How efficient is the Crookes radiometer? I remember many years ago, I think at 8th grade, seeing the teacher show us a Crookes radiometer. I remember it being very fascinating. Today I read the wiki article on it, after looking up what it was called, but the article wasn't very clear in my opinion. Essentially the molecules that hit the dark sides have more energy and thus exert pressure which causes the device to rotate. And something about Einstein...
One thing lacking in the article was that it didn't clarify how effective this device is. It seems to me that relatively little energy input results a rather striking output (at least subjectively). So if I heat such a device with, say, $100 \ W$, how many watts would the output be in comparison?
Could such a device be built in large scale and used to convert solar energy to movement?
 A: It is hard to make a precise determination of the efficiency, but we can make a Fermi estimate that it is of order $10^{-2}$.
Efficiency is the ratio of output to input, but even defining the input here is not easy.  A 100 watts of heat is mentioned, but what does that mean?   Is 100 watts of light shining on one of the radiometer vanes or is 100 watts of heat applied over the radiometer's glass envelope? (The current explanation for the radiometer rotation are thermal transpiration forces that can be driven by either heating or cooling
the radiometer.)
For optimistic simplicity, let's assume that the input energy is just the bright sunshine hitting the vanes. Sunshine on the earth's surface on a clear day with the sun directly overhead is about $1\, \textrm{kW/m}^2$, but since the vane is not perpendicular to the sunlight, let's call it $0.5\,\textrm{kW/m}^2$.  Typical vanes look a bit bigger than a centimeter in size, so let's assume an area of about $2\,\textrm{cm}^2$, and the power on one vane is $\sim 0.1\,\textrm{W}$. A typical radiometer has 4 vanes, however, but on average only two will be facing the sun, so we'll take the input power to be
$$P_{in}\sim 0.2\,\textrm{W}$$
When a radiometer is placed in the sunshine, it starts to rotate and speeds up until it reaches its terminal rotational velocity where the force driving the rotation matches the gas drag from the residual gas in the enclosure. We'll assume the frictional losses from the central axis pin are negligible.
The question asks about converting solar energy to movement, so a plausible efficiency is ratio of mechanical rotational power to input power.  The rotational energy of the light mill is roughly
$$E \approx 4 \frac{1}{2}m r^2 \omega^2$$
where $m$ is the mass of one of the 4 vanes, r is the radius of rotation of the centre-of-mass of the vanes, and $\omega$ is the angular velocity.  The vanes are thin metal, so $4m\sim 1\,\textrm{g}$ seems reasonable, and my eyeball estimate is that $r\sim 2\,\textrm{cm}$.  We just need the angular velocity to estimate the kinetic energy of the light mill.
There are many videos of Crookes radiometers online, and even though some of the narration is inaccurate, I will take this Radiometer Demonstration as a reference. (I have since found other videos such as this or this or this or
this or
this that are consistent.) When placed in the sunshine, you can can see the vanes accelerate, then apparently slowing down and reversing as they pass through the frame rate of the camera, which for a typical phone is $\sim 30\,\textrm{fps}$ (frames per second). This is consistent the claims from the many sales sites that radiometers can spin at 3000 rpm (revolutions per minute).
In 1876 Crookes himself reported even more impressive results:

The speed with which a sensitive radiometer will revolve in full sunshine is almost incredible; nothing is apparent but an undefined nebulous ring, which becomes at times almost invisible. The number of revolutions per second cannot be counted : but it must be several hundreds; for one candle will make the arms spin round forty times a second."

My guess is that the amazing speed for a single candle was because the candle was so close to the radiometer that it greatly heated the glass.
Let's assume a revolution speed of $\,\textrm{Hz}$, corresponding to angular velocity of $\sim 200\,\textrm{radians/second}$. The estimated rotational kinetic energy is then
$$E \sim 4 \frac{1}{2} (0.001\,\textrm{kg}) (0.02\,\textrm{m})^2 (200\,\textrm{rad/s})^2 = 0.03\,\textrm{J}$$
We can estimate the mechanical power by noting that in the video it took the order of 10 seconds for the sunshine to drive the vanes up to full speed. Dividing the final energy by the time it takes to reach that energy roughly gives us the driving power.  For this Fermi estimate, the details of the ramp (e.g. exponential vs linear, factors of 2 or e) are not significant, so our very rough estimate for the mechanical output power of the radiometer is
$$P_{out}\sim \frac{0.03\,\textrm{J}}{10\,\textrm{s}}=0.003\,\textrm{W}$$
so the efficiency estimate is
$$\frac{P_{out}}{P_{in}} \sim \frac{0.003\,\textrm{W}}{0.2\,\textrm{W}} \sim 1.5\%\sim 10^{-2}$$
This is not great, and the effective efficiency would be much, much worse if instead of only considering the sunshine hitting the vanes for $P_{in}$, we included all the light averaged over the radiometer enclosure.
This low efficiency may be why the various patents for actual Crookes generators such as Radiometer generator or
Solar Electric Generator, have not been commercially successful (as far as I am aware).
Although there is a large literature on Crookes radiometers, I had trouble finding papers with sufficient information (e.g. illumination, rotation speed, torque) to calculate an efficiency. The above estimate is, however, the same order-of-magnitude as one experimental result for a much smaller radiometer under weaker illumination.
For roughly $4\,\textrm{mm}^2$ vanes, $1.2\,\textrm{mW/cm}^2$ illumination (i.e. $P_{in}\sim 1\times 10^{-4}\,\textrm{W}$ using a 2 vane effective area), Han 2010 achieved about 1000 rpm and an output torque $\tau$ about 4 nN m, giving a rotational power of $P_{out}=\omega \tau \sim 4\times 10^{-7}\, \textrm{W}$ and an efficiency $\sim 0.4\%$.
