## anonymous one year ago problem attached inside!

1. anonymous

2. anonymous

@Michele_Laino do you understand this one perhaps? :)

3. Michele_Laino

yes! That is nuclear physics

4. anonymous

yay:)

5. Michele_Laino

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

6. Michele_Laino

Z is going from 88 to 89

7. Michele_Laino

Z is the atomic number or proton number

8. Michele_Laino

what is happened inside the nucleus of Ra?

9. Michele_Laino

here is the nuclear process occurred inside that nucleus:

10. Michele_Laino

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

11. anonymous

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

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

13. Michele_Laino

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

14. anonymous

ooh okay!!

15. Michele_Laino

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

16. anonymous

ahh okay:)

17. Michele_Laino

ok! :)

18. anonymous

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

19. Michele_Laino

complete equation is:

20. Michele_Laino

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

21. anonymous

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

radiation is $\Large {\beta ^ - }$ radiation

23. anonymous

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

24. Michele_Laino

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

25. anonymous

it is this, corret? ß^- ?

26. Michele_Laino

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

27. anonymous

ohhh okay :D

28. Michele_Laino

oops.. stands for "beta minus radiation"

29. 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?

30. anonymous

:)

31. 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: $\Large \begin{gathered} Title:{\text{Introduction To High Energy Physics}} \hfill \\ Author:{\text{D}}{\text{.H}}{\text{.Perkins}} \hfill \\ Publisher:{\text{Addison - Wesley}} \hfill \\ 3rd - Edition \hfill \\ \end{gathered}$

32. 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

33. Michele_Laino

oops..electronic anti-neutrino*

34. anonymous

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

35. Michele_Laino

the decay is the "beta-minus" decay

36. Michele_Laino

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

37. anonymous

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

38. 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

39. anonymous

oohh:O

40. Michele_Laino

yes! we have finished!

41. anonymous

yay!! thank you very much!! :D

42. Michele_Laino

thank you!! :D