Crooke’s Rradiometer
Recoiling Photoelectrons
In my contribution to CNPS 2016 “Physics without Paradoxes” I have made a statement that the effect in Crooke’s radiometer depends on a recoil effect from photoelectrons emitted from the black (most absorbing) side of the radiometer. However, an alternative interpretation has been suggested. Since I have problems accepting this alternative I will explain my motivations in more detail.
The alternative interpretation states that air (of very low pressure) has higher temperature in front of the black surface. This means that temperature rotates with the same frequency as the radiometer. However, I have 2 arguments against this interpretation. The first argument is that the temperature difference between 2 sides of a thin metal is small and also that both sides have almost the same emission factor at 300 K (or 10 mikro-meters). Therefore, thermal emission is almost symmetric. The second argument is the extremely good transmission of 300 K radiation in air. On a clear night the temperature is falling rapidly due to the good transmission through many kilometers in normal atmosphere. The transmission through a few centimeters of extremely low pressure air would be very good. These 2 arguments together means that temperature changes rotating inside the radiometer are not very plausible and this interpretation therefore is unrealistic.
I conclude therefore that Crooke’s radiometer is best explained by recoil from photoelectrons.
John-Erik Persson
Hi John-Erik, can you go in to a bit more detail about recoil from photoelectrons? Are you saying that the black side of the radiometer emits electrons stimulated by the incoming light? Does that imply that radiometers would not rotate in perpetuity if light was shone on them?
Rhoroderick
I think that the black surface absorbs about an order of magnitude more light than the white surface. The number of emitted photoelectrons is much larger, and this explains the direction of rotation much better than the official version. This explains also why the effect disappears for very low and very high pressures.
I think that if the pressure is suitable, probably this closed current can continue for ever.
It would be of interest if you could find a method to separate between this idea and the official explanation.
Regards ____________ John-Erik