How did Thomson calculate the momentum of electron when he didn't know the mass of electron?

This was proven in 1903 by J. J. Thomson who calculated that the momentum of the electrons hitting the paddle wheel would only be sufficient to turn the wheel one revolution per minute.
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J. J. Thomson measured the charge to mass ratio of electron in year 1897, and R. K. Millikan measured charge of electron in year nearly about 1910, i.e., Thomson knew only the charge to mass ratio of electron in 1903.

Then how did he was able to determine the momentum of electron when he didn't knew the mass of electron?

• physics.princeton.edu/~mcdonald/examples/thomson.pdf – Zhe Aug 7 '19 at 14:16
• Another way to put it is that the computed radiation pressure is too small: en.wikipedia.org/wiki/Radiation_pressure – Zhe Aug 7 '19 at 14:17
• Recall that momentum is the product of mass and velocity. You don't need to know the mass or velocity of the objects hitting something and transferring momentum to something else - you just need to know the mass and velocity of what is being hit. – Jon Custer Aug 7 '19 at 15:06
• Might be a good question to ask at physics SE. – Buck Thorn Aug 7 '19 at 19:30
• R. K. Millikan? Robert Andrews Millikan? – Rodrigo de Azevedo Jun 27 '20 at 10:10

Thomson described the mechanical effect produced by cathodic rays on the first edition of its book Conduction of Electricity Through Gases (1903, pp. 501–502) and again later on the second edition (1906, pp. 629-630).

As you mentioned it refers to the Crookes experiment with the paddle wheel in which the electron beam collides with the mill coausing it to move (fig.below).

The electron beam carries a current ($$I$$) that can be measured and is equal to the number of electrons hitting the paddle every second (N) multiplied by the charge of the electron (e). $$I =N\cdot e$$ Considering that the electrons are much smaller than the paddle you can consider that they bounce back with the same velocity (like when you hit a wall with a ball) and therfore the momentum transferred to the paddle per unit time ($$p_{tot}$$) is twice the momentum of the electron beam.

The momentum of the electron beam per unit time can be calculated with the well known formula $$m\cdot v\cdot N$$ with N to account for the number of collisions per unit time.

Therefore the total momentum per unit time transferred to the mill is: $$p_{tot} = 2m\cdot v\cdot N$$ This is equivalent to (multiplied and divided by the same quantity $$e$$): $$p_{tot} = 2\cdot \frac{m}{e}\cdot v\cdot (N\cdot e)= 2\cdot (\frac{m}{e})\cdot v\cdot I$$

Thompson considered an hypothetical best case scenario with a very high current of $$\pu{10^{-5}A}$$ carried by the electrons and a very high velocity of $$\pu{10^8 m/s}$$.

These are the highest reasonable values he could think could be applied in the Crookes' experiments (he even mentioned that the actual current measured was only $$\pu{10^{-7}A}$$.

The current can be easily measured while the velocity can be estimated from various methods using magnetic and electric fields (i.e. see section Charge-to-mass ratio of the Wikipedia where v is derived). And $$\frac{m}{e}$$ was already determined (around $$\pu{6e-12Kg/C}$$.

With those data, you can estimate that in the best case scenario (very high velocity and many particles hitting the paddle per unit time) the momentum transferred to the paddle per unit time is: $$p_{tot} = 2\cdot (\frac{m}{e})\cdot v\cdot I = 2\cdot 6\cdot 10^{-12}\cdot 10^8\cdot 10^{-5}= \pu{1.2e-8 N}$$ This corresponds to $$\pu{1.2e-3 Dyne}$$ in the cgs units (units used by Thompson).

If you consider a paddle of $$\pu{1cm2}$$ then the pressure on the paddle is $$\pu{1.2e-4 Pa}$$, that is extremely small (an atmosphere is around $$\pu{101 325 Pa}$$!). Because of this it is reasonable to attribute the movement to heating transfer.

As you see here, Thompson couldn't calculate the momentum defined as $$m\cdot v$$ but instead he calculated the momentum per unit time (that is a force)! He also considered the momentum carried by many electrons so that he could use the current in the formula of $$p_{tot}$$.