MILLIKAN EXPERIMENT: PROCEDURE, EXPLANATION, IMPORTANCE - PHYSICAL - 2025

The Millikan experiment, carried out by Robert Millikan (1868-1953) together with his student Harvey Fletcher (1884-1981), began in 1906 and aimed to study the properties of electric charge, analyzing the movement of thousands of drops of oil in the midst of a uniform electric field.

The conclusion was that the electric charge did not have an arbitrary value, but came in multiples of 1.6 x 10 ^-19 C, which is the fundamental charge of the electron. In addition, the mass of the electron was found.

Figure 1. On the left the original apparatus used by Millikan and Fletcher in their experiment. On the right a simplified diagram of it. Source: Wikimedia Commons / F. Zapata, Previously, the physicist JJ Thompson had experimentally found the charge-mass relationship of this elementary particle, which he called "corpuscle", but not the values ​​of each magnitude separately.

From this charge - mass relationship and the charge of the electron, the value of its mass was determined: 9.11 x 10 ^-31 Kg.

To achieve their purpose, Millikan and Fletcher used an atomizer that sprayed a fine mist of oil droplets. Some of the droplets were electrically charged due to friction in the sprayer.

The charged drops were slowly settling on parallel flat plate electrodes, where a few passed through a small hole in the upper plate, as shown in the diagram in figure 1.

Inside the parallel plates it is possible to create a uniform electric field perpendicular to the plates, whose magnitude and polarity were controlled by modifying the voltage.

The behavior of the drops was observed by illuminating the interior of the plates with bright light.

Explanation of the experiment

If the drop has a charge, the field created between the plates exerts a force on it that counteracts gravity.

And if it also manages to remain suspended, it means that the field exerts an upward vertical force, which exactly balances gravity. This condition will depend on the value of q, the charge of the drop.

Indeed, Millikan observed that after turning on the field, some drops were suspended, others began to rise or continued to descend.

By adjusting the value of the electric field - through a variable resistance, for example - a drop could be made to remain suspended within the plates. Although in practice it is not easy to achieve, should it happen, only the force exerted by the field and gravity act on the drop.

If the mass of the drop is m and its charge is q, knowing that the force is proportional to the applied field of magnitude E, Newton's second law states that both forces must be balanced:

The value of g, the acceleration of gravity is known, as well as the magnitude E of the field, which depends on the voltage V established between the plates and the separation between these L, as:

The question was to find the mass of the tiny drop of oil. Once this is accomplished, determining the charge q is perfectly possible. Naturally, m and q are respectively the mass and the charge of the oil drop, not the electron.

But… the drop is charged because it loses or gains electrons, so its value is related to the charge of said particle.

The mass of the oil drop

Millikan and Fletcher's problem was to determine the mass of a drop, not an easy task due to its small size.

Knowing the density of the oil, if you have the volume of the drop, the mass can be solved. But the volume was also very small, so conventional methods were of no use.

However, the researchers knew that such small objects do not fall freely, since the resistance of the air or the environment, intervenes, slowing down their movement. Although the particle, when released with the field turned off, experiences an accelerated vertical movement and downwards, it ends up falling with constant speed.

This speed is called "terminal velocity" or "limit velocity", which, in the case of a sphere, depends on its radius and the viscosity of the air.

In the absence of a field, Millikan and Fletcher measured the time it took for the drops to fall. Assuming that the drops were spherical and with the value of the viscosity of the air, they managed to determine the radius indirectly from the terminal velocity.

This speed is found by applying Stokes' law and here is its equation:

- v _t is the terminal velocity

- R is the radius of the drop (spherical)

- η is the viscosity of air

- ρ is the density of the drop

Importance

Millikan's experiment was crucial, because it revealed several key aspects in Physics:

I) The elemental charge is that of the electron, whose value is 1.6 x 10 ^-19 C, one of the fundamental constants of science.

II) Any other electrical charge comes in multiples of the fundamental charge.

III) Knowing the charge of the electron and the charge-mass relationship of JJ Thomson, it was possible to determine the mass of the electron.

III) At the level of particles as small as elementary particles, the gravitational effects are negligible compared to the electrostatic ones.

Figure 2. Millikan in the foreground on the right, alongside Albert Einstein and other notable physicists. Source: Wikimedia Commons.

Millikan received the Nobel Prize in Physics in 1923 for these discoveries. His experiment is also relevant because he determined these fundamental properties of electric charge, starting from a simple instrumentation and applying laws well known to all.

However, Millikan was criticized for having discarded many observations in his experiment, for no apparent reason, to decrease the statistical error of the results and make them more "presentable".

Drops with variety of charges

Millikan measured many, many drops in his experiment and not all of them were oil. He also tried mercury and glycerin. As stated, the experiment began in 1906 and lasted for a few years. Three years later, in 1909, the first results were published.

During this time, he obtained a variety of charged drops by striking X-rays through the plates, to ionize the air between them. In this way charged particles are released that the drops can accept.

Furthermore, he did not focus solely on the suspended droplets. Millikan observed that when the drops rose, the rate of rise also varied according to the load delivered.

And if the drop descended, this extra charge added thanks to the intervention of the X-rays, did not change the speed, because any mass of electrons added to the drop is miniscule, compared to the mass of the drop itself.

Regardless of how much charge it added, Millikan found that all the drops acquired charges that were integer multiples of a certain value, which is e, the fundamental unit, which as we have said is the charge of the electron.

Millikan initially obtained 1,592 x 10 ^-19 C for this value, slightly lower than the currently accepted value, which is 1,602 x 10 ^-19 C. The reason may have been the value that he gave to the viscosity of air in the equation for determine the terminal velocity of the drop.

Example

Levitating a drop of oil

We see the following example. An oil droplet has a density ρ = 927 kg / m ³ and is released in the middle of the electrodes with the electric field off. The droplet quickly reaches terminal velocity, whereby the radius is determined, the value of which turns out to be R = 4.37 x10 ^-7 m.

The uniform field turns on, is directed vertically upward, and has magnitude 9.66 kN / C. In this way it is achieved that the drop remains suspended at rest.

It asks:

a) Calculate the droplet charge

b) Find how many times the elemental charge is contained in the charge of the drop.

c) Determine if possible, the sign of the load.

Figure 3. An oil droplet in the middle of a constant electric field. Source: Fundamentals of Physics. Rex-Wolfson.

Solution to

Previously, the following expression was derived for a drop at rest:

Knowing the density and radius of the drop, the mass of the drop is determined:

Thus:

Therefore, the charge of the drop is:

Solution b

Knowing that the fundamental load is e = 1.6 x 10 ^-19 C, divide the load obtained in the previous section by this value:

The result is that the charge on the drop is approximately twice (n≈2) the elemental charge. It is not exactly double, but this slight discrepancy is due to the inevitable presence of experimental error, as well as rounding in each of the previous calculations.

Solution c

It is possible to determine the sign of the charge, thanks to the fact that the statement gives information about the direction of the field, which is directed vertically upwards, as well as the force.

Electric field lines always start with positive charges and end with negative charges, therefore the lower plate is charged with a + sign and the upper plate with a - sign (see figure 3).

Since the drop is directed towards the plate above, driven by the field, and since charges of opposite sign attract each other, the drop must have a positive charge.

Actually keeping the drop suspended is not easy to achieve. So Millikan used the vertical displacements (ups and downs) that the drop experienced by turning the field off and on, plus the changes in X-ray charge and travel times, to estimate how much extra charge the drop had acquired.

This acquired charge is proportional to the charge on the electron, as we have already seen, and can be calculated with the rise and fall times, the mass of the drop and the values ​​of g and E.

References

1. Open Mind. Millikan, the physicist who came to see the electron. Recovered from: bbvaopenmind.com
2. Rex, A. 2011. Fundamentals of Physics. Pearson.
3. Tippens, P. 2011. Physics: Concepts and Applications. 7th Edition. McGraw Hill.
4. Amrita. Millikan's oil drop experiment. Retrieved from: vlab.amrita.edu
5. Wake Forest College. Millikan's oil drop Experiment. Recovered from: wfu.edu