# Homework Help: Nuclear physics problem - energy of reaction

1. Feb 24, 2014

### Saitama

1. The problem statement, all variables and given/known data
To activate the reaction $(n,\alpha)$ with stationary $B^{11}$ nuclei, neutrons must have the activation kinetic energy $T_{th}=4\,MeV$. $(n,\alpha)$ means that $n$ is bombarded to obtain $\alpha$. Find the energy of this reaction.

(Ans: $Q=-11/12 T_{th}$)

2. Relevant equations

3. The attempt at a solution
The reaction taking place is:
$$n+B^{11}\rightarrow B^{12}\rightarrow A^{8}+\alpha$$
From conservation of energy
$$T_{th}=Q+T_A+T_{\alpha}\,\,\,\,\,(*)$$
where $T$ is used to denote the kinetic energy.
Also, from conservation of linear momentum, $p_{A}=p_{\alpha} \Rightarrow p_{A}^2=p_{\alpha}^2 \Rightarrow m_AT_A=m_{\alpha}T_{\alpha}\,\,\,(**)$.

From (*) and (**), I get:
$$T_{th}=Q+T_A\left(1+\frac{m_A}{m_{\alpha}}\right)$$
I still need one more equation.

Any help is appreciated. Thanks!

2. Feb 24, 2014

### BvU

You have 4 MeV plus the rest masses of the reactants on one side, and something comparable for the products on the other.

3. Feb 24, 2014

### Saitama

Sorry but I can't understand where are you trying to hint at.

4. Feb 24, 2014

### BvU

When I look up the masses of the reactants and the products, I get a nice imbalance. What happens with the 'disappearing' mass ?

5. Feb 24, 2014

### Saitama

Are you talking about the mass defect? But I don't have the nuclear masses of A and B, if I had the nuclear masses, won't it be trivial to find Q?

Last edited: Feb 24, 2014
6. Feb 25, 2014

### BvU

You have a point there. Didn't work for me either, anyway: I ended up with -5.6 MeV/c2 .

Other question: what about the initial momentum of the neutron ? Your conservation is only valid in the center of mass.

7. Feb 25, 2014

### voko

What is A here? B is boron, Z number = 5. Upon emitting the alpha particle, the Z number must be 3, which makes the element Lithium. Lithium-8 is unstable, with its half life less than 1 second.

I do not even think there is an element whose symbol is just A.

Secondly, should you not assume that the new element, whatever it is, is stationary?

8. Feb 25, 2014

### TSny

I wonder if it could be that they are asking for the "activation energy of the reaction" which is different from the "activation kinetic energy of the neutron".

When the neutron first enters the B11 nucleus, you get an excited B12 nucleus. Only a portion of the initial KE of the neutron goes into excitation energy, the rest is wasted as recoil KE of the B12. The activation energy of the reaction is the minimum excitation energy that will lead to the final reaction to Li8 and the alpha particle.

9. Feb 25, 2014

### Saitama

The new particle $B^{12}$ won't be stationary. But if I were to consider its motion in linear momentum conservation, I don't know about the motion of new particles i.e the question doesn't state in which direction the two new particles fly off.

And yes, I know its boron and lithium but I don't think its necessary to specify them for the given problem.

10. Feb 25, 2014

### TSny

Yes. That's why only some of the kinetic energy of the neutron is effective in initiating the reaction.

Not sure why you are worried about this. I think all you need to consider is how much of the kinetic energy of the neutron is converted into "internal energy" as the B11 nucleus absorbs the neutron to momentarily become a B12 nucleus.

You might want to imagine the collision in the center of mass frame.

OK. Basically all you need to know is that the target nucleus has 11 nucleons (if I am interpreting the problem correctly).

11. Feb 25, 2014

### Saitama

Do I have to find this internal energy? If so, how?

Are you talking about the conservation of linear momentum?

12. Feb 25, 2014

### TSny

Recall your happy days with collisions in elementary mechanics. If a block of mass m collides with a block at rest of mass M so that they become one block (at least momentarily), what fraction of the initial KE remains? What fraction of the initial KE is converted to some other form of energy of the two-block system?

Momentum is conserved in both the lab frame and the center of mass frame. But going to the center of mass frame makes it clear how much energy gets converted into internal energy. The center of mass frame is also the frame in which the total momentum is zero.

13. Feb 25, 2014

### Saitama

Since the momentum is conserved, I can write: $p_n=p_{B^{12}} \Rightarrow m_nT_{th}=12m_nT_{B^{12}}$
$$\Rightarrow T_{B^{12}}=\frac{T_{th}}{12}$$
This is the kinetic energy of $B^{12}$ nucleus. The energy lost is $(11/12)T_{th}$. Though this matches with the given answer but I don't see why this should the answer.

14. Feb 25, 2014

### TSny

I'm not positive I'm interpreting the problem correctly. But think of it this way. Suppose you had a block of mass m that is going to collide with a block at rest of mass 11m that has a spring as shown. Suppose that if the spring becomes fully compressed, the system explodes. Thus, there is a certain "activation energy" that the spring must absorb before we get an explosion ("reaction").

If you are told that the minimum kinetic energy that the small block must have initially to get an explosion is To, then you can work out the "activation energy" for the reaction.

#### Attached Files:

• ###### Collision.png
File size:
722 bytes
Views:
137
15. Feb 25, 2014

### Saitama

This "activation energy" is the potential energy stored in the spring, right?

But I really don't see how to carry this analogy in the given problem. I have the energy lost and final energy of the B^(12) nucleus but what to do with them.

16. Feb 25, 2014

### TSny

Yes.

If the initial KE of the neutron is below 4 MeV, then the specified reaction doesn't occur. This is because not enough energy is transferred into "internal energy" (analogy: potential energy of the spring). My interpretation of the question is that they are just asking for the minimum value of the increase in internal energy in order for the reaction to "go".

It is very much like activation energy for some chemical reactions. There must be a certain input of energy before the reaction will go. This is different from the energy released in the reaction. Two different chemical reactions could have the same activation energy, but release very different amounts of energy.

17. Feb 25, 2014

### TSny

Where does the "energy lost" go? Where did it go in the block analogy?

18. Feb 25, 2014

### Saitama

If there were no spring, then in an inelastic collision, energy wasted gets converted in heat or some other form of energy, I guess in the spring case, it gets converted into the potential energy of spring?

19. Feb 25, 2014

### TSny

Yes. Anyway, my interpretation is that they want this "spring" energy.

20. Feb 25, 2014

### Saitama

Thanks a lot TSny!

21. Feb 25, 2014

### BvU

Is everybody happy now with the original question to "find the energy of the reaction" being rephrased to: "find the activation energy of the reaction" ?

It would certainly make it possible to come near the "(Ans: $Q=-{11\over 12} \ T_{th}$)".
In that case: why is it considered negative ?

for 11B + n I have 11194.67 MeV/c2. Throw in 4 MeV and there is more than enough to make 12B (11191.3 MeV/c2). So this guy might be in some excited state or whatever.

For 8Li + $\alpha$ I have 11200.28 MeV/c2. Just the rest masses. Can't happen.

12B has a half-life of 20 ms. 98.4% $\beta$ decay to 12C, but also 1.6% $\beta , \alpha$ decay to 8Be. Much nicer: leaves some 9.88 MeV/c2 (not yet including the $4\times {11\over 12}$ MeV) for the fragments to fly off.

My current view:
1) They are asking for the net activation energy. (but: why the minus sign?)
2) It's about $\beta , \alpha$ decay to Be

Is there a nuclear wizard who can corroborate that (or shoot it off?)

22. Feb 26, 2014

### TSny

I think the minus sign is to indicate energy "taken in" rather than released.

Your analysis of the masses is nice. It does appear that Li-8 and an alpha particle cannot result if the neutron only has 4 Mev of kinetic energy.