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

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?

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.

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

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".

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.

Vanadium 50 said:

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) )

eneacasucci said:

but how can be "everything at rest"

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

Vanadium 50 said:

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

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.

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.

mfb said:

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)

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

mfb said:

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!

mfb said:

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)

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


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

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