Calculating disintegration energies of alpha emmision.

[Finaid]

Homework Statement

(a) Calculate the disintegration energy when 232/92U decays by alpha emission
into 228/90Th. Atomic masses of 232/92U and 228/90Th are 232.037156u and 228.028741u, respectively.

(b) For the 232/92U decay in part (a), how much of the disintegration energy will be carried off by the alpha particle?

Given: Mass of 4/2He = 4.002603u
c^2 = 931.5MeV

Homework Equations

E=mc^2

The Attempt at a Solution

Well for part (a), first I found the difference in the starting masses and the end masses ie,
232.037156u - (228.028741u + 4.002603u) = 0.005812u
I then put this into the equation and got 5.413878MeV.
I thought this was right until I read part (b) and now I'm starting to think this might be how I'm meant to do that part, not part (a). Could anyone tell me if I'm even on the right track with this question or should I be using different equations?

 

[da_nang]

I see nothing wrong in a). In b), consider what happens when U-232 disintegrates, if you look at it in the frame of reference of that isotope. What happens to the nuclei? And assuming no energy loss, where does that energy go?

 

[Finaid]

Does most or all of the energy get carried off by the alpha particle then?

 

[da_nang]

Not all of it, no. Think of the U-232 nucleus as a group of particles bundled together. In their frame of reference, they aren't moving. Then the nucleus gets split into two different nuclei, but they can't share the same space. So what happens then to rectify this?

 

[Finaid]

Well the alpha particle and the Th-228 have to move away from each other so the energy turns into kinetic energy. So if it isn't moving in the beginning, and momentum has to be conserved, then they move in opposite directions. So then since K = (p^2)/2m, the energy is divided according to their masses? So, 228/(228+4) = 0.9827. Then (0.9827)(5.413878MeV) = 5.3202179MeV goes to the alpha particle. Is that right?

 

[da_nang]

Yes. Just remember to use all of those decimals.

 

[Finaid]

Ok, thanks so much for the help!