I've been learning how beta decay works but the only thing i really dont know is how long it lasts in general like the fasted time and the slowest time. It would help a lot thanks.

It depends on the decay reaction.

The mass difference between the reactants and products will usually be assigned as kinetic energy to the emiitted particle.

If you are asking how long will a sample of Uranium 239 keep emitting beta radiation, then it depends on the amount that you are starting with and the decay constant.

You would need to solve for t in the following equation and provide a percentage that is low enough to imply that no uranium 239 is left over

ln(A_{t}/A_{0}) = -?t

If what you are asking is how long will an emitted beta particle last then it depends on the material through which it is traveling.

for Uranium 239 the energy of the beta particles is 1.21 MeV for about 80% of emitted particles and 1.28 for the other 20%.

The relation for the energy loss per unit path length for non-relativistic electrons is

dE/dx = (4?e^{4}z^{2}/mV^{2})NB

where V is the velocity of incident particle - which you calculate from the kinetic energy
ze is the charge of incident particle in coulombs
N is the number of absorber atoms/cm^{3}
B is the atomic stopping number.

It depends on the decay reaction.

The mass difference between the reactants and products will usually be assigned as kinetic energy to the emiitted particle.

Okay how long does beta decay last on uranium 239

You have to be a little more specific.

If you are asking how long will a sample of Uranium 239 keep emitting beta radiation, then it depends on the amount that you are starting with and the decay constant.

You would need to solve for t in the following equation and provide a percentage that is low enough to imply that no uranium 239 is left over

ln(A

_{t}/A_{0}) = -?tIf what you are asking is how long will an emitted beta particle last then it depends on the material through which it is traveling.

for Uranium 239 the energy of the beta particles is 1.21 MeV for about 80% of emitted particles and 1.28 for the other 20%.

The relation for the energy loss per unit path length for non-relativistic electrons is

dE/dx = (4?e

^{4}z^{2}/mV^{2})NBwhere V is the velocity of incident particle - which you calculate from the kinetic energy

ze is the charge of incident particle in coulombs

N is the number of absorber atoms/cm

^{3}B is the atomic stopping number.