Nuclear Reaction: 33Cl + 2n -> 15P + 2He

[FrostScYthe]

I don't know how to solve this problem... well here it goes...

33Cl(n,?)31p

this is what I did...

33 (1) 31 4
17 Cl + 2(0)n -----> 15 P + 2 He

Is this right ?

 

[Monique]

Hi FrostScYthe, did you find the correct answer yet?

I am not an expert, but I tried tackling the problem
If I remember correctly, an atom is represented as follows:

n+p p
n..X

I am not sure what you mean by the part in your equation between the + and the arrow, but this is the basic idea:

33...16....31..16
17 Cl...(neutron,?) 15 P + ?

It is clear that two protons are released from the Cl.
No neutrons are release, but a neutron is put into the reaction.
Another particle can be put into the reaction.

If the reaction is (n,n) you get a Helium atom (2p,2n) but I am not sure if that is allowed since my knowledge doesn't reach that far.
Maybe it is (n,gamma) where you've got something with (2p,n) at the other end of the reaction..

Let me know what you think and explain what is meant with your solution.

 

[Monique]

Ah! Wait!

I figured out your reaction, the upper line is a little hard to read and confused me. It seem we both came to the same conclusion, it is (n,n) and a stable He is produced.

 

[anuradha mitra]

33 1 31 4
Cl + 2 n = P He
17 0 15 2
so, here He is liberated as an ejected particle