Physics Teachers Forum

Hi ALL,

can anyone clarify for us how the varying energies of beta particles from beta minus decay is accounted for by the production of neutrinos?

All references seem to indicate only one neutrino is produced, and being a fundamental particle, shouldnt it also have a set energy?

Is there more than one neutrino produced or do they in fact each have different energies?

thanks

Matt,
In beta decay or any other nuclear decay reaction the total energy released in the process is fixed. The reason that you get the same energy for the decay products of many reactions is because there are only two particles in the products. Total energy and momentum have to be conserved. So that, if there are two products and, say, the original nucleus was stationary, then the resulting products will fly out in opposite directions each having a certain energy that is the same for any nucleus of that type that decays. This is similar to a father and a young child on ice (so no friction). If the father and child push on each other with a fixed amount of total energy, then they fly out in opposite directions with the same individual energies each time they play this game. You can calculate this from conservation of energy and momentum using mechanics.

Now imagine that the father had two children on the ice with him and they all push on each other with the same total energy each time they play. But this time, they have no control over the directions they fly out. Their individual share of that total energy will depend on the angle with which they fly out in order to conserve total momentum and energy. In the same way, a nucleus that decays into three particles has no control over the angle with which all three particles fly out. The angles are completely random with each decaying nucleus. This is in contrast to the two particle case where they always fly out 180 degrees with respect to each other. Since the angle was fixed, then so was their individual energies and momenta.

If you measure the energy of the electron in a beta minus decay reaction, you will find that this energy takes on a wide range of values. For example, the element Promethium-147 (Pm-147) decays to Samarium-147 (Sm-147) and gives out an electron. By having a solid of Pm-147 and measuring the energies of all the electrons coming out of it, you will find that there is a wide range of electron energies. This would not be the case if the only products of the reaction were Sm-147 and an electron. For this reason they realized that there was a third undetected particle being given out. To conserve charge, they also realized that this third particle must be neutral. From the resulting energies of the electron and daughter nucleus, they realized that this unseen particle must have a small mass. For this reason they called it “little neutral one” – neutrino.

Hi,

Just wondering if you could clarify something for me.

With the PE effect, if we have two sources of light of equal "intensity" but different wavelength, what effect will this have on photocurrent?

I commonly see the term "intensity" being misused, when i think another word, such as "irradiance" should be used. -- but aren't these also used interchangably in some aspects of optics?!?!

What would the definition of intensity be in this case?

Would it be the number of photons passing a point per unit time or the total radiant energy of these photons passing a point per unit time or something else?

I explained it to my students as "intensity" referring to the number of photons striking a metal surface (and causing photoelectron emission) per unit time.

So for equal intensity, there would be the same number of photoelectrons emitted, but the one with lower wavelength will have higher KE, hence there will be greater photocurrent due to the rate of flow of charge.

The vast majority of texts and reliable websites which I have used agree with my opinion, but the occasional website says one of two things

1. frequency has no effect on photocurrent, but does effect stopping voltage (Of course i agree with the latter part, just not the first part)

2. increasing frequency results in less photoelectrons being emitted as intensity = (#photons * energy of photons) and seeing as energy increases with frequency, then # of photons must decrease to cause less photoelectron emission, hence lower current

I disagree with the definition of intensity here and also disagree with the fact that the current would be lower anyway.
Firstly, yes it would be lower if the emitted photoelectrons were travelling at the same speed, but they won't be, due to higher energy of a photon of lower wavelength light.

Is there an equation which quantifies the photocurrent produced for a metal surface? or any surface for that matter

Cheers
-Chris