Turning of a paddle wheel when exposed to light So, we observe that in the discharge tube experiment when a very light paddle wheel is placed in the path of cathode rays, it starts turning due to the particle nature of the rays.

My question is if the same paddle wheel is placed in light, why does it not start turning ? 
Light is also of particulate nature (in some cases- made up of photons).
Is it due to the less mass of photons?
Thanks.

Edit:-

Thank you, particularly @Poutnik for giving me such great answers.

I just read an article while researching that the paddle wheel didn't turn because of the transfer of momentum by the cathode rays but due to changes in pressure of the gas around the paddles due to heating, which is much same as that to Crookes radiometer.
This fact was discovered later, much later after the experiment.
This solves my problem that even in the cathode ray experiment, it wasn't the trasfer of momentum that moved the paddles, because the actual pressure of the rays is quite weak (again as suggested by @Poutnik). And since the ray can't turn the paddles, the light no way can.
And yes, I calculated the momentum of sunlight and that transferred by cathode rays to see for myself that movement of the paddle by each of these is not possible.

Thanks everyone once again.
Not the complete answer, but a short analysis of expectable forces:
The energy and momentum of light are related by equation $E=p \cdot c $.
Then force caused by light $F=\frac{\mathrm{d}p}{\mathrm{d}t}=\frac{\mathrm{d}E}{c \cdot \mathrm{d}t} = \frac Pc$
Therefore the light power 30 W (some cca 50-60W fluorescent bulb or cca 35-40 W LED bulb ) would cause direct  force cca  100 nN, equivalent of cca 0.01 mg.
Or twice as much if reflected.
Crookes radiometer shows the wheel is turning in the opposite direction than it should be, if caused by the light pressure.
The motion is caused by uneven heating and expansion of gas around the wheel, causing pressure differences in the opposite sides of paddles.
When pressure is too low, the gas there is not enough and there remains just light pressure, that is too weak to turn the wheel.

Quoting Effect observations from article.
The effect begins to be observed at partial vacuum pressures of several hundred pascals (or a few torr), reaches a peak at around $\pu{1 Pa}$ ($\pu{7.5e−3 torr}$) and has disappeared by the time the vacuum reaches $\pu{1e−4 Pa}$ ( $\pu{7.5e−7 torr}$) (see explanations note 1). At these very high vacuums the effect of photon radiation pressure on the vanes can be observed in very sensitive apparatus (see Nichols radiometer) but this is insufficient to cause rotation.

Nichols radiometer can register the true light pressure by mirrors on torsion balance.

For cathode rays = electrons
If there is
potential difference $U$
current $I$
elementary charge $e$
electron mass $m_e$
charge $Q$
Area $A$
pressure $P$ (avoiding confusion)
momentum of 1 Coulomb  $p$
electron momentum $p_e$
then for kinematics:
$$eU=\frac 12 m_e.v^2$$
$$v=\sqrt{\frac{2eU}{m_e}}$$
for momentum:
$$p_e=m_e.v=\sqrt{2m_eeU}$$
$$p=\frac 1e \cdot p_e=Q \sqrt{\frac{2m_eU}{e}}$$
for force and pressure
$$F=dp/dt= I \sqrt{\frac{2m_eU}{e}}$$
$$P=F/A= \frac IA \sqrt{\frac{2m_eU}{e}}$$
for $\pu{380 V}$ and $\pu{1 A}$ it gives force about $\pu{66e-6 N}$, what is equivalent to weight of the mass about $\pu{6.6 mg}$.
This is a model case. For a particular scenario, the geometry and electrostatic forces would play significant role.