# Q-value calculation - nuclear physics

• Ene Dene
In summary, the conversation discusses the calculation of Q values for nuclear reactions, specifically the reactions D + p → ³He + γ and 3He + 3He → 4He + 2p. The values obtained using the table of masses from a given link are different from the values mentioned in lectures, causing confusion. In the case of the first reaction, Q is equal to the energy of the gamma photon, while in the second reaction, Q is equal to the difference between the initial and final energies. The discrepancy in the calculated Q value for the second reaction is likely due to using the wrong mass for the ³He nucleus.
Ene Dene
http://www.nndc.bnl.gov/masses/mass.mas03, and the missing data:
u=931.50MeV
p=1.00728u=938.28MeV
d=2.01355u=1875.63MeV
e=0.511MeV

A(a,b)B=a+A->b+B
Q=[(m(a)+m(A))-(m(b)+m(B)]MeV (1)

This can't be more simple, but in every example that I saw in my lectures different values occur. For example:
3He+3He->4He+2p (this is not 3 times He, but p+p+n), Q=12.859
Using the table of masses from a given link and (1) I get Q=13.83MeV.

The second thing I don't understand is when I have this situation:
d+p->3He+photon,
1875.63MeV+938.28MeV->2809.4MeV+photon. Is Q now equal to energy of photon?
In this case 1875.63 + 938.28 - 2809.4=4.51MeV+photon.
Btw, also in my lectures Q of this reaction is 5.493MeV.

I really can't understand these differences.

This question is 14+ years old at the time of answering. But maybe this will help someone who comes across it.

The link given in Post #1 not longer exists, so I can't check the values and calculation for the first question.

Let me address the second question.

deuteron + proton → helium-3 + gamma photon
D + p → ³He + γ

For this reaction, Q= 5.493MeV. ‘Q’ is the gamma photon energy.

(For information, this is the second reaction in the p-p chain, e.g. see https://en.wikipedia.org/wiki/Proton–proton_chain)

I suspect your calculated value of Q (4.51Mev) is wrong because you have used the mass of a ³He atom. You should have used the mass of a ³He nucleus. That would account for most of the discrepancy.

Minor edits.

Last edited:

## 1. What is a Q-value in nuclear physics?

The Q-value in nuclear physics refers to the change in energy that occurs during a nuclear reaction. It is calculated by subtracting the total mass of the reactants from the total mass of the products, and then converting the difference into energy using Einstein's famous equation, E=mc^2.

## 2. How is the Q-value calculated?

The Q-value is calculated using the mass-energy equivalence equation, E=mc^2. The total mass of the reactants is subtracted from the total mass of the products to determine the change in mass, which is then converted into energy. This energy is the Q-value of the reaction.

## 3. What is the significance of the Q-value in nuclear reactions?

The Q-value is a crucial factor in determining whether a nuclear reaction is energetically favorable or not. If the Q-value is positive, it means that the reaction will release energy, while a negative Q-value indicates that energy must be supplied for the reaction to occur. The Q-value also determines the type of reaction that will take place, such as fusion or fission.

## 4. How does the Q-value affect the stability of a nucleus?

The Q-value is related to the binding energy of a nucleus, which is the energy that holds the nucleus together. A higher Q-value indicates a more stable nucleus, as more energy is required to break apart the nucleus. This is why many nuclear reactions strive to produce nuclei with higher Q-values, as they are more stable and less likely to undergo radioactive decay.

## 5. Can the Q-value be negative?

Yes, the Q-value can be negative. This means that energy must be supplied for the reaction to occur, and the reaction is not energetically favorable. This is often the case for nuclear reactions involving heavier nuclei, where a lot of energy is required to overcome the strong repulsive forces between protons.

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