Negative Q-value for 7Li(alpha,n) reaction

[eneacasucci]

I see that the Q value for the reaction 7Li + α -> 10B + neutron
is Q=-2789.909 keV, which is a negative value. I knew that Q<0 means that the reaction is not spontaneous.

The "Glenn F. Knoll - Radiation detection and Measurement" says: "In particular, the 7Li (α, n) reaction with its highly negative Q-value leads to a neutron spectrum with a low 0.5 MeV average energy that is especially useful in some applications." talking about neutron sources with Am241. My question is: how is it possible for the reaction to happen given the Q<0? because for example the reaction with 9Be and α has a Q>0 and is used for neutron source too.

where does the energy come from to overcome the negative Q-value and make the 7Li reaction happen?

 

[mfb]

From the kinetic energy of the reaction partners, generally the alpha particle. Americium-241 has a decay energy of 5.5 MeV, almost all of that goes into alpha particle.

 

[Vanadium 50]

It may be helpful to know that all alphas are around 5 MeV or so, irrespective of source.

 

[eneacasucci]

Thank you for both answers, what I am not able to understand then is what the Q-value represents, I know it can be defined in this way ##Q=K_f-K_i = (m_i - m_f)c^2##, and that given the reaction
7Li + α -> 10B + neutron
the Q-value associated is a fixed value, of course I'd say, since the masses are fixed.
The Q-value is -2789.909 keV (ref. https://www.nndc.bnl.gov/qcalc/ )
But how is it calculated considering the difference in kinetic energies? I mean, what is the physical system in which the Q value is -2789.909 keV? (alpha and Li at rest with respect to the reference system?)
I got the point that in the particular case of Am-241 decay, the alpha has nearly 5,5 MeV that compensate the negative Q-value giving the total energy balance a positive value, this situation is the one as "a reaction with a negative Q value is endothermic, i.e. requires a net energy input".

 

[Vanadium 50]

The Q value is defined for everything at rest. How can it be otherwise, as you would need a different one for every source of reactants.

 

[eneacasucci]

The Q value is defined for everything at rest. How can it be otherwise, as you would need a different one for every source of reactants.

That's what I thought because the Q value is fixed for the particular reaction, but how can be "everything at rest"? If everything is at rest the Q-value should be 0 since it is also defined as ##Q=K_f-K_i## where K is the kinetic energy of the initial particles and final particles, and if everything is at rest ##K_f=K_i=0##

(for the Q-value I'm referring to the def. found here https://en.wikipedia.org/wiki/Q_value_(nuclear_science) )

 

[Vanadium 50]

but how can be "everything at rest"

Officer, how can I be going 60 mph when I have only been driving for 10 minutes?

 

[eneacasucci]

Officer, how can I be going 60 mph when I have only been driving for 10 minutes?

I would be grateful if you could answer my question more clearly because I cannot understand this answer

 

[Vanadium 50]

Reporting speeds in miles per hour is a convention. It doesn't mean you have to be driving for an hour. Reporting pressure in pounds per square inch is a convention. It doesn't mean you have exactly one square inch.

Reporting Q as the mass differences - i.e. not assuming any particular value of kinetic energy - is a convention. A sensible one, to be sure, but still a convention.

 

[mfb]

If both initial particles are at rest, then you are missing the Q value as energy for the reaction and it cannot happen. If your initial particles have that much kinetic energy in the center of mass frame, then you have exactly enough energy for the reaction. If they have more than that, then they can react and there will be some extra energy - which can become kinetic energy of the reaction products, but you can also have some energy emitted as a photon for example. Saying the Q value is the difference in kinetic energies doesn't work for that reason.

 

[eneacasucci]

Saying the Q value is the difference in kinetic energies doesn't work for that reason.

Thank you. The first part of your answer is very clear and understandable to me (it makes me think of the Q-value, if it is negative, as a potential step to be overcome).

I'd like to dig deeper in the meaning of Q-value, to be sure I understood it.
Let's write the definition as it appears for example in Wikipedia:

Given the simple reaction
##a+b \rightarrow c+d##
the law of conservation of energy can be expressed as
##m_a c^2 + K_a + m_b c^2 + K_b = m_c c^2 +K_c + m_d c^2 + K_d##
where K is the kinetic energy and m is the rest mass. Rewriting this relationship in the form
##(m_a + m_b - m_c - m_d) c^2 = K_c + K_d - K_a - K_b##
then the Q-value is defined as
##Q= (m_a + m_b - m_c - m_d) c^2=K_c + K_d - K_a - K_b##

Now, I'd say that ##(m_a + m_b - m_c - m_d) c^2## is a constat value and so should be ##K_c + K_d - K_a - K_b## .
If we provide more energy to the initial system (considering that all the energy provided goes as kinetic energy) then the same amount of energy should go in the K energy of the products, to comply with the conservation of energy.
What should we conclude? That the Q-value i.e. ##K_c + K_d - K_a - K_b## is constant, but of course ##K_a## and ##K_b## can vary, and, in the case of Q<0, if ##K_a+K_b>Q## then the reaction can occur? (I don't know... I was just trying to make an effort to answer, sorry If i write stupid things)

 

[mfb]

You need ##K_a + K_b > Q## in the center of mass frame. Having it in a different frame is not sufficient.

 

[eneacasucci]

You need ##K_a + K_b > Q## in the center of mass frame. Having it in a different frame is not sufficient.

Ok I was missing this point. Now the meaning of Q-value is clearer I guess, thank you so much!

 

[eneacasucci]

You need ##K_a + K_b > Q## in the center of mass frame. Having it in a different frame is not sufficient.

The Q value is constant and the same also when I have the condition ##K_a + K_b > Q## (in the center of mass frame), right? (due to the conservation of energy)

 

[eneacasucci]

I also don't understand why I have this energy threshold:

(ref: https://www.nndc.bnl.gov/qcalc/ )

 

[mfb]

It's the helium energy in the lab frame. The lab frame is not the center of mass frame.

 

[Sargon38]

Actually, it can be fun to look at the reaction in the opposite direction. The interaction of thermal neutrons with B-10 gives you a Li-7 and a He-4 nucleus, and this reaction has of course the inverse Q value. Sometimes you also obtain a photon, but if we concentrate on the B-10 (n,alpha) Li-7 reaction, you have your reaction but backwards in time. It is a reaction that is used in thermal neutron detection, but what's more important in your case, it is spontaneous, and hence can happen with negligible kinetic energies. There is no threshold (as is often the case with positive Q value neutron reactions, as there's no Coulomb barrier). So essentially, the lab frame is the center of mass frame here.
You get a slow neutron combining with a slow B-10 nucleus, and an emission of a He-4 and Li-7 nucleus "back to back¨. The total kinetic energy of the emitted particles amounts to the Q value.
Once you understand the reaction in this direction, you can also understand it in the opposite direction, if you do a mental time-reversal. That said, the reaction will not happen if you would do this in time-reversal, you wouldn't obtain a B-10 and a *thermal* neutron. You'd have to overcome a Coulomb barrier, that's why there is a threshold energy that is higher than the Q value. You will get the surplus back of course, in kinetic energies of the B-10 and the neutron, but if you don't provide it in the first place the reaction will (almost) not happen. I say "almost" because there is theoretically of course always an infinitesimal tunneling probability for it to happen, but FAPP it doesn't until you provide the threshold. I'm not sure the threshold I'm talking about here is the value you showed - this might actually just be the lab-frame kinetic energy needed to attain the c.o.m. Q value. But there will be an extra amount of energy needed.
This is exactly the same as activation energy in chemistry.

 

[mfb]

The Coulomb barrier is small compared to the Q value here. If you get the energy just right you could get a slow neutron, although the cross section might still be pretty small.

 

[Sargon38]

The Coulomb barrier is small compared to the Q value here. If you get the energy just right you could get a slow neutron, although the cross section might still be pretty small.

Yes, indeed, I realised that when I looked at the differential cross sections: non-zero values are actually already present not much above the kinetic cutoff.