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anonymous

  • one year ago

problem attached inside!

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Michele_Laino do you understand this one perhaps? :)

  3. Michele_Laino
    • one year ago
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    yes! That is nuclear physics

  4. anonymous
    • one year ago
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    yay:)

  5. Michele_Laino
    • one year ago
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    as we can see the atomic number is increased, so the number of protons is increased

  6. Michele_Laino
    • one year ago
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    Z is going from 88 to 89

  7. Michele_Laino
    • one year ago
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    Z is the atomic number or proton number

  8. Michele_Laino
    • one year ago
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    what is happened inside the nucleus of Ra?

  9. Michele_Laino
    • one year ago
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    here is the nuclear process occurred inside that nucleus:

  10. Michele_Laino
    • one year ago
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    \[\Large n \to p + {e^ - } + {\bar \nu _e}\]

  11. anonymous
    • one year ago
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    i am not sure.... :/ so for the first part, the complete equation is what you wrote above? or is that just the formula?

  12. Michele_Laino
    • one year ago
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    namely a neutron is transformed itself in a proton plus an electron plus an anti-neutrinos

  13. Michele_Laino
    • one year ago
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    you have to write e^-, namely electron, and the radiation involved is the \[\Large {\beta ^ - }\] radiation

  14. anonymous
    • one year ago
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    ooh okay!!

  15. Michele_Laino
    • one year ago
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    since \[\Large {\beta ^ + }\] radiation is associated to an emission of positrons

  16. anonymous
    • one year ago
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    ahh okay:)

  17. Michele_Laino
    • one year ago
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    ok! :)

  18. anonymous
    • one year ago
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    so the first part asks for the complete equation, right? what is that? :/

  19. Michele_Laino
    • one year ago
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    complete equation is:

  20. Michele_Laino
    • one year ago
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    \[\Large {}_{88}^{228}Ra \to {}_{ - 1}^0{e^ - } + {}_{89}^{228}Ac\]

  21. anonymous
    • one year ago
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    ohhh okay! awesome! so the next part asks for type of radiation given off in this reaction? how can we find that?

  22. Michele_Laino
    • one year ago
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    radiation is \[\Large {\beta ^ - }\] radiation

  23. anonymous
    • one year ago
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    what is that symbol called again? is it just radiation? I forgot :(

  24. Michele_Laino
    • one year ago
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    we can conclude that the radiation involved is the \[\Large {\beta ^ - }\] radistion, since in a nuclear reaction electric charge is conserved

  25. anonymous
    • one year ago
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    it is this, corret? ß^- ?

  26. Michele_Laino
    • one year ago
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    this symbol: \[\Large {\beta ^ - }\] stands for "beta minus radistion"

  27. anonymous
    • one year ago
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    ohhh okay :D

  28. Michele_Laino
    • one year ago
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    oops.. stands for "beta minus radiation"

  29. anonymous
    • one year ago
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    and so the last part asks where the particle comes from and the process of this kind of decay? how do i find that?

  30. anonymous
    • one year ago
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    :)

  31. Michele_Laino
    • one year ago
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    it is a classic nuclear reaction, you can find it into the most nuclear or particle physics textbooks. Here is my particle physics textbooks: \[\Large \begin{gathered} Title:{\text{Introduction To High Energy Physics}} \hfill \\ Author:{\text{D}}{\text{.H}}{\text{.Perkins}} \hfill \\ Publisher:{\text{Addison - Wesley}} \hfill \\ 3rd - Edition \hfill \\ \end{gathered} \]

  32. Michele_Laino
    • one year ago
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    here is the involved nuclear reaction: \[\Large n \to p + {e^ - } + {{\bar \nu }_e}\] n=neutron p= proton \[{{\bar \nu }_e}\] = electronic anti-neutrinos \[{e^ - }\] is the electron

  33. Michele_Laino
    • one year ago
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    oops..electronic anti-neutrino*

  34. anonymous
    • one year ago
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    ahh okay!! :D and so that is where the particle comes from, right? and what about the decay?

  35. Michele_Laino
    • one year ago
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    the decay is the "beta-minus" decay

  36. Michele_Laino
    • one year ago
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    the decay type comes from the type of the emitted radiation from that decay

  37. anonymous
    • one year ago
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    oh okay!! and so is that all for this problem? :O

  38. Michele_Laino
    • one year ago
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    we have many types of decays, namely gamma decay, if there is an emission of photons, alpha decay, if there is an emission of alpha particles

  39. anonymous
    • one year ago
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    oohh:O

  40. Michele_Laino
    • one year ago
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    yes! we have finished!

  41. anonymous
    • one year ago
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    yay!! thank you very much!! :D

  42. Michele_Laino
    • one year ago
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    thank you!! :D

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