Millikan’s oil drop experiment: task 3

Next, V is turned up slightly so that the oil droplet rises with a new terminal velocity v₂.

We have:

$F_{e}+F_{u}=F_{d}+F_{g}\; \; \; \; \; \; \; (11)$

Note that the electric force is acting upwards (since the oil droplet is rising); upthrust is also acting upwards (since the pressure is still greater below the object than above it due to the exponential distribution of air); drag is now acting downwards since the oil droplet is rising and weight remains acting downwards due to the effect of gravitational force. Substituting eq2, eq3 (with v₂instead of v₁), eq4 and eq7 in eq 11, we have

$qE+\frac{4\pi }{3}r^{3}\rho _{air}g=6\pi \eta rv_{2}+\frac{4\pi }{3}r^{3}\rho g$

$qE-\frac{4\pi }{3}r^3(\rho -\rho _{air})g=6\pi \eta rv_{2}\; \; \; \; \; \; \; (12)$

Substituting eq5 in eq12:

$qE-6\pi r\eta v_{1}=6\pi \eta rv_{2}\; \; \; \; \; \; \; (13)$

Substituting eq8 in eq13 and rearranging:

$q=\frac{6\pi r\eta (v_{1}+v_{2})d}{V}\; \; \; \; \; \; \; (14)$

Once again, Millikan recorded the distance travelled by the rising oil droplet over a period of time and calculated the new terminal velocity, v₂.

Substituting the values of v₁, v₂and the calculated value of r (from eq6) in eq14, Millikan determined a value for q. He repeated the experiment multiple times, each time varying the strength of X-rays ionizing the air, resulting in a varying number of electrons attaching to the oil droplet. He then obtained various values of q and found them to be multiples of 1.5924 x 10^-19C. Millikan concluded that the value of 1.5924 x 10^-19C is equal to the charge of a single electron. Subsequent determinations refined that value to 1.602176487 x 10⁻¹⁹ C.

Using the Faraday constant and the value of the charge of a single electron that he determined, Millikan calculated the Avogadro constant and proved that one Faraday is the quantity of charge for one mole of electrons.

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