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anonymous
 one year ago
problem attached inside!
anonymous
 one year ago
problem attached inside!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino do you understand this one perhaps? :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! That is nuclear physics

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2as we can see the atomic number is increased, so the number of protons is increased

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2Z is going from 88 to 89

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2Z is the atomic number or proton number

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2what is happened inside the nucleus of Ra?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here is the nuclear process occurred inside that nucleus:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large n \to p + {e^  } + {\bar \nu _e}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am not sure.... :/ so for the first part, the complete equation is what you wrote above? or is that just the formula?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2namely a neutron is transformed itself in a proton plus an electron plus an antineutrinos

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2you have to write e^, namely electron, and the radiation involved is the \[\Large {\beta ^  }\] radiation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2since \[\Large {\beta ^ + }\] radiation is associated to an emission of positrons

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the first part asks for the complete equation, right? what is that? :/

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2complete equation is:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large {}_{88}^{228}Ra \to {}_{  1}^0{e^  } + {}_{89}^{228}Ac\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhh okay! awesome! so the next part asks for type of radiation given off in this reaction? how can we find that?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2radiation is \[\Large {\beta ^  }\] radiation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is that symbol called again? is it just radiation? I forgot :(

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we can conclude that the radiation involved is the \[\Large {\beta ^  }\] radistion, since in a nuclear reaction electric charge is conserved

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is this, corret? ß^ ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2this symbol: \[\Large {\beta ^  }\] stands for "beta minus radistion"

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2oops.. stands for "beta minus radiation"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and so the last part asks where the particle comes from and the process of this kind of decay? how do i find that?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2it 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
 one year ago
Best ResponseYou've already chosen the best response.2here is the involved nuclear reaction: \[\Large n \to p + {e^  } + {{\bar \nu }_e}\] n=neutron p= proton \[{{\bar \nu }_e}\] = electronic antineutrinos \[{e^  }\] is the electron

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2oops..electronic antineutrino*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ahh okay!! :D and so that is where the particle comes from, right? and what about the decay?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the decay is the "betaminus" decay

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the decay type comes from the type of the emitted radiation from that decay

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay!! and so is that all for this problem? :O

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! we have finished!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay!! thank you very much!! :D
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