# Charge of electron-Millikan’s oil drop experiment, Observation, Conclusion, Practice problems and FAQs

Imagine, if you were in 1897 and the charge to mass ratio of electrons was derived and the whole world was in the race of finding the value of charge of electrons, what would be your approach ?

Thomson’s experiments confirmed that electrons are thousand times lighter than the hydrogen atom and exist inside of atoms, but these experiments added nothing to the numerical value of the charge of electrons. Millikan developed a new set of experiments in which he applied an electric field and monitored the effect of electric field on motion of oil drops. He observed that charge on oil drops is always an integral multiple of charge of electron (e).

Millikan earned the Nobel Prize in physics (1923) for his work on the elementary charge of electricity and on the photoelectric effect.

## Experimental Setup

• It consists of two metal plates at a distance separated by an insulated rod.
• An atomizer dispersed very small oil droplets in the upper chamber.
• A telescope is placed to observe the droplets in the chamber.
• High voltage is applied across the plates.

## Milliken’s Oil Drop Experiment Procedure

• An atomizer sprays a fine cloud of oil drops into the upper section of the chamber.
• Due to the effect of air resistance and gravity, few oil droplets pass through a small hole in the top metal plate.
• X - rays were used to ionize the air present in the chamber.
• After ionization of air, electrons from the air attach with falling oil droplets, causing them to acquire a negative charge.
• With the help of a microscope downward motions of oil drops are observed and then terminal velocity acquired by oil drops are measured.
• The electric field is set up between the two plates and so the speed of charged oil droplets can be adjusting the potential difference, or voltage.
• The charged droplet remains stationary, when the upward electric force equals the downward gravitational force.
• At equilibrium, charges present on the oil drops are calculated. Which depends on the strength of the electric field and mass of the oil drops..

## Charge on electron:

• Through repeated application of this method, the values of the electric charge (q) on individual oil drops are always an integral multiple of the electrical charge (e), that is,

• Electric charges on individual oil drops are always whole number multiples of a lowest value, that value being the elementary electric charge itself (about -1.602 × 10-19 coulomb).

## Mass of electron:

The mass of the electron (me) was determined by combining these results with Thomson’s value of ratio.

## Practise Problems

Q1. The following charges (in units) were found on a series of oil droplets during an oil drop experiment: 7.5 × 10-18, 5 × 10-18, 10 × 10-18, 12.5 × 10-18, 15 × 10-18. calculate charge on electron (in mentioned unit) should be

(A) 5 × 10-18
(B) 15 × 10-18
(C) 2.5 × 10-18
(D) 1.6 × 10-19

Answer: (C), Magnitude of electrical charge, q, on the droplets is always an integral multiple of the electrical charge, e, that is, q = ne, where n = 1, 2, 3... .

q1= 7.5 × 10-18 unit = 3 × 2.5 × 10-18 unit

q2= 5 × 10-18 unit = 2 × 2.5 × 10-18 unit

q3= 10 × 10-18 unit = 4 × 2.5 × 10-18 unit

q4= 12.5 × 10-18 unit = 5 × 2.5 × 10-18 unit

q4= 15 × 10-18 unit = 6 × 2.5 × 10-18 unit

So, e = 2.5 × 10-18 unit

Q2. Calculate charge on one mole of electron

(A) -96472 C
(B) -1.602 × 10-19C
(C) 96472 C
(D) None of these

Solution: charge on 1 electron = -1.602 × 10-19 C

Charge on 1 mole of electron = -1.602 × 10-19C 6.022 × 1023

= -96472 C

Q3. During Millikan's oil drop experiment the charge on an oil drop is found to be -144.2 ×10-20C. The number of electrons present in single oil drop is

A. 3
B. 6
C. 9
D.  12

Solution: charge on 1 electron = -1.602 × 10-19 C

Q4. Which value of electron was calculated as by product after cathode ray discharge tube experiment and millikan’s oil drop experiment

A. Charge of electron
B. Specific charge of electron
C. Mass of electron’
D. All of these

Solution: from cathode ray discharge experiment, value of electrons was calculated and by Millikan oil drop experiment, charge of electrons was calculated. From the above values, mass of electrons was calculated.

Question 1. How do they ensure there is only one electron on an oil drop in the Millikan oil-drop experiment?
Answer: In the initial experiment there was an atomizer which could divide the drops to the smallest possible. Though it literally did not produce oil molecules. It all works on probability that atleast one drop would contain a single ion. At the beginning of experiment the starting electric fields are kept at their minimum powers. And with time the voltage is increased till a few drops start to rise. Then by adjusting voltages different terminal velocities can be achieved and e/m is calculated. Also there are chances that one would reach a point where e/m would turn out to be multiples of each other.

Repeating the experiment gave charge of electron to the accuracy of within 1% of the currently accepted value.

Question 2. What is the charge of an electron electrostatic unit?
Answer: esu is the unit of charge in the CGS unit. Magnitude of charge of electron =  4.8 × 10-10 esu

Question 3. Can charge be created or destroyed?
Answer: Electric charge is conserved. It can neither be created nor be destroyed.

Question 4. In Millikan's experiment, why did some oil drops rise and fall faster than others?
Answer: In a vacuum, all objects fall at the same speed when dropped from the same height. However, the mass and the charge are not uniformly distributed - so when you apply an electric field, the downward acceleration of all the drops is the same, but the upward acceleration (due to the field) is different. This means that all the drops have slightly different accelerations, determined by the magnitude of the electric field you are applying.

Related Topic

 Discovery of electron Thomson atomic model Atomic Number and Mass Number Rutherford atomic model Discovery of neutron Electronic Configuration