8, 2009 #2


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    1 Mt is 4.1 x10^15 J so the Tsar bomb (in it's tested config) at 42MT = 1.75 x10^17 J
    2 kg of E=mc^2 = 1.79 x10^17 J So yes pretty similair.

    It wouldn't necessarily be clean, the intense x-ray emission would irradiate anything around it.
    Last edited: Jan 8, 2009
  4. Jan 8, 2009 #3
    Wow. In day to day life, we should probably think about existing nuclear weapons more often than we do.
  5. Jan 8, 2009 #4
    thank you mgb phys that was exactly what I was looking for
  6. Jan 8, 2009 #5
    I didn't cheked if the calculations above are corect,but I saw at discovery that a teaspoon of antimatter is equivalent to 100 hidrogen bombs...of course,I realy don't trust discovery anymore...in a magasine they stated that at - 30 kelvin degrees(yes,with the simbol of degrees) life can be barely posible...first of all,there's no such thing as negative temperature on kelvin scale,second,degrees are only used on celsius or faranheit(don't know the exact riting) and mabie some more,but you don't use degrees on kelvins...so discovery(my sourse) is not that shore...
    also,we can create antimatter(you to can have 1 gram of it if you can pay for it...just 30 trilion us dollars:) or 60 trilion,I'm not shore)
    so the scientists have very phew of it...also you can store it in very powerfull electromagnetic fields
  7. Jan 8, 2009 #6


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    Who say's we cannot create or store antimatter?

    The thing is that even 1gram of it would take million of years to create..
  8. Jan 8, 2009 #7


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    The calculation isn't difficult and the equation is rather well known ( e=mc2 )
    One teaspoon of matter (5g) is 4.5x1015 J,

    If you were exploding a teaspoon of anti-matter the total energy would be twice this since you are also using up a teaspoon of matter.

    One gram of TNT is 4184 J
    So 1MegaTon (10 12g ) is 4.148 x 1015 J

    Depends how big your hydrogen bomb is of course, biggest is the Tsar device at 50Mt. A typical modern fusion weapon is 100-200Kt so a teaspoon of antimatter would be 10-20 of these.
    It would be about 100x the yield of a Hiroshima (fission) bomb - perhaps this is what the TV program was thinking of?
    Last edited: Jan 8, 2009
  9. Jan 8, 2009 #8

    How much antimatter would be required to trigger a fission reaction?
  10. Jan 8, 2009 #9


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    Not sure that is really a meaningfull question.
    To trigger a fission reaction in say U235 you only need a single low speed neutron with an energy of 0.5ev.
    So if you are asking how much anti-matter do you need to accelerate a neutron to 0.5ev.
    Not very much at all, a fraction of a neutrino's worth
  11. Jan 8, 2009 #10
    super-deformation spontaneous fission...

    More theoretical, firing an anti-proton or anti-neutron at a isotope nucleus, transforming the isotope into a super-deformation spontaneous fission nucleus.

    The lightest spontaneous fission reaction:
    [tex]\boxed{^9 _4 \text{Be} + ^1 _0 \overline{\text{n}} \rightarrow ^8 _4 \text{Be} ( 3 \cdot 10^{-16} \; \text{s} ) + \text{E} \rightarrow 2 [^4 _2 \text{He} (0.09 \; \text{Mev})]}[/tex]

    Mathematically, the criterion for whether spontaneous fission can occur is approximately:
    [tex]\boxed{\frac{Z^2}{A} \ge \frac{92^2}{235.0439299} \geq 36.0103}[/tex]

    Where Z is the atomic number and A is the mass number (e.g., 235 for U-235).

    [tex]\boxed{^{236} _{92} \text{U} + ^1 _0 \overline{\text{n}} \rightarrow ^{235} _{92} \text{U}( \text{super-deformation} ) + \text{E} \rightarrow \text{spontaneous fission}}[/tex]

    Spontaneous fission - Wikipedia
    Superdeformation - Wikipedia
    Nuclear Isotope Half-lifes
    Tsar bomba - computer simulation
    Last edited: Jan 8, 2009
  12. Jan 8, 2009 #11
    You don't get 42 MT of useful energy on of a kilogram of antimatter, because more than half of that is carried away by neutrinos.

    It would not be perfectly clean, but the amount of radioactive waste per megaton of energy would likely be much smaller than in a regular nuke.
  13. Jan 9, 2009 #12
    no,it said hidrogen bombs...but again,that TV chanel has becomed a very bad one...numbers don't lie,so yes,you are right...
  14. Jan 10, 2009 #13
    antimatter has definitely been created, but creating substantial amounts of it is difficult.
  15. Jan 10, 2009 #14


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    How did you get your number of "more than half" ?
  16. Jan 10, 2009 #15
    I read it somewhere a while ago, I can't find the source now. It's not hard to understand why it's possible. Contrary to public opinion, main direct product of a pp or a nn annihilation is not a couple of 938 MeV photons, but rather a large number of various pions. Neutral pions decay into two gammas, and charged pions ultimately decay into an electron/positron and three neutrinos. Unless you have a lot of "stuff" around the point of explosion to decelerate pions and intermediate muons via brehmsstrahlung, considerable part of the overall energy will be in neutrinos.
  17. Jan 11, 2009 #16
    You need to go back and review your particle physics notes and take a look at the PDG.

    Charged pions decay to muon and appropriate neutrino 99.99% of the time. The decay into an electron (or positron) is about 6 orders of magnitude less likely. And why would you get 2 extra neutrinos in that decay?

    And do you have a reference to the p-antip -> multiple pion channel? Because it seems to me the cross section for direct annihilation (or mediated by a single virtual neutral pion) should be much more likely then the creation of multiple pions and their off products.
  18. Jan 11, 2009 #17
    You get 2 extra neutrinos because the muon decays into an electron and 2 neutrinos. :)

    According to this article, p-antip annihilation results, on average, in 3 charged and 2 neutral pions.

  19. Jan 11, 2009 #18
    I read over your "ultimately" qualifier in your original statement. The 3 neutrinos made me wonder if that was what you meant, but I was a little taken aback. :)

    Re: the pion multiplicity in the p-antip annihilation, I did find one paper that called into question the average of 5 statement. You can find it here: http://arxiv.org/abs/hep-ph/9409343.

    Either way, thanks for indulging me.
  20. Jan 9, 2010 #19
    antimatter is posible

    Anti-protons, can be obtained in small quantities from high-energy accelerators slamming particles into solid targets. They are then collected and held in a magnetic bottle.
    The Bottle has liquid nitrogen and helium to keep about a trillion anti-protons this is about one nanogram.
  21. Jan 9, 2010 #20
    All muons decay into an electron (or positron) and two neutrinos within a few microseconds, so including the neutrino associated with charged pion decay, there are 3. See
    p-bar p annihilation is a strong interaction process, and the most likely channel is lots of pions (maybe ~6 total), probably 1/3 of which are neutral pions, and 2/3 are charged.
    Bob S
    [added] Se Fig.1 in
    The average number is ~5 pions per annihilation, not ~6.
    Last edited: Jan 9, 2010
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