Radiation and Bremstrahlung in Measurements of Planck's Constant

Most recent answer: 11/09/2014

Q:
Why the energy of photon radiation emitted by the electrons during their acceleration from relative rest to a certain velocity is ignored in the energy equations of both famous experiments devised a century ago for measuring the Planck’s constant (Millikan in 1914, respectively Duane, Hunt and others in 1917-1920), and why the energy of bremsstrahlung emitted by the accelerated electron when they come again at relative rest is ignored in Millikan’s experiment, although it is considered in the other experiment?
- Sava (age 17)
Belgravistan
A:

The second part of your question is easy. In Millikan's photoelectric experiment, Planck's constant is estimate from the maximum voltage hill that the electrons can climb and still reach the anode. Since those electrons are reaching the anode with no energy to spare, they lose essentially no Bremstrahlung energy on being absorbed. 

Why isn't it necessary to consider radiative loss in the photoelectric experiment as the electron leaves the metal? If the electron gradually sped up as it absorbed the incoming radiation it would emit some predictable outgoing radiation as it accelerated. A proper calculation would then take that into account. That's not what happens, however. The transition from the electron state in the metal plus an incoming photon to the state without the photon and with the electron heading away is an intrinsically quantum process. In that process, it's true that there are other ways that some of the original photon's energy can leave. Some can go off as sound waves (phonons), jiggling of atoms in the metal. Some can go off as lower energy photons, the process you're thinking of. These are random quantum events so some of the electrons leave without emitting phonons or photons. Thus the most energetic electrons haven't lost energy to those processes. Planck's constant is inferred from the voltage needed to stop all the electrons, including the most energetic ones, not just to reduce the current by stopping less energetic ones. 

You may wonder why you don't see the reverse process- the electron picking up a little extra energy from the phonons and photons that are around anyway at room temperature. That would give a little tail of electrons with extra energy beyond what they should have from the photon from your light source. These processes do indeed occur, but they make little difficulty for experiment. The reason is that the energies of the incoming photons are many times the characteristic thermal energy at room temperature. So they hardly fuzz up the current vs. photon frequency curve at all.

Mike W.

p.s. Now we'll have to get Google analytics to keep track of questions from fictional countries as well as real ones.


(published on 11/09/2014)