How does the recoil of the nucleus affect the speed of the alpha particle?

[thebomb]

HELP! URGENT! momentum!

The nucleus of 238U decays by emitting an alpha particle, 4He. The velocity of the alpha has been measured to be v. The mass of the original nucleus is 238 u (atomic mass units) and the mass of the alpha is 4 u. Do not convert atomic mass units into kilograms.

find what speed we would expect to measure for the alpha if the nucleus is not free to move?



m1u1 + m2u2 = m1v1 + m2v2



my work:
m1u1 + m2u2 = 0
in this equation final velocity of nucleus (u1) is zero because the nucleus is not moving. so, when I try to solve for the velocity of the alpha particle (u2), the answer comes out to be zero, which is not correct

 

[Stonebridge]

The nucleus of 238U decays by emitting an alpha particle, 4He. The velocity of the alpha has been measured to be v. The mass of the original nucleus is 238 u (atomic mass units) and the mass of the alpha is 4 u. Do not convert atomic mass units into kilograms.

find what speed we would expect to measure for the alpha if the nucleus is not free to move?



m1u1 + m2u2 = m1v1 + m2v2



my work:
m1u1 + m2u2 = 0
in this equation final velocity of nucleus (u1) is zero because the nucleus is not moving. so, when I try to solve for the velocity of the alpha particle (u2), the answer comes out to be zero, which is not correct

You need to make sure you have accounted for the different masses of the alpha and the nucleus.
If you call the alpha mass m1, then the undecayed nucleus, m2, is 238/4 times m1.
m2 = 59.5m1
After emission, the nucleus is 234/4 times the mass of the alpha, as it has lost 4 mass units.

 

[Borek]

the answer comes out to be zero, which is not correct[/b]

I can be wrong, but seems to me that this is the correct answer. For the momentum to be conserved nucleus has to have exactly the same momentum as the alpha particle, it just has to move in the exactly opposite direction. If the nucleus stays in place, there is no recoil, so - absurd how it looks - the only way momentum can be conserved is that alpha particle has also no momentum at all, that means speed of zero.

Then, my faith in my understanding of basic physics has been seriously undermined in the last months, so I can be missing something

 

[ehild]

The statement "the nucleus is not free too move" means that it is attached to something (other atoms in a crystal for example) that hinders it motion. In this case the nucleus and alpha particle is not a closed system any more. You can imagine that the nucleus has infinite mass, in this case its recoil speed is zero, but the product of the infinite mass and zero speed can be anything. Nothing can be done with conservation of momentum, but there is an other thing: conservation of energy.
During alpha decay, the binding energy among the protons and neutrons inside the nucleus decreases and this energy will be shared as KE between the new nucleus and the alpha particle if the nucleus is free to move. The change of the binding energy is the same either the nucleus can move or not. If it does not move, the alpha particle gets all this energy.

ehild

 

[thebomb]

the answer to this problem is greater than part a. and part a is following.

a) Assuming the uranium nucleus was initially at rest, what is the recoil velocity of the remaining nucleus after decay? Explain the physics.
the answer i got for part (a) is vf = 3/5v ; where v is initial velocity.

 

[thebomb]

thanks for trying though!

I need this in few hours. Please help! I am not sure what i am doing wrong. maybe my equation is wrong? I have no idea.
HELP please!

 

[Stonebridge]

I can be wrong, but seems to me that this is the correct answer. For the momentum to be conserved nucleus has to have exactly the same momentum as the alpha particle, it just has to move in the exactly opposite direction. If the nucleus stays in place, there is no recoil, so - absurd how it looks - the only way momentum can be conserved is that alpha particle has also no momentum at all, that means speed of zero.

Then, my faith in my understanding of basic physics has been seriously undermined in the last months, so I can be missing something

Strictly speaking this is correct. However, I'm assuming that the idea behind this is that you calculate the velocity of the alpha and the nucleus after the split when both move - using momentum considerations.
From this you can find the total kinetic energy released.
In the case when the nucleus doesn't move, you assign all this K.E, to just the alpha, and thus find its velocity from half mv squared.
It's a strange question. Maybe there is some information missing?

 

[thebomb]

A big thanks to all you guys for replying so quickly!

 

[masterice]

yes thank you all who replied! you will definitely be paid back in full soon in your life I promise!

 

[Borek]

Nothing can be done with conservation of momentum, but there is an other thing: conservation of energy.

Problem is, in the real world particles don't select which law of conservation they obey this time, they have to obey each one every time. Using just kinetic energy to calculate velocity is equivalent to assuming momentum is not conserved. This is - to say the least - not pedagogical.

 

[ehild]

Problem is, in the real world particles don't select which law of conservation they obey this time, they have to obey each one every time. Using just kinetic energy to calculate velocity is equivalent to assuming momentum is not conserved. This is - to say the least - not pedagogical.

Particles do no need to obey all conservation laws.

For the momentum: The time derivative of the momentum of a particle is equal to the force acting on it. The time derivative of the momentum of a system of particles is equal to the sum of all external forces, acting on the system. Only in case of closed systems , when there are no external forces, is the momentum conserved.
The same for the angular momentum: it is conserved when the resultant torque is zero.
For mechanical energy: it is conserved only when all forces are conservative (have potential).
You have to select very carefully the system before applying a conservation law.

ehild

 

[Stonebridge]

Problem is, in the real world particles don't select which law of conservation they obey this time, they have to obey each one every time. Using just kinetic energy to calculate velocity is equivalent to assuming momentum is not conserved. This is - to say the least - not pedagogical.

From the educational viewpoint, the question is attempting to separate the two conservation laws - applying momentum first, and then kinetic energy.
There would have to be a recoil of the nucleus in the real world, but by stating that in the second part the nucleus is held so that it does not move, we have entered the world of fantasy. Such a world only exists in physics pedagogy, and as such, it is legitimate to apply only K.E. conservation. I would expect any good physics teacher to explain this to the student.

 

[Borek]

You have to select very carefully the system before applying a conservation law.

My understanding of conservation laws is that things that have to be conserved are conserved always. External forces that change momentum of the system at the same time act on something external (changing its momentum) - and total momentum is conserved again. It is not a matter of selecting system in such a way things will be conserved, quite the opposite - as long as something is not conserved we know system is incorrectly selected (incomplete).

This is not necesarilly practical when it comes to solving problems, but that's another story.

In this case looks like question asks to use momentum conservation to calculate velocity and then to ignore momentum conservation to calculate velocity. Something is put on the head. Probably it is just a matter of poor wording of the question. Say - emitting atom is a part of large crystal. It is easy to show then that velocity of the crystal can be neglected.

 

[Borek]

Such a world only exists in physics pedagogy

You may call it a pet peeve of mine, but I hate questions that introduce wrong concepts/facts. Questions are important part of the learning process and they should be constructed in such a way they don't introduce misconceptions. Obviously they have to simplify the world sometimes, but these simplifications have to be reasonable, and not suggest things that are simply wrong.

 

[Stonebridge]

You may call it a pet peeve of mine, but I hate questions that introduce wrong concepts/facts. Questions are important part of the learning process and they should be constructed in such a way they don't introduce misconceptions. Obviously they have to simplify the world sometimes, but these simplifications have to be reasonable, and not suggest things that are simply wrong.

If the student is aware of what conservation of momentum actually means, he would/should realize that the statement "the nucleus is not free to move" means that conservation of momentum is not used to solve this part of the question.
If I gave you a problem to calculate the height to which a rubber ball rebounds when dropped onto the ground, given h and the coefficient of restitution of the rubber; would you be at all interested in using momentum to solve this, even though, as you say, it is always conserved? Would you be unhappy if I said "ignoring air resistance", even though this is "simply wrong"?

 

[kuruman]

For all it's worth, recoil-free emission of gamma rays (not alphas) by nuclei is a physical reality, has been observed and has been named the Mossbauer effect. Gamma rays are emitted by single nuclei and the recoil is taken up by the entire lattice, in which case it is negligible and the gamma has (to a ridiculous accuracy) the energy corresponding to the energy difference between the initial and final nuclear states.

A similar model of recoil-free emission (for alphas) is what ehild suggested in posting #4 and I believe is how the problem-giver expects the solution to go: Let the recoil be taken up by the lattice atoms, all 10²³ (more or less) of them, conserve momentum and energy and see what you get for the recoil speed and change in kinetic energy of the lattice. We don't bother with the Earth's recoil velocity when we solve projectile problems, do we? Do we selectively ignore momentum conservation? You betcha!

 

[ehild]

Similar problem: collision of a particle with a wall. We can consider the particle and the wall a closed system, where the particle gives momentum to the wall, but the wall does not move as its mass is practically infinitive.
Or we can investigate the particle separately, as subjected to by a force from the wall, and the impulse of this force changes its momentum.

ehild

 

[Borek]

Let the recoil be taken up by the lattice atoms, all 10²³ (more or less) of them, conserve momentum and energy and see what you get for the recoil speed and change in kinetic energy of the lattice.

the particle gives momentum to the wall, but the wall does not move as its mass is practically infinitive.ehild

Funny thing, this is post you are answering:

Say - emitting atom is a part of large crystal. It is easy to show then that velocity of the crystal can be neglected.

Deja vu?

If the student is aware of what conservation of momentum actually means, he would/should realize that the statement "the nucleus is not free to move" means that conservation of momentum is not used to solve this part of the question.

Well, obviously student was not aware and obviously wording of the question was not helpful, that was our starting point

I am perfectly aware that there are many ways of dealing with questions and that there are many approximations that we use to simplify the reality and make questions possible to solve. However, stating things like "particles do no need to obey all conservation laws" is a grave mistake. Students are very clever in remembering such simple, catchy and wrong statements. The more wrong the statement is, the better they remember it.

 

[Stonebridge]

I can be wrong, but seems to me that this is the correct answer. For the momentum to be conserved nucleus has to have exactly the same momentum as the alpha particle, it just has to move in the exactly opposite direction. If the nucleus stays in place, there is no recoil, so - absurd how it looks - the only way momentum can be conserved is that alpha particle has also no momentum at all, that means speed of zero.

Then, my faith in my understanding of basic physics has been seriously undermined in the last months, so I can be missing something

Well, obviously student was not aware and obviously wording of the question was not helpful, that was our starting point

I am perfectly aware that there are many ways of dealing with questions and that there are many approximations that we use to simplify the reality and make questions possible to solve. However, stating things like "particles do no need to obey all conservation laws" is a grave mistake. Students are very clever in remembering such simple, catchy and wrong statements. The more wrong the statement is, the better they remember it.

It was not I who stated that.

The question should have said something along the lines that the recoil of the nucleus "can be ignored". On that I agree.