anonymous
  • anonymous
problem attached inside!
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
@Michele_Laino do you understand this one perhaps? :)
Michele_Laino
  • Michele_Laino
yes! That is nuclear physics

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anonymous
  • anonymous
yay:)
Michele_Laino
  • Michele_Laino
as we can see the atomic number is increased, so the number of protons is increased
Michele_Laino
  • Michele_Laino
Z is going from 88 to 89
Michele_Laino
  • Michele_Laino
Z is the atomic number or proton number
Michele_Laino
  • Michele_Laino
what is happened inside the nucleus of Ra?
Michele_Laino
  • Michele_Laino
here is the nuclear process occurred inside that nucleus:
Michele_Laino
  • Michele_Laino
\[\Large n \to p + {e^ - } + {\bar \nu _e}\]
anonymous
  • anonymous
i am not sure.... :/ so for the first part, the complete equation is what you wrote above? or is that just the formula?
Michele_Laino
  • Michele_Laino
namely a neutron is transformed itself in a proton plus an electron plus an anti-neutrinos
Michele_Laino
  • Michele_Laino
you have to write e^-, namely electron, and the radiation involved is the \[\Large {\beta ^ - }\] radiation
anonymous
  • anonymous
ooh okay!!
Michele_Laino
  • Michele_Laino
since \[\Large {\beta ^ + }\] radiation is associated to an emission of positrons
anonymous
  • anonymous
ahh okay:)
Michele_Laino
  • Michele_Laino
ok! :)
anonymous
  • anonymous
so the first part asks for the complete equation, right? what is that? :/
Michele_Laino
  • Michele_Laino
complete equation is:
Michele_Laino
  • Michele_Laino
\[\Large {}_{88}^{228}Ra \to {}_{ - 1}^0{e^ - } + {}_{89}^{228}Ac\]
anonymous
  • anonymous
ohhh okay! awesome! so the next part asks for type of radiation given off in this reaction? how can we find that?
Michele_Laino
  • Michele_Laino
radiation is \[\Large {\beta ^ - }\] radiation
anonymous
  • anonymous
what is that symbol called again? is it just radiation? I forgot :(
Michele_Laino
  • Michele_Laino
we can conclude that the radiation involved is the \[\Large {\beta ^ - }\] radistion, since in a nuclear reaction electric charge is conserved
anonymous
  • anonymous
it is this, corret? ß^- ?
Michele_Laino
  • Michele_Laino
this symbol: \[\Large {\beta ^ - }\] stands for "beta minus radistion"
anonymous
  • anonymous
ohhh okay :D
Michele_Laino
  • Michele_Laino
oops.. stands for "beta minus radiation"
anonymous
  • anonymous
and so the last part asks where the particle comes from and the process of this kind of decay? how do i find that?
anonymous
  • anonymous
:)
Michele_Laino
  • Michele_Laino
it is a classic nuclear reaction, you can find it into the most nuclear or particle physics textbooks. Here is my particle physics textbooks: \[\Large \begin{gathered} Title:{\text{Introduction To High Energy Physics}} \hfill \\ Author:{\text{D}}{\text{.H}}{\text{.Perkins}} \hfill \\ Publisher:{\text{Addison - Wesley}} \hfill \\ 3rd - Edition \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
oops..electronic anti-neutrino*
anonymous
  • anonymous
ahh okay!! :D and so that is where the particle comes from, right? and what about the decay?
Michele_Laino
  • Michele_Laino
the decay is the "beta-minus" decay
Michele_Laino
  • Michele_Laino
the decay type comes from the type of the emitted radiation from that decay
anonymous
  • anonymous
oh okay!! and so is that all for this problem? :O
Michele_Laino
  • Michele_Laino
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
anonymous
  • anonymous
oohh:O
Michele_Laino
  • Michele_Laino
yes! we have finished!
anonymous
  • anonymous
yay!! thank you very much!! :D
Michele_Laino
  • Michele_Laino
thank you!! :D

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