Conservation of momentum uranium atom problem

[BrainMan]

Homework Statement

A uranium atom of mass 238 u decays by emitting an alpha particle (the nucleus of a helium atom) of mass 4 u at a speed of 2 x 10⁷ m/s. (Note that "u" is the symbol for atomic mass units and 1 u = 1.67 x 10^-27 kg.) What is the recoil speed of the resulting nucleus immediately after the decay?

Homework Equations

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

The Attempt at a Solution

The first thing I did was realize that the total momentum before the uranium emittes the alpha particle is zero so the momentum after the alpha particle is released is zero. So m₁v₁ + m₂v₂ = 0. I plugged in the masses of the two particles and the speed of the smaller particle and solved for the velocity of the larger particle to get the recoil speed.

 

[abitslow]

That is correct (in the inertial rest frame of the nucleus).
It would be a tiny bit better to have said it this way:
"Since (linear) momentum is conserved, the total momentum before the decay is exactly the same as after it."
This avoids the problem of which rest frame you are referring to. Nothing much wrong with claiming momentum (or velocity) is zero, aside from the fact that it is only zero for one particular set of coordinates. What you can calculate is the difference in velocity of the two particles after the decay, compared to before it. That is
MuVu = MthVth+MaVa (where V are the velocity vectors, and u=uranium, th= thorium and a= alpha particle). Choosing Vu to be zero greatly simplifies the math. Picking the right coordinates is a very important consideration in solving Physics problems.
Oops!
2E7 m/s ?? 2E7 m/c ÷ 3E8 m/s ~ 0.067c...you *might* want to add the relativistic correction to the momentum of the alpha particle - meaning its momentum is actually more than just ma*va. The Th nucleus will have a velocity substantially lower, so you don't need to worry about it's relativistic momentum. √(1-(v/c)²) ~
√(1-0.067²) ~ 0.998; meaning given velocity with one place precision, no relativistic correction is necessary even for the alpha particle.

 

[BrainMan]

That is correct (in the inertial rest frame of the nucleus).
It would be a tiny bit better to have said it this way:
"Since (linear) momentum is conserved, the total momentum before the decay is exactly the same as after it."
This avoids the problem of which rest frame you are referring to. Nothing much wrong with claiming momentum (or velocity) is zero, aside from the fact that it is only zero for one particular set of coordinates. What you can calculate is the difference in velocity of the two particles after the decay, compared to before it. That is
MuVu = MthVth+MaVa (where V are the velocity vectors, and u=uranium, th= thorium and a= alpha particle). Choosing Vu to be zero greatly simplifies the math. Picking the right coordinates is a very important consideration in solving Physics problems.

Where are you getting Thorium?

 

[SammyS]

Where are you getting Thorium?

That's element 90 .

 

[BrainMan]

That's element 90 .

Is it the same thing as the alpha particle?

 

[SammyS]

Is it the same thing as the alpha particle?

No.

What is an alpha particle's composition?

 

[BrainMan]

I don't know I haven't learned anything about particles on an atomic level.

 

[BrainMan]

No.

What is an alpha particle's composition?

I haven't learned anything about particles of that size.

 

[SammyS]

I haven't learned anything about particles of that size.

What size ?


From your OP:

... emitting an alpha particle (the nucleus of a helium atom) of mass 4 u ...

 

[BrainMan]

Ok so the only particles involved are the alpha particle that has a mass 4u and the uranium particle with mass 238? If so what did I do wrong in my original calculations?

 

[SammyS]

What is an alpha particle's composition?

You didn't really answer this. You could easily look it up, but here's the answer.

The alpha particle is composed of 2 protons & 2 neutrons.

Ok so the only particles involved are the alpha particle that has a mass 4u and the uranium [STRIKE]particle[/STRIKE] nucleus with mass 238? If so what did I do wrong in my original calculations?

If a Uranium (atomic number, 92) nucleus loses 2 protons (and 2 neutrons), what will be the resulting element ?




.

 

[BrainMan]

You didn't really answer this. You could easily look it up, but here's the answer.

The alpha particle is composed of 2 protons & 2 neutrons.



If a Uranium (atomic number, 92) nucleus loses 2 protons (and 2 neutrons), what will be the resulting element ?




.

OK it is Thorium. So what do I do after I realize that it is Thorium?

 

[haruspex]

OK it is Thorium. So what do I do after I realize that it is Thorium?

The fact that the other resultant particle is Thorium was an interesting discussion, but not relevant to solving the question. You know its atomic mass, and that's all that matters.
In the OP, you didn't indicate what answer you got or whether it was known to be wrong. Are you now saying your answer was wrong? If so, either you made a calculation error or you need to take into account the relativistic momentum of the alpha particle, as abitslow suggested. Please post your answer and the one given as correct.

 

[BrainMan]

The fact that the other resultant particle is Thorium was an interesting discussion, but not relevant to solving the question. You know its atomic mass, and that's all that matters.
In the OP, you didn't indicate what answer you got or whether it was known to be wrong. Are you now saying your answer was wrong? If so, either you made a calculation error or you need to take into account the relativistic momentum of the alpha particle, as abitslow suggested. Please post your answer and the one given as correct.

OK here is what I did to calculate my answer
238u(v) = -(4u)(2 x 10⁷ m/s)
v = -3.36 x 10⁵ m/s
the correct value for v should be 3.42 x 10⁵ m/s

 

[BrainMan]

Sorry for the confusion

 

[tms]

OK here is what I did to calculate my answer
238u(v) = -(4u)(2 x 10⁷ m/s)
v = -3.36 x 10⁵ m/s
the correct value for v should be 3.42 x 10⁵ m/s

With that small a discrepancy, I'd suggest just using the mass of thorium instead of uranium. Since your initial data have only one significant figure, you can probably ignore the binding energy difference and any relativistic corrections due to the speed of the α, and also just take the mass of thorium to be $238 - 4$. As for the sign, I'd guess they just chose a different positive direction.

 

[haruspex]

With that small a discrepancy, I'd suggest just using the mass of thorium instead of uranium.

BrainMan, in case it's not clear, you should have used 234 in the first place. The m1 and m2 of your equation here map to the 234u and 4u, as though two separate objects from the start.

As for the sign, I'd guess they just chose a different positive direction.

The OP asks for speed, not velocity, so an unsigned answer is required.

 

[tms]

The OP asks for speed, not velocity, so an unsigned answer is required.

So it does.

 

[BrainMan]

Ok I get it now. Thanks!