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 (because the pressure is still greater below the object than above it due to the exponential distribution of air); drag is now acting downwards because the oil droplet is rising, and weight continues to act downwards due to the effect of gravitational force. Substituting eq2, eq3 (with v₂instead of v₁), eq4 and eq7 in eq 11 yields

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

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

Substituting eq5 in eq12 gives

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

Substituting eq8 in eq13 and rearranging results in

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

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) into eq14, Millikan determined a value for q. He repeated the experiment multiple times, each time varying the strength of the X-rays used to ionise 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 this value represents the fundamental unit of charge, e. Subsequent determinations refined that value to 1.602176487 x 10⁻¹⁹ C.

Because eq. 14 is derived from force balance conditions that determine only the magnitude of charge on the droplets, the computed values of q are positive. However, electrons by convention carry a negative charge, q_e= –e, consistent with their attraction to positive electrodes.

Using the Faraday constant and the value of the magnitude of the charge of a single electron that he determined, Millikan calculated the Avogadro constant and showed that one Faraday is the total quantity of charge carried by one mole of electrons.

Question

Why is the magnitude of the charge of a single electron used to calculate the Avogadro constant?

Answer

The Avogadro constant represents the number of particles in a mole, which is a count and therefore positive. Since the Faraday constant is defined as the magnitude of charge per mole of electrons, the magnitude of the charge of a single electron is used in the calculation.

Millikan’s oil drop experiment: task 3

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