Millikan's experiment

MILLIKAN'S EXPERIMENT

- Computer programm - experiment simulation
In the year 1896 John Thomson discovered electron and gave the value of its charge to its mass. However a year later he tried to reckon elementary charge (a bit earlier Townsend also tried to calculate it and the result he got was 1*10^-19 coulombs). Thomson's experiment was conducted using Wilson's cloud chamber (this instrument due to the process of condensation let us observe small particles, even ions) and let him reckon the e quantity (elementary charge's value) but with quite a big error. According to his experiment e had to be equal about 2,2*10^-19 coulombs. After conducting a modificated experiment he got new issue e equals about 1,1*10^-19 coulombs.
The precise measurement of e, which made possible to prove that all electrons had the same charge wasn't done till 1911. That was gained by Robert Andrews Millikan. To achieve this he constructed a special device (picture no. 1).
The diagram of Millikan's experiment

    Using a atomiser he generated oil droplets above two parallel plates. The droplets were falling through a small hole in the upper, isolated plate to the space between the plates. Droplets could be observed at one side through a short focal distance telescope. To make droplets more visible he illuminated them from one side. The field of view was intersected by two horizontal, parallel, thin lines. Droplets of oil could electrify themselves owing to friction when they were atomised. They could also get the charge by X-raying (picture no. 1). Millikan put some electric potential difference on plates generating electric field between them.
    If the drop has its own mass m and the charge q then some forces effects on it - downward Earth gravitation m*g, electrical force E*q (where E is the intensity of the electric field between the plates) downwarder or upwarder depending on field's direction. Millikan assumed that edge effect occurring on plates decide about the intensity of the electric field. Millikan could change that potential difference using voltage divider (picture no. 2).
    Where there is no electric field then drop fells down with a constance speed v₀ depending on medium's resistance. According to Stokes's formula v₀ depends on the radius a drop's density f, the intensity of the gravitation field g, medium's density f₁ and medium's viscosity x:

v0 = (2/9)*(g*a^2*(f-f1))/x

(1)
or

(2)
where k is constant for given drop:

k = v0/(m*g)

(3)
If one applies such electric field that its force increase the velocity of the drop then that velocity v₁ equals:

v1 = k*(m*g + E*q)

(4)
Millikan determined both velocities measuring times in which the drop covered the distance between two horisontal lines (table 1).

Table 1

v_e		Number
0,0417
	+0,0003	1
0,0420
	-0,0181	2
0,0239
	-0,0090	3
0,0149
	-0,0087	4
0,0062
	-0,0001	5
0,0061
	+0,0090	6
0,0151
	-0,0001	7
0,0150
	+0,0176	8
0,0326
	-0,0176	9
0,0150
	+0,0001	10
0,0150
	+0,0187	11
0,0238

Comparing the values of v₁ from different measurements it's clear that some of them are different, so the forces of the electric field must have changed. And so the chart of the drop must have changed. Writing two equations for velocities of two drops:

v11 = k*(m*g + E*q1)

(5)

(6)
and substracting them by sides we get:

v11 - v21 = k*(E q1 - E q2)

(7)
and after transformation:

q2 - q1 = (1/(k*E))*(v21 - v11)

(8)
Looking at the table no.1 one may see that in cases ordered by numbers 1, 5, 7, 10 drop's charges didn't change. Numbers 3, 4 and 6 had identical changes of charges. The quantity of the change is the elementary charge - e. In cases 2 and 9 the quantity of the charge decrease by 2*e, and in cases 8 and 11 it increase by 2*e. And so, the charge of the drop always changed abrupt by e. This observation Millikan confirmed experimented with hundreds of drops. He never found the change of charge which would equal to fractional e. The experiment proved that the charge on the oil drop changed always by ne quantity where n is a positive or negative integer. He experimented using different kinds of drops in different mediums. And the dependence was always the same. Knowing that dependence and the formula (3) it's possible to write the formula (8) in the form:

((m*g)/(E*v0))((v21 - v11) = n*e

(9)
therefore:

e = ((m*g)/(E*v0*n))((v21 - v11)

(10)
Millikan knew all components of the right side of the equation. All he had to do was substituting symbols by numerical values he got in the experiment. Unfortunately after calculation it appeared that for different drops e was also different. For smaller drops and with less air pressure e was bigger. Millikan analysed precisely his equations. He came to the opinion that Stoke's formula should be modified. It described well balls moving in viscous, continuos medium but it wasn't proper for describing air medium. Millikan decided to put some corrections into Stoke's formula. They refereed the radius of the drop a which in relation to used before radius a₁ changed in compliance with the equation:

a = a1*pierw(1/(1 + A(l/a))

(11)
where A is some constance equal 0,874; the way of finding A will be shown later.
If we put the equation m = 4/3((

)a³f)v (drop's mass) into the equation (10) we get:

e = (4*(pi)*a^3*f*g)/(3E)*((v21 - v11)/v0)*(1/n)

(12)
putting it into a equation calculated by equation (11) we get:

e = (4*(pi*a1^3*f)*g)/(3E)*((v21 - v11)/v0)*(1/n)*(pierw(1 + A(l/a)))^33

(13)
and so finally:

e = e1*(1 + A(l/a))^(-3/2)

(14)
where e₁ is a charge got from uncorrected formula.

Transforming no. 14 we get:

e1^(2/3) = (1 + A(l/a))*e^(2/3)

        (15)
    IIt issues, from that formula, that evaluating e₁^2/3 for different drops with the medium under different pressure one can plot a graph of the dependence of e^2/3 to (l/a) and the graph would be a straight line. Ae₁^2/3 equals to the tangens of the line inclination. The line cross the axis of ordinates in the point e^2/3. Thanks to that graph e and A may be calculated. But Millikan, wanting to calculate A, had to know the velocity of a, which dependence on A. He solved that problem finding the approximate velocity of A by plotting e₁^2/3 of l/a and putting an approximate, uncorrected quantity of the drop radius to the formula. Then he used that quantity to calculate the most exact quantity of a and e₁. Then he plot a graph basing on that corrected values and calculated a more exact quantity of A. The whole procedure he did twice and he got quite an exact value of A. Finally he calculated an approximate value of e (1,59*10^-19 coulombs). In 1914 he did the experiment again trying to find the most exact value of the elementary charge. It was o be equal 1,592+/-0,0017*10^-19 coulombs.
    Today we know that Millikan and his collaborates did an systematic error by calculating air viscosity and the exact e velocity is 1,6021773*10^-19 coulombs.
    Millikan proved that the change of the charge on oil droplets is always equal ne, where n and e is constant. The change of a drop charge, also positive and negative was always equal e. There are no two elementary charges smaller than e but he also didn't prove there are no such charges. The charges discovered by Millikan could for example consisted of the charges (1/2)e occurring in pairs.

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