
In Millikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is $1.282\times {{10}^{-18}}C$ Calculate the number of electrons present on it.
Answer
505.8k+ views
Hint: A force is exerted by a matter when it is placed in an electromagnetic field. In an electromagnetic field there are two types of electric charge present named as positively and negatively charged. Like charges repel each other while unlike charges attract each other.
Complete Solution :
- In the year of 1909 Millikan gave an experiment which is known by the name oil drop experiment. In this experiment a simple apparatus is used in which actions of all forces i.e. electric, gravitation and air drag forces balances each other and this is used to determine the electric charge present on many of the droplets in an oil mist. The force on any electric charge in an electric field is given by the product of the charge and the electric field.
Charge carried by the oil drop = $1.282\times {{10}^{-18}}C$
Charge of electron = $1.6022\times {{10}^{-19}}C$
\[\therefore \] Electrons present on the oil drop carrying $1.282\times {{10}^{-18}}C$ charge
= $\dfrac{Charge~carried~by~the~oil~drop}{Charge~carried~by~an~electron}$
= $\dfrac{1.282\times {{10}^{-18}}}{1.6022\times {{10}^{-19}}}$
= $0.8001\times 10$
= $8.0$
Thus the number of electrons on the surface is $8.0$
Note: Electric charge is said to be conserved in nature which means the net charge of the isolated system will not change. The SI unit of electric charge is coulomb which is represented by C. The main advantage of Millikan’s oil experiment is it tells us that it is necessary for establishing the charge of an electron. In this experiment rather than other liquids oil is chosen because it resists its mass over a while and exposes to higher temperature.
Complete Solution :
- In the year of 1909 Millikan gave an experiment which is known by the name oil drop experiment. In this experiment a simple apparatus is used in which actions of all forces i.e. electric, gravitation and air drag forces balances each other and this is used to determine the electric charge present on many of the droplets in an oil mist. The force on any electric charge in an electric field is given by the product of the charge and the electric field.
Charge carried by the oil drop = $1.282\times {{10}^{-18}}C$
Charge of electron = $1.6022\times {{10}^{-19}}C$
\[\therefore \] Electrons present on the oil drop carrying $1.282\times {{10}^{-18}}C$ charge
= $\dfrac{Charge~carried~by~the~oil~drop}{Charge~carried~by~an~electron}$
= $\dfrac{1.282\times {{10}^{-18}}}{1.6022\times {{10}^{-19}}}$
= $0.8001\times 10$
= $8.0$
Thus the number of electrons on the surface is $8.0$
Note: Electric charge is said to be conserved in nature which means the net charge of the isolated system will not change. The SI unit of electric charge is coulomb which is represented by C. The main advantage of Millikan’s oil experiment is it tells us that it is necessary for establishing the charge of an electron. In this experiment rather than other liquids oil is chosen because it resists its mass over a while and exposes to higher temperature.
Watch videos on
In Millikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is $1.282\times {{10}^{-18}}C$ Calculate the number of electrons present on it.

Structure of Atom NCERT EXERCISE 2.39 | Class 11 Chemistry | CBSE 2024
Subscribe
likes
2K Views
1 year ago
Recently Updated Pages
NCERT Solutions For Class 6 Maths Prime Time Exercise 5.6

NCERT Solutions For Class 6 Maths Lines And Angles Exercise 2.3

NCERT Solutions For Class 11 Maths Sets Exercise 1.2

NCERT Solutions For Class 12 Maths Linear Programming Exercise 12.1

Science Models For Class 8

NCERT Solutions For Class 2 English Marigold - The Wind And The Sun

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE
