Kinetic energy of electron after beta decay

1. Aug 6, 2012

Xavius

Is that a constant? If not, is there an equation that allows us to calculate its kinetic energy based on the properties of the atom it came from?

2. Aug 6, 2012

Staff: Mentor

The total energy available in the decay is fixed, according to the difference in mass between the initial and final nuclei. However, that energy is distributed between two outgoing particles, the electron and the (anti)neutrino, and that distribution varies randomly from one decay to another. So beta-decay electrons from a particular isotope have different energies, ranging from zero up to the total energy available.

http://hyperphysics.phy-astr.gsu.edu/Hbase/nuclear/beta2.html

3. Aug 6, 2012

QuantumPion

It is not a constant, it depends on the nuclide. Tritium decays with a 5 keV beta. C-14 decays with a 50 keV beta. Co-60 decays with a 300 keV beta, etc. A description of the beta energy spectrum can be found here: http://hyperphysics.phy-astr.gsu.edu/Hbase/nuclear/beta2.html

4. Aug 6, 2012

Xavius

Ok, I think I see. So is this why a tritium decay into He4 isn't very energetic but a decay in a heavier element could be quite energetic?

5. Aug 6, 2012

QuantumPion

heh you beat me to it :)

6. Aug 6, 2012

Xavius

Sorry, was typing that question as you were typing the answer :)

7. Aug 6, 2012

Emilyjoint

the energy in beta decay is shared between the electron (positron) and a neutrino or antineutrino so jtbell answer is best

8. Aug 6, 2012

Xavius

Another newb question, the total energy resulting from a tritium atom undergoing beta decay is 18.59 keV. According to wiki the average kinetic energy of the electron is 5.7 keV leaving 12.89 keV on average attributed to the antineutrino.

Are those the only particles emitted during beta decay?

Maybe I did the maths incorrectly but the 12.89 keV kinetic energy translated to relativistic velocity for a particle with the rest mass range of an electron neutrino (< 2.2 eV) puts the upper velocity limit at > 2.998*10^8 m/s. How is that possible?

9. Aug 6, 2012

Staff: Mentor

We can't tell you whether your equations are correct if you don't tell us what they are.

Even if your equations are correct, you have to be very very careful about the arithmetic in a calculation like this, because roundoff errors in intermediate steps can bite you hard.

For K = 12890 eV and mc2 = 2.2 eV I get v/c = 0.999999985.

10. Aug 6, 2012

RocketSci5KN

Small correction, T decays into He3, not He4.

11. Aug 6, 2012

Xavius

That's the problem, I don't know the equations, I'm a total amateur :) I was plugging the values into wolfram and that's what it spit out. I think the math it did was bugged because I put in a value of 1 MeV kinetic energy and it was less than 1c.