# Homework Help: Nuclear Decay - Semi Empirical Formula

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1. Apr 18, 2015

### unscientific

1. The problem statement, all variables and given/known data
(a) What processes changes atomic number by 1? What are the favourable conditions? How do you tell a neutrino is involved? How can we use this to understand the mass of this particle?
(b) Use semi empirical mass formula to explain why odd-odd isobars are unlikely, while even-even isobars are possible. Why is $Tc$ the lightest odd-odd isobar? Show by mass difference $Mo$ is stable.
(c) What is the other possible process and its $Q$ value?
(d) What background is suppressed and what other background exists? What's the significance of T=Q?

2. Relevant equations

3. The attempt at a solution
This question completely stumped me, as I have no idea how to proceed.

Part (a)
I suppose the two processes are neutron capturing: $_a ^b X + n \rightarrow _a ^{b+1} Y + \gamma$ and fission: $_a ^b Z + n \rightarrow _a ^{b-1} W + 2n + \gamma$.
I think favourable conditions are an odd-even nucleus? Not sure what other conditions there are.
When a neutrino is involved, I suppose we find leptons? How do we use leptonic decay to figure out the mass?

Part (b)
For an odd-odd configuration, $\delta_P < 0$ so lower binding energy compared to an even-even configuration where $\delta_P = 0$. Not sure why $(Z,N) = (43,57)$ is the lightest odd-odd isobar. What about $(41,59)$?
It seems that $Mo$ has a higher binding energy than $Tc$, so I suppose it is more stable.

Not sure for the other parts of this question..

2. Apr 18, 2015

### SteamKing

Staff Emeritus
For part a), it's the atomic number which changes. Which particle determines the atomic number of an atom?

Does fission change the atomic number by 1? Or does fission do something entirely different to a nucleus?

3. Apr 18, 2015

### unscientific

Ok, the atomic number is the proton number. So instead of a neutron it would be a proton then.

Fission actually breaks the nucleus into many parts + some other neutrons, so I don't think it will be applicable in this case.

4. Apr 18, 2015

### SteamKing

Staff Emeritus
Is capturing a proton the only process which increases the atomic number of an atom? Think about the different types of radioactivity (there are three types).

5. Apr 18, 2015

### unscientific

Ok, I can think of 3 radiations:
Beta decay: $_a X~ \rightarrow~ e^- + _{a+1}Y$
Alpha decay: Won't this change the atomic number by $2$?
Gamma decay: $_a X~ \rightarrow~ _{a+1}Y + e^- + \nu_e + \gamma$.

6. Apr 18, 2015

### SteamKing

Staff Emeritus
Does gamma emission always lead to a change in atomic number? What about beta decay?

7. Apr 18, 2015

### unscientific

Gamma emission doesn't always, but beta decay has to (conservation of charge).

8. Apr 18, 2015

### SteamKing

Staff Emeritus
Can you answer question a) now?

9. Apr 18, 2015

### unscientific

What are the favourable conditions? How do we tell if a neutrino is involved? Even if a neutrino is involved, I can't see how that can be used to figure out the mass of the reactants...

10. Apr 18, 2015

### SteamKing

Staff Emeritus

The question is not asking for the mass of the reactants ...

11. Apr 18, 2015

### unscientific

I suppose one of the experimental signature that the sum of the partial widths is less than it would be a peak/resonance in the cross-section where $E=m_Z$ due to the production of the $Z^0$ boson. The "missing" or "invisible" width belongs to the neutrino?

12. Apr 19, 2015

### unscientific

I suppose the mass of the neutrino can be calculated from the width $\Gamma_{\nu}$?

13. Apr 19, 2015

### SteamKing

Staff Emeritus
The question is a bit vague. IMO, "understand" is not the same as "calculate", w.r.t. the mass of the neutrino.

I think the question here is looking for a qualitative estimate of whether neutrinos are relatively massive particles, or if there is not much mass to one ...

14. Apr 19, 2015

### unscientific

True, I read it has a very low but non-zero mass.

For part (b), why is the $Tc$ the lightest isobar? Why can't $Z=41, N=59$ be formed?

15. Apr 19, 2015

### SteamKing

Staff Emeritus
I'm afraid my knowledge of atomic physics is pretty much exhausted after being able to tell the difference between atomic number and atomic mass ...

16. Apr 20, 2015

### unscientific

No problem, thanks alot for spending time on this problem.

Would appreciate help from anyone else who have done atomic physics!

17. Apr 22, 2015

### unscientific

bumpp

18. Apr 25, 2015

### unscientific

bumpp

19. Apr 30, 2015

### unscientific

bumpp

20. May 4, 2015

### unscientific

bumpp would appreciate any help to start

21. May 7, 2015

### unscientific

bumpp

22. May 10, 2015

### unscientific

bumppp

23. May 14, 2015

### unscientific

bumpp

24. May 14, 2015

### Brage

Think of conservation rules, I.e. total lepton number and specific lepton number are always conserved in interactions!
a) Beta pluss and minus decay as you said is the only really viable option, (you can have emission of a proton but this is very rare so i doubt it is what the question asks for). So if we have then dont have a neutrino present we would have a system with $\Delta L_e = \pm 1$. Consider also if decay of a neutron or proton would result in higher binding energy, and how this relates to the most favorable condition.
b)Look at the last term in the semi emperical mass formula (SEMP), and remember the less mass you have the more energy you need to break the nucleus appart
c)The answer is hinted to in part d, what do we look for in very deep mines? specifically the interaction of what particle?
d)Basically they are asking which particles will be blocked out by a LOT of matter, and what can go pretty much undesterbed through matter? Also if T=Q then all the energy released from a reaction is in the kinetic energy of 1 particle.. so if we know the kinetic energy of the emitted charged particles (ofc you need to figure out what particles are emitted but this is related to part c), what does this tell us about energy of the incoming particle?

This should be enough to help you out with answering the questions, soz it took so long for you to get a reply.