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 v2 instead 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 v1v2 and 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 10-19 C. Millikan concluded that the value of 1.5924 x 10-19 C 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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