Finding the amount of a uranium-238 atom's internal energy that is released

[volcore]

Homework Statement

A uranium-238 atom can break up into a thorium-234 atom and a particle called an alpha particle, α-4. The numbers indicate the inertias of the atoms and the alpha particle in atomic mass units (1 amu = 1.66 × 10^−27 kg). When an uranium atom initially at rest breaks up, the thorium atom is observed to recoil with an x component of velocity of -2.5 × 10^5 m/s.

How much of the uranium atom's internal energy is released in the breakup?
Express your answer to three significant digits and include the appropriate units.

Relevant Equations

Ke = 1/2 mv^2
0 = m1v1f + m2v2f

I assume this interaction is an explosive separation one, and as such, I thought the problem gave me the numbers to satisfy the second equation. I converted the 238 amu and 234 amu and plugged the results into the second equation of 0 = m1v1f + m2v2f. I got 0 = (3.95*10^-25kg)v1f + 3.88*10^-25kg (-2.5 * 10^5m/s). Solving for the final velocity, I got 2.46 * 10^5m/s. I then plugged this velocity into the equation Ke = 1/2 mv^2 to get Ke = 1/2 (3.95*10^-25kg)(2.46*10^5m/s)^2. From this, I got a Ke of 1.1935*10^−14 J, but my homework program said this is wrong. What am I missing?

 

[kuruman]

0 = (3.95*10^-25kg)v1f + 3.88*10^-25kg (-2.5 * 10^5m/s)

This equation says that the mass of the parent U238 nucleus multiplied by some velocity plus the the mass of the daughter Th234 nucleus multiplied by some other velocity is equal to zero. This equation does not make sense. After the decay there is no U238.

 

[neilparker62]

Does the homework program give the right answer - then we'll figure out how to get there ?

 

[volcore]

Does the homework program give the right answer - then we'll figure out how to get there ?

Sorry for the delay, but the correct answer is, according to the homework site. 7.22×10^-13 J Hopefully this helps.

 

[neilparker62]

Ok thanks - all good. Begin by sorting out your equation for conservation of momentum. The problem is as indicated in post #2. You can probably simplify things a bit by not worrying about the conversion of mass to kg in this equation.

 

[volcore]

Ok thanks - all good. Begin by sorting out your equation for conservation of momentum. The problem is as indicated in post #2. You can probably simplify things a bit by not worrying about the conversion of mass to kg in this equation.

Sorry, what exactly do you mean? Should I use this equation? 0 = (3.95*10^-25kg)v1f + 3.88*10^-25kg (-2.5 * 10^5m/s)

 

[kuruman]

Sorry, what exactly do you mean? Should I use this equation? 0 = (3.95*10^-25kg)v1f + 3.88*10^-25kg (-2.5 * 10^5m/s)

Please, please, please: Write the equation in symbolic form and then substitute the numbers. It is very difficult to sort out what you were saying especially if there is some mistake.

 

[volcore]

Please, please, please: Write the equation in symbolic form and then substitute the numbers. It is very difficult to sort out what you were saying especially if there is some mistake.

Sorry about that, the original equation is 0 = m1v1f + m2v2f. Hope this helps.

 

[kuruman]

It helps somewhat. The next step is to identify what the symbols stand for in terms of the items in the problem. What do m1, v1f, etc. stand for?

 

[volcore]

It helps somewhat. The next step is to identify what the symbols stand for in terms of the items in the problem. What do m1, v1f, etc. stand for?

m1 and m2 correspond to the inertia/mass of the 2 atoms, and v1f and v2f represent the final velocities of the 2 atoms.

 

[kuruman]

m1 and m2 correspond to the inertia/mass of the 2 atoms, and v1f and v2f represent the final velocities of the 2 atoms.

You can do better than that. Be more specific. Which atom is 1 and which atom is 2?

 

[volcore]

You can do better than that. Be more specific. Which atom is 1 and which atom is 2?

The first atom is the Uranium 238 atom, and the second one is the thorium atom.

 

[kuruman]

The first atom is the Uranium 238 atom, and the second one is the thorium atom.

So you are saying that the momentum of the uranium atom plus the momentum of the thorium atom is zero. What basis do you have to say that? If it's momentum conservation, the idea is that the total momentum before something happens is the same as the total momentum after that same something happens. How does this idea apply to what you wrote down?

 

[volcore]

So you are saying that the momentum of the uranium atom plus the momentum of the thorium atom is zero. What basis do you have to say that? If it's momentum conservation, the idea is that the total momentum before something happens is the same as the total momentum after that same something happens. How does this idea apply to what you wrote down?

Now that you mention it, I guess there really is no application for the equation. My textbook listed the equation as one that was applicable in explosive separation collisions, and this one seemed like one, so I assumed it would fit.

 

[kuruman]

Now that you mention it, I guess there really is no application for the equation. My textbook listed the equation as one that was applicable in explosive separation collisions, and this one seemed like one, so I assumed it would fit.

You assumed correctly. Explosions and collisions both conserve momentum. A collision in which two objects stick together is an explosion with time running backwards. Neither conserves kinetic energy, of course. So back to my question in #13. How does momentum conservation apply here? All you need to do is define what the "something that happens" is, find the total moment before and after the happening and set the two equal. Can you do that?

 

[volcore]

You assumed correctly. Explosions and collisions both conserve momentum. A collision in which two objects stick together is an explosion with time running backwards. Neither conserves kinetic energy, of course. So back to my question in #13. How does momentum conservation apply here? All you need to do is define what the "something that happens" is, find the total moment before and after the happening and set the two equal. Can you do that?

The event in question is the uranium atom breaking up, so wouldn't its momentum = the combined momenta of the thorium and alpha particle?

 

[kuruman]

The event in question is the uranium atom breaking up, so wouldn't its momentum = the combined momenta of the thorium and alpha particle?

Now you got it! Can you write two separate expressions for P_before and P_after? Do that first on two separate lines, then set the two equal. It may take a little more time but it helps sort out what you are doing and why.

 

[volcore]

Pinitial = m * v, so initially, since the uranium 238 atom doesn't move, its 0?

As a result, Pfinal = 0, meaning 0 = m1v1f + m2v2f, where m1 represents the alpha particle's inertia, v1f is the alpha particle's final velocity, and m2 represents the thorium atom's inertia with its final velocity represented by v2f?

 

[kuruman]

Absolutely correct. Now you see where you went wrong initially. It's a matter of thinking through what you will be doing and why before you put in the numbers.

 

[volcore]

Absolutely correct. Now you see where you went wrong initially. It's a matter of thinking through what you will be doing and why before you put in the numbers.

Thanks for that, it's starting making sense now,

so once I solve for alpha particle's velocity, I'd calculate its kinetic energy with k=0.5mv^2 and combine it with the kinetic energy of the thorium atom for the total kinetic energy of the system, which would be equal to the system's (uranium 238 atom) internal energy due to energy conservation?

 

[kuruman]

Thanks for that, it's starting making sense now,

so once I solve for alpha particle's velocity, I'd calculate its kinetic energy with k=0.5mv^2 and combine it with the kinetic energy of the thorium atom for the total kinetic energy of the system, which would be equal to the system's (uranium 238 atom) internal energy due to energy conservation?

I would think so. Just to be sure, does the term "binding energy per nucleon" mean anything to you? Is $E=mc^2$ part of the chapter where this question belongs?

 

[volcore]

I would think so. Just to be sure, does the term "binding energy per nucleon" mean anything to you? Is $E=mc^2$ part of the chapter where this question belongs?

Not yet, we haven't discussed relativity as of now.

 

[kuruman]

OK then, just calculate the sum of the kinetic energies of thorium and alpha and call that the answer.

 

[volcore]

OK then, just calculate the sum of the kinetic energies of thorium and alpha and call that the answer.

Thanks for all of your help, I was able to calculate the correct answer!