# What is the energy of the alpha particle and recoiling nucleus?

1. May 15, 2017

### says

1. The problem statement, all variables and given/known data
The Q-value of alpha decay of Pu-239 is 5.244 MeV.
Calculate the energy of the alpha particle in MeV.
Calculate the energy of the recoil nucleus decay product in MeV.
(answer correct to 3 decimal places)

2. Relevant equations
The Q value of a nuclear reaction is the difference between the sum of the masses of the initial reactants and the sum of the masses of the final products.
Neutron (Energy equivalent) = 939.57 MeV
Proton (Energy equivalent) = 938.28 MeV
Electron (Energy Equivalent) = 0.511 MeV
3. The attempt at a solution

Pu-239 > U-235 + alpha particle
Q = Pu-239 - (U-235 + alpha particle) = 5.244 MeV
Pu-239 = (n=145, p=94, e=94)
U-235 = (n=143, p=92, e=92)

Q = (145 * 939.57 + 94 * 938.28 + 94 * 0.511) - ((143 * 939.57 + 92 * 938.28 + 92 * 0.511) - alpha particle = 5.244 MeV
Q = 3756.72 MeV - alpha particle = 5.244 MeV
Energy of Alpha Particle = 3751.476 MeV

I'm not sure how to calculate the energy of the recoil nucleus decay product though...

2. May 15, 2017

### Staff: Mentor

The mass of a nucleus is not the sum of its protons and neutrons (otherwise the decay wouldn't release energy!), and there are no electrons involved.
5.244 MeV is the sum of the kinetic energy of alpha particle and nucleus. Why do both move after the reaction? Think about conservation laws you know. One of them will tell you how the 5.244 MeV get distributed over the two particles.

3. May 15, 2017

### says

So If I calculate the kinetic energy of the nucleus I can say 5.244 MeV - kinetic energy of nucleus = energy of the alpha particle?

4. May 15, 2017

### Staff: Mentor

That is right. It is one of the two equations you'll need to find the energies.

5. May 16, 2017

### says

mass of a nucleus = (number of protons * atomic weight of protons) + (number of neutrons * atomic weight of neutrons) - binding energy of the nucleus.

Mass of alpha particle = (2 * 938.3084221246 + 2 * 939.5985407812) - 28.3 = 3727.51392581 MeV
Mass of U-235 = (92 * 938.3084221246 + 143 * 939.5985407812) - 1783.870285 = 218903.095882 MeV
Mass of Pu-239 = (94 * 938.3084221246 + 145 * 939.5985407812) - 1806.921509 = 222635.858584 MeV

Q = 222635.858584 - (218903.095882 + 3727.51392581) = 5.24877619001 MeV

So the energy of the alpha particle = 3727.51392581 MeV
I'll have to fix up the sig figs a little bit for the final answer because it's wanted to 3 decimal places...

6. May 16, 2017

### Staff: Mentor

That is the mass of it, not its energy, and not its kinetic energy (=what the answer wants) either.

You are not even close to the final answer, and the binding energy calculations do not help you here.

7. May 16, 2017

### says

I'm a bit lost then...
The question says the Q-value of alpha decay of Pu-239 is 5.244 MeV.

Pu239 decays to U235 + alpha particle + 5.244 MeV
if the q-value = the difference between the sum of the masses of the initial reactants and the sum of the masses of the final products.
Q=Pu239-(U235+alpha particle)=5.244 MeV

I thought in asking for the energy they were asking for the mass but in terms of MeV...

8. May 17, 2017

### says

I'd they were asking for kinetic energy I think they would have given more than just the Q value of the reaction

9. May 17, 2017

### says

Ok, I tried a new approach. Can't believe I didn't use this from the start. Conservation of energy.

Energy of alpha particle = Q/(1+(mass of alpha particle / mass of daughter particle)

5.244/ (1+(4.001506179127 / 235.0439299) = 5.15621795008 MeV = 5.156 MeV (3 decimal places)

energy of recoil nucleus = Q/(1+(mass of daughter / mass alpha)

5.244 / (1 + (235.0439299 / 4.001506179127) = 0.08778204992 MeV = 0.088 MeV (3 decimal places)

10. May 17, 2017

Right.