The Student Room Group

Physics a level help please!

so the question is :

For a helium-3 nucleus and a helium-4 nucleus to fuse they need to be separatedby no more than 3.5 × 10–15 m:
(i) Calculate the minimum total kinetic energy of the nuclei required for them to
reach a separation of 3.5 × 10–15 m.


I get that the Ek=Ep but what i dont get is why we dont double the Ek at the end since each nuclei must have the energy that we calculated right?
Reply 1
The kinetic energy of a given particle is in general arbitrary, since you can always add a constant velocity to the whole system and nothing changes (relativity of inertial frames). However, the total kinetic energy of the system in the centre-of-momentum frame (i.e. when the total momentum is zero) is unique and minimal across all possible inertial frames, and this what matters for the fusion to occur. Try to think of kinetic energy as a property of the system, rather than individual component particles.
Reply 2
Original post by lordaxil
The kinetic energy of a given particle is in general arbitrary, since you can always add a constant velocity to the whole system and nothing changes (relativity of inertial frames). However, the total kinetic energy of the system in the centre-of-momentum frame (i.e. when the total momentum is zero) is unique and minimal across all possible inertial frames, and this what matters for the fusion to occur. Try to think of kinetic energy as a property of the system, rather than individual component particles.


😵😵, I think i get what you mean but could you elaborate more in how that specifically relates to this question please?
Reply 3
Sorry if I wasn't clear.

If my understanding is correct, you were asking why the kinetic energy needed to fuse the helium nuclei (Ek = Ep) wasn't doubled due to their being two of them. The answer is because Ek is the total kinetic energy of the system (viewed from a frame in which the total momentum of the system is zero). In fact, the total kinetic energy is partitioned out between the two nuclei in a way we don't really need to worry about because it is not the KE of each particle that matters, but the total available to the system.

Does that make sense?
Reply 4
Original post by lordaxil
Sorry if I wasn't clear.

If my understanding is correct, you were asking why the kinetic energy needed to fuse the helium nuclei (Ek = Ep) wasn't doubled due to their being two of them. The answer is because Ek is the total kinetic energy of the system (viewed from a frame in which the total momentum of the system is zero). In fact, the total kinetic energy is partitioned out between the two nuclei in a way we don't really need to worry about because it is not the KE of each particle that matters, but the total available to the system.

Does that make sense?

so surely both nuclei will have their own electrical potential energy which adds together to give the Ek of the system right? so you would have to double the Ep?
Original post by FM1/FP1
so surely both nuclei will have their own electrical potential energy which adds together to give the Ek of the system right? so you would have to double the Ep?


No, the electrical potential energy is shared between the 2 nuclei. It is a result of them being close together and the strong repulsive forces needed to be overcome to get them there. It is the PE of the system. As has already been said.
The question then asks you what is the total KE of the nuclei needed to do this. Not what the KE is of the individual nuclei.
This is just an application, as you rightly originally stated, of conservation of energy.
The total ke of the nuclei (combined) has been reduced to zero, and this energy is now stored as the (total) potential energy of the 2 nuclei.
Reply 6
Original post by Stonebridge
No, the electrical potential energy is shared between the 2 nuclei. It is a result of them being close together and the strong repulsive forces needed to be overcome to get them there. It is the PE of the system. As has already been said.
The question then asks you what is the total KE of the nuclei needed to do this. Not what the KE is of the individual nuclei.
This is just an application, as you rightly originally stated, of conservation of energy.
The total ke of the nuclei (combined) has been reduced to zero, and this energy is now stored as the (total) potential energy of the 2 nuclei.


oh i see, so the PE of each nuclei is the PE of the system which would then be shared out as kinetic energy between the nuclei?
Original post by FM1/FP1
oh i see, so the PE of each nuclei is the PE of the system which would then be shared out as kinetic energy between the nuclei?

It seems that you are a “victim” of the misconception of potential energy that usually originated from the GCSE level.

At the GCSE level, students are usually asked to compute the gain in gravitational potential energy of an object with mass m that is raised from a height of h1 to h2 near the surface of the Earth, where h2 > h1. As a result, students have the tendency to think that potential energy belongs solely to a particular object. This is very unfortunate.

Consider another similar problem: A stone is held above the Earth’s surface and the stone is let go to fall under the influence of gravity. We often say the gravitational potential energy of the stone is converted to the kinetic energy of the stone at the GCSE level. Why?
When the stone is let go, both the stone and the Earth move toward each other, but the Earth’s motion is too small to detect (or Earth does not move significantly) due to the immense mass of the Earth relative to that of the stone. This means that the change in the separation of the stone-Earth system comes about largely due to the motion of the stone.
So, we often associate the gravitational potential energy of the stone-Earth system with the stone alone at the GCSE level.
In reality, gravitational potential energy is actually shared by both the stone and the Earth and does not belong to the stone alone.

Potential energy is always associated with a system of two or more interacting objects, so PE is a shared property among the two or more interacting objects. Thus, it is incorrect to call the electric potential energy of a nucleus or the potential energy of a particle.
Original post by FM1/FP1
oh i see, so the PE of each nuclei is the PE of the system which would then be shared out as kinetic energy between the nuclei?


Yes as Eimmanuel and Lordaxil have both said- you have to think of it as the energy of the complete system. The total original KE (combined) of the 2 nuclei before they get to that separation, is equal to the total PE of the system when the 2 nuclei have come to rest at that close separation.
You can think of it in reverse too. Hold the 2 nuclei together (find the total PE they have) - let them spring apart. The total KE they will get is equal to the PE they once had. The KE is shared between them. By the way- if (and only if) the 2 masses were equal, then they would share that KE equally, with each getting half. Hope this helps.
Reply 9
Thanks @Stonebridge, @Eimmanuel and @lordaxil I get what we are actually working out when we find the PE of "one" nuclei. your explanations were all much more detailed than I expected

Quick Reply

Latest