problem attached inside!

- anonymous

problem attached inside!

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- anonymous

##### 1 Attachment

- anonymous

@Michele_Laino do you understand this one perhaps? :)

- Michele_Laino

yes! That is nuclear physics

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## More answers

- anonymous

yay:)

- Michele_Laino

as we can see the atomic number is increased, so the number of protons is increased

- Michele_Laino

Z is going from 88 to 89

- Michele_Laino

Z is the atomic number or proton number

- Michele_Laino

what is happened inside the nucleus of Ra?

- Michele_Laino

here is the nuclear process occurred inside that nucleus:

- Michele_Laino

\[\Large n \to p + {e^ - } + {\bar \nu _e}\]

- 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

namely a neutron is transformed itself in a proton plus an electron plus an anti-neutrinos

- Michele_Laino

you have to write e^-, namely electron, and the radiation involved is the
\[\Large {\beta ^ - }\] radiation

- anonymous

ooh okay!!

- Michele_Laino

since \[\Large {\beta ^ + }\] radiation is associated to an emission of positrons

- anonymous

ahh okay:)

- Michele_Laino

ok! :)

- anonymous

so the first part asks for the complete equation, right? what is that? :/

- Michele_Laino

complete equation is:

- Michele_Laino

\[\Large {}_{88}^{228}Ra \to {}_{ - 1}^0{e^ - } + {}_{89}^{228}Ac\]

- 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

radiation is \[\Large {\beta ^ - }\] radiation

- anonymous

what is that symbol called again? is it just radiation? I forgot :(

- Michele_Laino

we can conclude that the radiation involved is the \[\Large {\beta ^ - }\] radistion, since in a nuclear reaction electric charge is conserved

- anonymous

it is this, corret? ß^- ?

- Michele_Laino

this symbol:
\[\Large {\beta ^ - }\] stands for "beta minus radistion"

- anonymous

ohhh okay :D

- Michele_Laino

oops.. stands for "beta minus radiation"

- 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

:)

- 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

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

oops..electronic anti-neutrino*

- anonymous

ahh okay!! :D and so that is where the particle comes from, right? and what about the decay?

- Michele_Laino

the decay is the "beta-minus" decay

- Michele_Laino

the decay type comes from the type of the emitted radiation from that decay

- anonymous

oh okay!! and so is that all for this problem? :O

- 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

oohh:O

- Michele_Laino

yes! we have finished!

- anonymous

yay!! thank you very much!! :D

- Michele_Laino

thank you!! :D

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