# Find the kinetic energy of the alpha particle

1. Apr 26, 2013

### Kyrios

1. The problem statement, all variables and given/known data

For the decay " Uranium-232 ---> Thorium-228 + alpha particle ",

What is the alpha particle's kinetic energy in the U-232 atom's rest frame?
We are given the masses for each one.

2. Relevant equations

E = mc^2

3. The attempt at a solution

I tried mass(U-232) - (mass(Th-228) + mass(α) ) = 5.9 x 10^ -3 u
And multiplied this by 931.5 to get E = 5.5 MeV.

Is this all there is to it? I was unsure what it meant by "in rest frame of U-232"

2. Apr 26, 2013

### Staff: Mentor

To conserve momentum, both Thorium and the alpha particle will move afterwards (in the rest frame of the initial nucleus). Therefore, Thorium will get some fraction of the released energy.

3. Apr 26, 2013

### vela

Staff Emeritus
It's conceivable, for example, that the uranium atom would be moving in the lab frame, and the question could have asked for the kinetic energy of the alpha particle that an observer at rest in the lab frame would see.

4. Apr 26, 2013

### Kyrios

I'm trying to work out the Kinetic energy of the alpha particle using this equation now:

$$KE_α = \frac{Q}{1+ \frac{m_α}{m_x}}$$

Where Q is the energy I had before, the 5.5 MeV
m_α is the mass of the alpha particle and m_x is the thorium mass

I get a similar but slightly smaller answer, 5.4 MeV. Is that the right way of going about it?

5. Apr 26, 2013

### Staff: Mentor

I don't know where the equation comes from, but it looks reasonable, and the result (slightly below 5.5 MeV) looks right.