In this article, we shall study to solve problems on the calculations of the number of electrons, protons, and neutrons in atoms, molecules, and species.

Example 01:

Calculate the charge and mass of 1 mole of electrons.

Solution:

1 mole of electron corresponds to 6.023 x 10²³ electrons

Mass of 1 electron = 9.1 x 10^-31 kg

Mass of one mole of electron = 9.1 x 10^-31 x 6.022 x 10²³ = 5.48 x 10^-7 kg

Charge of 1 electron = 1.602 x 10^-19 C

Charge on one mole of electron = 1.602 x 10^-19 x 6.022 x 10²³ = 9.65 x 10⁴ C

Ans: The mass of one mole of electrons is 5.48 x 10^-7 kg and the charge on one mole of electrons is 9.65 x 10⁴ C

Example 02:

Calculate the total number of electrons in 1 mole of ammonia (NH₃).

Solution:

Nitrogen (N):

Atomic number = Z = 7

Number of electrons = Atomic number = 7

Hydrogen (H):

Atomic number = Z = 1

Number of electrons in 1 atom of hydrogen = Atomic number = 1

Number of electrons in 3 hydrogen atoms = 1 x 3 = 3

Number of electrons in 1 molecule of ammonia = 7 + 3 = 10

1 mole of ammonia contains 6.022 x 10²³ molecules of ammonia

Number of electrons in 1 mole of ammonia = 10 x 6.022 x 10²³ = 6.022 x 10²⁴

Ans: The number of electrons in 1 mole of ammonia = 6.022 x 10²⁴

Example 03:

Calculate the total number of electrons in 1 mole of methane (CH₄).

Solution:

Carbon (C):

Atomic number = Z = 6

Number of electrons = Atomic number = 6

Hydrogen (H):

Atomic number = Z = 1

Number of electrons in 1 atom of hydrogen = Atomic number = 1

Number of electrons in 4 hydrogen atoms = 1 x 4 = 4

Number of electrons in 1 molecule of ammonia = 6 + 4 = 10

1 mole of ammonia contains 6.022 x 10²³ molecules of ammonia

Number of electrons in 1 mole of ammonia = 10 x 6.022 x 10²³ = 6.022 x 10²⁴

Ans: The number of electrons in 1 mole of ammonia = 6.022 x 10²⁴

Example 04:

Find the total number and the total mass of neutrons in 7 m of ¹⁴C. Mass of neutron = 1.675 x 10^-27 kg.

Given mass of carbon = 7 mg = 7 x 10^-3 g

Molecular mass of carbon = 14 g

Number of moles of carbon = (7 x 10^-3)/14 = 5 x 10^-4  mol

Number of carbon atoms in 7 mg of ¹⁴C = 5 x 10^-4 x 6.022 x 10²³ = 3.011 x 10²⁰

Number of neutrons in 1 atom of carbon = A – Z = 14 – 6 = 8

Number of neutrons in 7 mg of ¹⁴C = 8 x 3.011 x 10²⁰ = 2.4088 x 10²¹

Mass of 1 neutron = 1.675 x 10^-27 kg.

Mass of neutron in 7 mg of ¹⁴C = 1.675 x 10^-27 x 2.4088 x 10²¹

Mass of neutron in 7 mg of ¹⁴C = 4.0347 x 10^-6 kg

Example 05:

Find total number and total mass of proton in 34 mg of ammonia at STP. Will the answer change if temperature and pressure are changed?

Given mass of ammonia = 34 mg = 34 x 10^-3 g

Molecular mass of ammonia = 14 + 3 = 17 g

Number of moles of ammonia = (34 x 10^-3)/17 = 2 x 10^-3 mol

Nitrogen (N):

Atomic number = Z = 7

Number of protons = Atomic number = 7

Hydrogen (H):

Atomic number = Z = 1

Number of protons in 1 atom of hydrogen = Atomic number = 1

Number of protons in 3 hydrogen atoms = 1 x 3 = 3

Number of protons in 1 molecule of ammonia = 7 + 3 = 10

Number of molecules in 34 mg of ammonia = 2 x 10^-3 x 6.022 x 10²³ = 1.2044 x 10²¹

Number of protons in 34 mg of ammonia = 10 x 1.2044 x 10²¹= 1.2044 x 10²²

Mass of 1 proton = 1.673 x 10^-27 kg.

Mass of protons in 34 mg of ammonia = 1.673 x 10^-27 x 1.2044 x 10²²

Mass of protons in 34 mg of ammonia = 2.015 x 10^-5 kg

Due to change in temperature and pressure, there is no change in number of moles of the gas. Hence there is no effect o change of temperature and pressure.

Example 06:

Calculate the number of electrons which will together weigh 1 gram.

Mass of 1 electron = 9.1 x 10^-31 kg = 9.1 x 10^-28 g

Number of electrons in 1 gram = 1/(9.1 x 10^-28) = 1.098 x 10²⁷ electrons

Example 07:

2 x 10⁸ atoms of carbon are arranged side by side, calculate the radius of carbon atom, if the length of this arrangement is 2.4 cm

Diameter of carbon atom = 2.4/(2 x 10⁸) = 1.2 x 10^-8  cm

Radius of carbon atom = (1.2 x 10^-8)/2 = 6 x 10^-9 cm = 6 x 10^-11 m = 0.6 x 10^-10  m = 0.6 angstrom

Example 08:

Find the total number and total mass of neutrons in 18 mL of water. The specific gravity of water = 1.

Given volume of water = 18 mL = 18 x 10^-3 L

Mass of water = Volume x density = 18 x 10^-3 L x 1 kg/L = 18 x 10^-3 kg = 18 g

Molecular mass of water= 2 + 16 = 18 g

Number of moles of water = 18/18 = 1 mol

Oxygen (O):

Atomic number = Z = 8

Atomic mass number = A = 16

Number of neutrons = 16 – 8 = 8

Hydrogen (H):

Atomic number = Z = 1

Atomic mass of hydrogen = A = 1

Number of neutrons in 1 atom of hydrogen = 1 – 1 = 0

Number of neutrons in 2 hydrogen atoms = 0 x 2 = 0

Number of neutrons in 1 molecule of water = 8 + 0 = 8

Number of molecules in 18 mL of water = 1  x 6.022 x 10²³ = 6.022 x 10²³

Number of neutrons in 18 mL of water = 8 x 6.022 x 10²³= 4.8176 x 10²⁴

Mass of 1 neutron = 1.675 x 10^-27 kg.

Mass of neutrons in 18 mL of water = 1.675 x 10^-27 x 4.8176 x 10²⁴

Mass of neutrons in 18 mL of water = 8.0695 x 10^-3 kg

Due to change in temperature and pressure, there is no change in number of moles of the gas. Hence there is no effect o change of temperature and pressure.

Science > Chemistry > Atomic Structure > Problems on Calculation of Number of Electrons, Protons, and Neutrons

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In this article, we shall study to solve problems based on the calculation of atomic number, atomic mass number, and neutron number

Atomic number (Z) :

The number of protons (positive charge) present in the nucleus of an atom of a particular element is called the atomic number of that element. It is denoted by letter ‘Z’.

Neutron number (N): 

The number of neutrons present in the nucleus of an atom is known as neutron number. It is denoted by ‘N’

Mass number (A): 

The total number of protons and neutrons present in the nucleus of an atom of the element is called mass number. The mass number is denoted as ‘A’.

A  =   Z + N

Example 01:

Calculate the number of electrons, protons, and neutrons in the following atoms.

Atomic Number = Z = 15

Atomic Mass Number = A = 31

Number of protons = atomic number = 15

Number of electrons = atomic number = 15

Neutron number = Number of neutrons = A – Z = 31 – 15 = 16

Number of nucleons = Atomic mass number = 31

Atomic Number = Z = 12

Atomic Mass Number = A = 24

Number of protons = atomic number = 12

Number of electrons = atomic number = 12

Neutron number = Number of neutrons = A – Z = 24 – 12 = 12

Number of nucleons = Atomic mass number = 24

Atomic Number = Z = 17

Atomic Mass Number = A = 37

Number of protons = atomic number = 17

Number of electrons = atomic number = 17

Neutron number = Number of neutrons = A – Z = 37 – 17 = 20

Number of nucleons = Atomic mass number = 37

Atomic Number = Z = 18

Atomic Mass Number = A = 40

Number of protons = atomic number = 18

Number of electrons = atomic number = 18

Neutron number = Number of neutrons = A – Z = 40 – 18 = 22

Number of nucleons = Atomic mass number = 40

Atomic Number = Z = 35

Atomic Mass Number = A = 80

Number of protons = atomic number = 35

Number of electrons = atomic number = 35

Neutron number = Number of neutrons = A – Z = 80 – 35 = 45

Number of nucleons = Atomic mass number = 80

Atomic Number = Z = 20

Atomic Mass Number = A = 40

Number of protons = atomic number = 20

Number of electrons = atomic number = 20

Neutron number = Number of neutrons = A – Z = 40 – 20 = 20

Number of nucleons = Atomic mass number = 40

Atomic Number = Z = 6

Atomic Mass Number = A = 13

Number of protons = atomic number = 6

Number of electrons = atomic number = 6

Neutron number = Number of neutrons = A – Z = 13 – 6 = 7

Number of nucleons = Atomic mass number = 13

Atomic Number = Z = 8

Atomic Mass Number = A = 16

Number of protons = atomic number = 8

Number of electrons = atomic number = 8

Neutron number = Number of neutrons = A – Z = 16 – 8 = 8

Number of nucleons = Atomic mass number = 16

Atomic Number = Z = 26

Atomic Mass Number = A = 56

Number of protons = atomic number = 26

Number of electrons = atomic number = 26

Neutron number = Number of neutrons = A – Z = 56 – 26 = 30

Number of nucleons = Atomic mass number = 56

Atomic Number = Z = 38

Atomic Mass Number = A = 88

Number of protons = atomic number = 38

Number of electrons = atomic number = 38

Neutron number = Number of neutrons = A – Z = 88 – 38 = 50

Number of nucleons = Atomic mass number = 88

Example 02:

Calculate the number of electrons, protons, and neutrons in the following species.

Atomic Number = Z = 15

Atomic Mass Number = A = 31

Number of protons = atomic number = 15

P^3- → P + 3 e^–

Number of electrons = 15 + 3 = 18

Number of neutrons = A – Z = 31 – 15 = 16

Number of nucleons = Atomic mass number = 31

Atomic Number = Z = 35

Atomic Mass Number = A = 80

Number of protons = atomic number = 35

Br^– → Br + e^–

Number of electrons = 35 + 1= 36

Number of neutrons = A – Z = 80 – 35 = 45

Number of nucleons = Atomic mass number = 80

Atomic Number = Z = 20

Atomic Mass Number = A = 40

Number of protons = atomic number = 20

Ca²⁺ → Ca – 2e^–

Number of electrons = 20 – 2= 18

Number of neutrons = A – Z = 40 – 20 = 20

Number of nucleons = Atomic mass number = 40

H₃PO₄

Hydrogen (H):

Atomic Number = Z = 1

Atomic Mass Number = A = 1

Number of protons per atom= atomic number = 1

Number of electrons per atom = Z = 1

Number of neutrons per atom = A – Z = 1 – 1 = 0

There are 3 hydrogen atoms in the molecule

Total number of protons in 3 hydrogens = 1 x 3 = 3

Total number of electrons in 3 hydrogens = 1 x 3 = 3

Total number of neutrons in 3 hydrogens = 0 x 3 = 0

Phosphorous (P):

Atomic Number = Z = 15

Atomic Mass Number = A = 31

Number of protons per atom= atomic number = 15

Number of electrons per atom = Z = 15

Number of neutrons per atom = A – Z = 31 – 15 = 16

There is 1 phosphorous atom in the molecule

Total number of protons in 1 phosphorous = 15 x 1 = 15

Total number of electrons in 1 phosphorous = 15 x 1 = 15

Total number of neutrons in 1 phosphorous = 16 x 1 = 16

Oxygen (O):

Atomic Number = Z = 8

Atomic Mass Number = A = 16

Number of protons per atom= atomic number = 8

Number of electrons per atom = Z = 8

Number of neutrons per atom = A – Z = 16 – 8 = 8

There are 4 oxygen atom in the molecule

Total number of protons in 4 oxygen= 8 x 4 = 32

Total number of electrons in 4 oxygen = 8 x 4 = 32

Total number of neutrons in 4 oxygen = 8 x 4 = 32

Ans:

Total number of protons in H₃PO₄ = 3 + 15 + 32 = 50

Total number of electrons in H₃PO₄ = 3 + 15 + 32 = 50

Total number of neutrons in H₃PO₄ = 0 + 16 + 32 = 48

NH₄⁺

Nitrogen (N):

Atomic Number = Z = 7

Atomic Mass Number = A = 14

Number of protons per atom= atomic number = 14

Number of electrons per atom = Z = 14

Number of neutrons per atom = A – Z = 14 – 7 = 7

Hydrogen (H):

Atomic Number = Z = 1

Atomic Mass Number = A = 1

Number of protons per atom= atomic number = 20

Number of electrons per atom = Z = 1

Number of neutrons per atom = A – Z = 1 – 1 = 0

There are 4 hydrogen atoms in the species

Total number of protons in 3 hydrogens = 1 x 4 = 4

Total number of electrons in 3 hydrogens = 1 x 4 = 4

Total number of neutrons in 3 hydrogens = 0 x 4 = 0

Total number of protons in NH₄⁺ = 7 + 4 = 11

NH₄⁺ → NH₄ – e^–

Total number of electrons in NH₄⁺ = 7 + 4 – 1 = 10

Total number of neutrons in NH₄⁺ = 7 + 4 = 11

ClO₃^–

Chlorine (Cl):

Atomic Number = Z = 17

Atomic Mass Number = A = 37

Number of protons per atom= atomic number = 17

Number of electrons per atom = Z = 17

Number of neutrons per atom = A – Z = 37 – 17 = 20

Oxygen (O):

Atomic Number = Z = 8

Atomic Mass Number = A = 16

Number of protons per atom= atomic number = 8

Number of electrons per atom = Z = 8

Number of neutrons per atom = A – Z = 16 – 8 = 8

There are 3 oxygen atom in the species

Total number of protons in 4 oxygen= 8 x 3 = 24

Total number of electrons in 4 oxygen = 8 x 3 = 24

Total number of neutrons in 4 oxygen = 8 x 3 = 24

Total number of protons in ClO₃^– = 17 + 24 = 41

ClO₃^– → ClO₃ + e^–

Total number of electrons in ClO₃^– = 17 + 24 + 1 = 42

Total number of neutrons in ClO₃^– = 20 + 24 = 44

Example 03:

Complete the table

Solution:

Zn

Atomic Number = Z = 30

Atomic Mass Number = A = 64

Number of protons = atomic number = 30

Number of electrons = atomic number = 30

Number of neutrons = A – Z = 64 – 30 = 34

Sr²⁺

Atomic Number = Z = 38

Atomic Mass Number = A = 90

Number of protons = atomic number = 38

Sr²⁺ → Sr – 2e^–

Number of electrons = 38 – 2 = 36

Number of neutrons = A – Z = 90 – 38 = 52

Te

Number of protons = 43

Number of Neutrons = N = 56

Atomic number =Number of protons = Z = 43

Atomic mass number = Z + N = 43 + 56 = 99

Number off electrons = Atomic number = 43

Br^–

Number of neutrons = N = 44

Number of electrons = 36

Br^– → Br + e^–

Number of protons = 36 – 1 = 35

Atomic Number = Number of protons = Z = 35

Atomic mass number = Z + N = 35 + 44 = 79

N

Number of neutrons = N = 7

Number of electrons = 7

Number of protons = 7

Atomic Number = Number of protons = Z = 7

Atomic mass number = Z + N = 7 + 7 = 14

Ca²⁺

Atomic Number = Z = 20

Number of neutrons = N = 20

Atomic Mass Number = Z + N = 20 + 20 = 40

Number of protons = atomic number = 20

Ca²⁺ → Ca – 2e^–

Number of electrons = 20 – 2 = 18

O

Atomic Number = Z = 8

Atomic Mass Number = A = 16

Number of protons = atomic number = 8

Number of electrons = atomic number = 8

Number of neutrons = A – Z = 16 – 8 = 8

The completed table is as follows:

Example 04:

The numbers of electrons, protons, and neutrons in a monoatomic species are equal to 36, 35, and 45 respectively. Assign proper symbol.

Solution:

Number of electrons = 36

Number of protons = Atomic number = Z = 35

Number of neutrons = N = 45

Atomic mass number = A = Z + N = 35 + 45 = 80

Charge on species = Z – Number of electrons = 35 – 36 = -1

The species is

Example 05:

The numbers of electrons, protons, and neutrons in a monoatomic species are equal to 18, 16, and 16 respectively. Assign proper symbol.

Solution:

Number of electrons = 18

Number of protons = Atomic number = Z = 16

Number of neutrons = N = 16

Atomic mass number = A = Z + N = 16 + 16 = 32

Charge on species = Z – Number of electrons = 16 – 18 = -2

The species is

Example 06:

Find the number of electrons in fluorine atom, fluorine molecule, and fluoride ion. The atomic number and mass number of fluorine are 9 and 19 respectively.

Solution:

Atomic number = Z = 9

Atomic mass number A = 19

Fluorine Atom (F):

Number of protons = Atomic number = 9

Number of Protons = Atomic number = 9

Number of electrons = Atomic number = 9

Number of neutrons = a – z = 19 – 9 =10

Fluorine Molecule (F₂):

There are 2 atoms in a molecule of fluorine

Number of protons in fluorine molecule= 9 x 2 = 18

Number of electrons in fluorine molecule= 9 x 2 = 18

Number of neutrons in fluorine molecule= 10 x 2 = 20

Fluoride Ion (F^–):

Number of Protons = Atomic number = 9

F^– → F + e^–

Number of electrons = 9 + 1 = 10

Number of neutrons = a – z = 19 – 9 =10

Example 07:

An isotope of atomic mass 24 had 12 neutrons in its nucleus. What is its atomic number? Represent the isotope in symbolic form.

Solution:

Atomic mass number = A = 24

Number of neutrons = N = 12

Number of protons = A – Z = 24 – 12= 12

Atomic number = Number of protons = Z = 12

The element is

Isotopes, Isotones, and Isobars

• Different atoms of the same element having the same atomic number but having different mass numbers are known as isotopes.
• Atoms of the different elements having a different atomic number but having the same mass numbers are known as isobars.
• Atoms of the different elements having the different atomic number, different mass number but having the same neutron number are known as isotones.

Example 08:

Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A):

• Z = 17 and A = 35

• Z = 92, A = 233

• Z = 4, A = 9

Example 09:

Give an isobar, an isotone, and an isotope of

Isobar
Isotone
Isotope

Science > Chemistry > Atomic Structure > Problems Based on Atomic Number, Mass Number, and Neutron Number

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Light and other electromagnetic radiations have dual nature viz: the particle nature and the wave nature.

Wave Nature of Radiations:

Radiation is the form of energy, which can be transferred from one point to another point in space. Radiations are considered to be transmitted by wave motion. A wave consist of crests and troughs.

• The distance between two consecutive troughs or consecutive crests is called as the wavelength (λ). its S.I. unit is m. and practical unit is angstrom (Å).
• The number of waves passing through any point in unit time (1 sec) is known as frequency (υ). It is Expressed in cycles per second (cps) or Hz. 
• Distance travelled by the wave in one second is called the velocity of the wave. denoted by ‘c’. Expressed in m/sec.

Electromagnetic Nature of Radiations:

Electromagnetic nature of radiations is explained by James Maxwell (1870). He suggested that when electrically charged particles move with an acceleration alternating electrical and magnetic fields are produced and transmitted. These fields are transmitted in the forms of waves called electromagnetic waves or electromagnetic radiation. The oscillating electric and magnetic fields produced by oscillating charged particles are perpendicular to each other and both are perpendicular to the direction of propagation of the wave.

The arrangement of different electromagnetic radiations in order of increasing wavelength is called electromagnetic spectrum. Unlike sound waves or water waves, electromagnetic waves do not require medium and can move in vacuum.

Particle Nature of Light or Planck’s Quantum Theory:

Quantum theory was given by Max Planck in 1900. Its important postulates are

• Absorption and emission of radiant energy does not takes place continuously but it takes place in the form of packets of energy
  called quanta. Quantum of light is called a photon.

• Each quanta has definite amount of energy which depends upon frequency of radiation. The relation is E = hυ, Where E is energy of photon, h is Planck’s constant and υ is the frequency of the radiation.
• Quantum is the smallest denomination of energy. The quantum of energy is always an integer.
• Energy less than quantum can never be absorbed or emitted. It can be emitted as whole number multiple of quantum I.e. 1hυ, 2hυ, 3hυ etc

Particle nature could explain the black body radiation and photoelectric effect satisfactorily but on the other hand, it was not consistent with the known wave behaviour of light which could account for the phenomena of interference and diffraction. The only way to resolve the dilemma was to accept the idea that light possesses both particle and wave-like properties, i.e., light has dual behaviour.

Dual Nature of Radiations:

Depending on the experiment, we find that light behaves either as a wave or as a stream of particles. Whenever radiation interacts with the matter, it displays particle like properties in contrast to the wavelike properties (interference and diffraction), which it exhibits when it propagates. Thus light has dual nature. Some microscopic particles like electrons also exhibit this wave-particle duality.

De-Broglie Equation:

de Broglie in 1924 proposed that matter, like radiation, should also exhibit dual behaviour i.e., both particle and wave like properties. This means that just as the photon has momentum as well as wavelength, electrons should also have momentum as well as wavelength.

De Broglie, from this analogy, gave the following relation between the wavelength (λ) and momentum (p) of a material particle.

where m is the mass of the particle, v its velocity and p its momentum.

Notes:

• De Broglie’s prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction, a phenomenon characteristic of waves. This fact has been put to use in making an electron microscope, which is based on the wave-like the behaviour of electrons just as an ordinary microscope utilizes the wave nature of light. An electron microscope is a powerful tool in modern scientific research because it achieves a magnification of about 15 million times.
• It needs to be noted that according to de Broglie, every object in motion has a wave character. The wavelengths associated with ordinary objects are so short (because of their large masses) that their wave properties cannot be detected.
• The wavelengths associated with electrons and other subatomic particles (with very small mass) can, however, be detected experimentally.

Science > Chemistry > Atomic Structure > Dual Nature of Radiations

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In this article, we shall study Heisenberg’s uncertainty principle, the concept of quantum numbers, and the model of an atom based on the quantum numbers.

Heisenberg’s Uncertainty Principle:

Werner Heisenberg a German physicist in 1927, stated the uncertainty principle which is the consequence of dual behaviour of matter and radiation. It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Mathematically, it can be given as

Where Δx is the uncertainty in position, Δp is the uncertainty in momentum and Δv is the uncertainty in velocity of the particle.

If the position of the electron is known with a high degree of accuracy (Δx is small), then the velocity of the electron will be
uncertain [Δ(vx) is large]. On the other hand, if the velocity of the electron is known precisely (Δ(vx ) is small), then the
position of the electron will be uncertain (Δx will be large).

Thus, if we carry out some physical measurements on the electron’s position or velocity, the outcome will always depict a
fuzzy or blur picture.

Notes:

The uncertainty principle can be best understood with the help of an example. Suppose you are asked to measure the thickness of a sheet of paper with an unmarked metre scale. Obviously, the results obtained would be extremely inaccurate and meaningless. In order to obtain any accuracy, you should use an instrument graduated in units smaller than the thickness of a sheet of paper. Analogously, in order to determine the position of an electron, we must use a scale calibrated in units of smaller than the dimensions of the electron ( the electron is considered as a point charge).

To observe an electron, we can illuminate it with “light” or electromagnetic radiation. The “light” used must have a wavelength smaller than the dimensions of an electron. The high momentum photons of such light would change the energy of electrons by collisions. In this process, we, no doubt, would be able to calculate the position of the electron, but we would know very little about the velocity of the electron after the collision.

Significance of Uncertainty Principle:

One of the important implications of the Heisenberg Uncertainty Principle is that it rules out existence of definite paths or
trajectories of electrons and other similar particles.

The trajectory of an object is determined by its location and velocity at various moments. If we know where a body is at a
particular instant and if we also know its velocity and the forces acting on it at that instant, we can tell where the body would
be sometime later. We, therefore, conclude that the position of an object and its velocity fix its trajectory.

Since for a subatomic object such as an electron, it is not possible simultaneously to determine the position and velocity at
any given instant to an arbitrary degree of precision, it is not possible to talk about the trajectory of an electron.

The effect of Heisenberg Uncertainty Principle is significant only for the motion of microscopic objects and is negligible for that of macroscopic objects.

Thus the classical picture of electrons moving in Bohr’s orbits (fixed) cannot hold good. It, therefore, means that the precise statements of the position and momentum of electrons have to be replaced by the statements of probability, that the electron has at a given position and momentum. This is what happens in the quantum mechanical model of an atom.

Schrodinger’s Wave Equation:

Schrodinger independently studied the nature of electron and gave equation which is known as Schrodinger wave equation.

Where, E = Total energy of the system
V = Potential energy of the system
m = Mass of electron
h = Planck’s constant
∇ = Operator
ψ = Wave function and is the amplitude of the electron wave.

ψ ² gives the probability of finding the electron at various places in a given region of space to another. Thus probabilities of finding the electron in different regions are different. This is in accordance with uncertainity principle.

The acceptable values of wave functions provide the regions around the nucleus in which probability of finding the electron is
maximum. These regions are called orbitals. Thus solving Schrodinger wave equation we can get the shape of the orbitals.This equation also gives certain specific numbers called quantum numbers which specify the location of an electron in an atom.

Concept of Orbitals:

The classical picture of electrons moving in Bohr’s orbits (fixed) cannot hold good. It, therefore, means that the precise statements of the position and momentum of electrons have to be replaced by the statements of probability, that the electron has at a given position and momentum. This is what happens in the quantum mechanical model of an atom.

After solving Schrodinger wave equation, the acceptable values of wave functions provide the regions around the nucleus in which probability of finding the electron is maximum. These regions are called orbitals. Thus solving Schrodinger wave equation we can get the shape of the orbitals.

• Thus electron is a cloud of negative charge with different shapes.
• The three-dimensional region in the space around the nucleus in which the probability of finding the electron is maximum is called orbital.
• In each orbital, the electron has a definite energy. The energy of an orbital is lower if it is concentrated near the nucleus.
• It is to be noted that orbital is not same as that of orbit.

There are four types of orbitals known to us till now, they are s, p, d and f orbitals

Science > Chemistry > Atomic Structure > Heisenberg’s Uncertainty Principle

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In this article, we shall study the concept of quantum numbers and the model of an atom based on the quantum numbers.

Concept of Quantum Numbers:

In 1926, Erwin Schrodinger put forward a theory of atom called as a quantum mechanical theory. Quantum mechanical theory is based on two principles. a) Wave-particle duality of subatomic particles and b) Uncertainty principle.

Wave-particle duality principle of subatomic particles was put forward by L. DeBroglie it states that “All the subatomic particles electrons, protons, and neutrons possess both particle and wave properties”.

Uncertainty principle was put forward by W. Heisenberg it states that “It is impossible to determine the position and momentum of a particle simultaneously”. It means that if the position of the electron in an atom is known then there is uncertainty about its momentum and when the momentum of an electron in an atom is known there is uncertainty about the position of the electron.

According to this theory, Each electron shell of an atom is composed of several subshells. To specify the position of the electron in an atom each electron is assigned four quantum numbers and these numbers give location and energy of the electron.

The numbers which assign the position and energy to an electron are called quantum numbers. There is a set of four quantum numbers associated with each electron of an atom. The four quantum numbers are a) Principal quantum number ‘n’  b) Azimuthal quantum number ‘l’ c) Magnetic quantum number ‘m’  d) Spin quantum number ‘s’

Principal Quantum Number (n):

This quantum number was introduced by Neil Bohr. It is denoted by the letter ‘n’. It gives the main energy level (shell) to which the electron belongs. It defines the distance of the electron from the nucleus and energy level. These energy levels are also designated by letters as follows:

Azimuthal Quantum Number (l):

It was introduced by Somerfield.  This quantum number refers to the energy sub-level (subshell).  It is denoted by the letter ‘l’. It can have any integral value from 0 to (n – 1). This quantum number indicates the subshell in which the electron is present and the number of subshells present in a principal shell.

It is also responsible for the particular shape of the particular shell because it defines the spatial distribution of the electron cloud around the nucleus. Thus the subshell (Orbital) with l = 0 is called ‘s’ orbital. The subshell (Orbital) with l = 1 is called ‘p’ orbital. The subshell (Orbital) with l = 2 is called ‘d’ orbital. The subshell (Orbital) with l = 3  is called ‘f’ orbital

It is used to calculate angular momentum of an electron in particular shell using the formula

Each subshell contains maximum 2 (2l +1) electrons. It also decides the number of nodal planes of the orbital. For s orbital l = 0, hence s orbital there is no nodal plane. For p orbital l = 1, hence p orbital there is one nodal plane. For d orbital l = 2, hence d orbital there are two nodal planes.

Notes:

For the first orbit:

n = 1, then  l = 0 to (1 – 1) = 0 to 0,  then  l = 0. Thus in the first main energy level, there is only one subshell i.e. ‘s’ orbital.

For the second orbit:

n = 2 then  l = 0 to (2 – 1) = 0 to 1, then l = 0, 1. Thus in the second main energy level, there are two subshells i.e. ‘s’ orbital and ‘p’ orbital.For the Third orbit n = 3 Then  = 0 to (3 – 1) = 0 to 2 then             = 0, 1, 2. Thus in the third main energy level there

For the Third orbit:

n = 3 then  l = 0 to (3 – 1) = 0 to 2 then l = 0, 1, 2. Thus in the third main energy level, there are three subshells i.e. ‘s’ orbital, ‘p’ orbital and ‘d’ orbital.

For the fourth orbit:

n = 4 Then l = 0 to (4 – 1) = 0 to 3 then l  = 0, 1, 2, 3. Thus in the fourth main energy level, there are four subshells i.e. ‘s’ orbital, ‘p’ orbital, ‘d’ orbital, and ‘f’ orbital.

Magnetic Quantum Number (m):

This quqntum number is denoted by letter ‘m’. This number was introduced to explain the splitting of atomic spectra in a magnetic field (Zeeman effect).

For Azimuthal number ‘l’ possible values of magnetic quantum numbers are – l  to + l   through 0. i.e. m =  , …., -3, -2, -1, 0, 1, 2, 3, ….., . This quantum number gives the possible orientation of the orbital in the space. For a given value of l, there are (2l +1) values of m.

This number is also known as the orientation quantum number because it gives the distribution of electron clouds in space around the nucleus.

The magnetic quantum number tells to which orbital the electron belongs. It also tells the number of orbitals which corresponds to a particular shell. For example: s subshell l = 0 therefore m = 0. This means that s subshell contains only one orbital.

p subshell l = 1 therefore m = -1, 0, +1. This means that p subshell contains three orbitals. They are designated as Px, Py and Pz,  along the x-axis, along the y-axis and along the z-axis respectively. Thus p orbitals have three orientations in space along co-ordinate axes. In absence of magnetic field, these orbitals are equivalent in energy and are said to have three fold degenerate.

d subshell l = 2 therefore m = -2, -1, 0, +1, +2. This means that d subshell contains five orbitals. They are designated as dxy, dyz, dzx, dx2-y2, and dz2.   Thus d orbitals have five orientations in space. In absence of magnetic field, these orbitals are equivalent in energy and are said to have five fold degenerate.

f subshell l = 3 therefore m = -3, -2, -1, 0, +1, +2, +3. This means that f subshell contains seven orbitals and have seven orientations in space. In absence of magnetic field, these orbitals are equivalent in energy and are said to have seven fold degenerate.

Spin Quantum Number (s):

It is denoted by letter ‘s’. The electron in orbital motion can spin clockwise or anticlockwise about its own axis, hence two values of quantum number are possible.  These values are taken as + 1/2 and – 1/2.

Opposite spins are called antiparallel spins. The same spin is called the parallel spin. The spin quantum number indicates a magnetic moment associated with the electron.

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In this article, we shall study the concept of quantum numbers and different types of orbitals and their shapes.

Difference Between Shell, Subshell, and Orbital:

• All electrons that have the same value for n (the principal quantum number) are in the same shell.Within a shell (same n), all electrons that share the same  (the angular momentum quantum number, or orbital shape) are in the same subshell. The electrons in subshell have same principal quantum number, same azimuthal quantum number and  differ  in magnetic and spin quantum number
• When electrons share the same n, l, and m, we say they are in the same orbital (they have the same energy level, shape, and orientation). The electrons in orbital differ only in spin quantum number.

Shapes of Orbitals:

• Node: There is a region where the probability of finding the electron is zero. It is called a node.
• Nodal Plane: A plane passing through the nucleus on which the probability of finding an electron is zero is called the nodal plane. The number of nodal planes in an orbital is equal to the azimuthal quantum number.
• Spherical or Radial Node: A spherical surface within an orbital on which the probability of finding the electron is zero is called a spherical or radial node.

Types of Orbitals:

s – orbital:

• For s orbital Azimuthal quantum number  = 0 and the magnetic quantum number m = 0 hence s orbitals have unique orientation in space. Thus s orbital corresponds to spherical shape with the atomic nucleus at its centre.
• For every value of ‘n’, there is one ‘s’ orbital i.e. s orbitals are present in all principal energy levels.  Its radius depends on the value of n. As the value of n increases the size of the s orbital also increases.
• It has no nodal plane.

p – orbital:

• For p orbital Azimuthal quantum number l = 1 and the magnetic quantum number  m = -1, 0, +1. Hence p orbitals have three orientations in space.
• Thus p orbital corresponds to dumb-belled shape with the atomic nucleus at its center.
• p orbitals have two lobes directed on opposite sides of the nucleus. P-orbitals are orientated in three different directions along X, Y and Z axis of the usual coordinate system.  These orbitals are designated as P_x, P_y & P_z orbitals.
• p-orbital have one nodal plane.

d – orbital:

• For d orbital Azimuthal quantum number l = 2 and the magnetic quantum number  m = -2, -1, 0, +1, +2. Hence d orbitals have five orientations in space.
• Thus d orbital corresponds to 4 double dumb-belled shapes (d_xy, d_yz, d_zx, d_x²_y²) with the atomic nucleus at its centre and one dumb belled with dough nut shaped (d_z²).
• d orbital has two nodal planes.

f – orbital:

• For f orbital Azimuthal quantum number l = 3 and the magnetic quantum number m = -3. -2, -1, 0, +1, +2, +3. Hence f orbitals have seven orientations in space.
• f orbital has complex shapes with the atomic nucleus at its centre.
• f orbital has three nodal planes.

Quantum Numbers of Electrons in First Three Shells (Orbits)

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The distribution of electrons in various shells and orbitals of an atom is known as electronic configuration. Generally electronic configuration is given by nl^x. Where n is shell number, l is subshell and x are the number of electrons.

Aufbau Principle:

The arrangement of the electrons in an atom is called the electronic configuration of an atom. The method of filling up or building up a sequence of energy levels for electrons in an atom is based on Aufbau principle. In German, the word aufbau means building up.

According to this principle, orbitals are filled in the order of increasing energy. The lowest energy level is filled first. The higher energy levels are then filled up one after another. This gives the atom the maximum stability and the atom is said to be in ground state or the normal state.

The order of increasing energy levels is 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p < 6f Diagrammatically it is represented as follows

Pauli’s Exclusion Principle:

“No two electrons. in an atom can have the same values for all four quantum numbers”. i.e. to exist in the same atom the two electrons can have three quantum numbers identical but fourth must be different.

Example: Let us consider electrons in the first orbit of helium atom

Here we can see that the three values of quantum number are the same but fourth value is different. Thus Pauli’s exclusion principle helps in calculating the maximum number of electrons present in any energy level (Orbit).

Hund’s Rule of Maximum Multiplicity:

In a given subshell, electrons occupy all the available orbitals singly, before they pair up in any orbital.

Example 1:

Consider electronic configuration of phosphorous. The atomic number of phosphorous is 15. Its basic electronic configuration is  2, 8, 5. The detailed electronic configuration is 1s², 2s² 2p⁶, 3s² 3p³. The box diagram for final orbit configuration for P is as below

Here we can see that the electrons in p orbitals occupy the position singly till all orbitals are singly filled.

Example – 2:

Consider electronic configuration of sulphur. The atomic number of sulphur is 16. Its basic electronic configuration is 2, 8, 6. The detailed electronic configuration is 1s², 2s² 2p⁶, 3s² 3p⁴. The box diagram for final orbit configuration for S is as below

Here we can see that the electrons in p orbital occupy the position singly till all orbitals are singly filled and then the pairing starts.

Stability of Completely Filled and Half Filled Subshells:

The ground state electronic configuration of the atom of an element always corresponds to the state of the lowest total electronic energy.

The electronic configurations of most of the atoms follow the basic rules viz. Aufbau principle, Pauli’s exclusion principle, and Hund’s rule. However, in certain elements such as Cu, or Cr, where the two subshells (4s and 3d) differ slightly in their energies, an electron shifts from a subshell of lower energy (4s) to a subshell of higher energy (3d), provided such a shift results in all orbitals of the subshell of higher energy getting either completely filled or exactly half filled.

The valence electronic configurations of Cr and Cu, therefore, are 3d⁵ 4s¹ and 3d¹⁰ 4s¹ respectively and not 3d⁴ 4s² and 3d⁹ 4s². It has been found that there is extra stability associated with these electronic configurations.

Electronic Configuration of copper (Z = 29):

Expected configuration: 1s², 2s² 2p⁶, 3s² 3p⁶, 4s² 3d⁹.

Actual configuration: 1s², 2s² 2p⁶, 3s² 3p⁶, 4s¹ 3d¹⁰.

Electronic Configuration of chromium (Z = 24):

Expected configuration:  1s², 2s² 2p⁶, 3s² 3p⁶, 4s² 3d⁴.

Actual configuration:  1s², 2s² 2p⁶, 3s² 3p⁶, 4s¹ 3d⁵.

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In this article, we shall study Thomson’s model of an atom and its deficiencies and Rutherford’s atomic model and its deficiencies.

Thomson’s Model of an Atom:

J. J. Thomson, in 1898, proposed that an atom possesses a spherical shape (radius approximately 10^–10 m) in which the positive charge is uniformly distributed. The electrons are embedded into it in such a manner as to give the most stable electrostatic arrangement. Many different names are given to this model, for example, plum pudding, raisin pudding, or watermelon. It can be visualized as a positive charge with plums or seeds (electrons) embedded into it (watermelon). An important feature of this model is that the mass of the atom is assumed to be uniformly distributed over the atom.

Although this model was able to explain the overall neutrality of the atom but was not consistent with the results of later experiments.

Basis of Rutherford’s Atomic Model:

Alpha particles from a radioactive substance were made incident on the thin foil of gold of thickness 10^-7 m. After passing through the foil, the alpha particles were detected at various places on the ZnS screen or photographic plate. The following four types of particles were detected.

• Most of the alpha particles just passed through without any deviation as if there is empty space.
• A few alpha particles were deflected through smaller angles.
• A few alpha particles deviated through larger angles.
• Very few rebounded back This larger deflection is possible only if α-particles collide with heavy and positively charged particles inside the atom because like charges only repel each other.  This massive +ve charge is at centre of an atom and called nucleus.
• Very few alpha particles were rebounded i.e. they deviated through 180°. This concludes that the nucleus is very small as compared to the volume of the atom.

Conclusions :

• Based on the experiment Rutherford concluded that most of the alpha particles having positive charge went through the foil undeflected. Hence there must be empty space in the atom.
• Some of the particles were deflected.  It is due to the positive charge present in a very small space within the atom. He called this centrally cored and positively charged region of an atom as a nucleus. The rest of the portion of the atom should be empty.

Rutherford’s Atomic Model:

From the observations of his experiment, Rutherford put forward the concept of his atomic model.

The atom consists of the centrally located positively charged nucleus. The whole mass of an atom is concentrated in the nucleus. Around the nucleus, there is empty space in which the negatively charged electrons revolve in different orbits. Rutherford’s model of an atom is also called as a planetary model of an atom.

Limitations Of Rutherford’s Atomic Model:

• The negatively charged electrons revolve around the nucleus in a circular orbit, hence they possess centripetal acceleration. According to the classical theory of electromagnetism, accelerated charge radiates energy continuously. Therefore, the
  electron should radiate energy while going round the nucleus losing its energy continuously. therefore, it should approach nearer the nucleus while going round emitting radiations of increasing frequency and finally falling in the nucleus. Thus it should move in a spiral path and should emit continuous spectrum and thus structure atom is not stable. Actually, the spectrum observed is line spectrum of definite frequency, and hence a modification of Rutherford’s atom model was necessary.

• This model of an atom fails to explain the distribution of electrons in different orbit around the nucleus.
• According to Rutherford’s model of an atom, the atomic spectrum should be continuous. But atomic spectrum is found to be discontinuous. Rutherford model fails to explain the discontinuity of the atomic spectrum.
• This model also fails to explain the line spectra of atoms, which show discrete lines, each line corresponds to a fixed frequency.

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In last article, we have discussed, the discovery of electron and characteristics of electron. In this article, we shall discuss the discovery of proton and neutron and their characteristics.

Discovery of Proton:

Proton was discovered by E. Goldstein in 1886. He performed the same experiment as performed by J.J. Thomson but used perforated cathode. He found that on passing an electric discharge through a gas under reduced pressure, rays containing positive particles move towards the cathode.  As they appear to arise from the anode they are called anode rays or canal rays. They are found to contain positively charged particles called protons.

Origin of Positive Rays:

When an electric discharge is passed through gas at very low pressure in the discharge tube cathode rays are produced. The cathode rays consist of a stream of high-speed electrons. When these fast-moving electrons strike the atoms or molecules of the gas present in the discharge tube, they remove one or more electrons from the neutral atoms or molecules. Thus positive ions of gas are formed. These ions which move towards the perforated cathode kept midway in the tube and constitute the positive rays coming through the perforated cathode.

Characteristics of Canal Rays:

• They travel in a straight line in the opposite direction to that of cathode rays.
• Canal rays produce fluorescence when incident on zinc sulphide screen.
• Unlike cathode rays, the positively charged particles depend upon the nature of gas present in the cathode ray tube.
• These are simply the positively charged gaseous ions.
• The charge to mass ratio of the particles is found to depend on the gas from which these originate.
• Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.
• The behaviour of these particles in the magnetic or electrical field is opposite to that observed for electron or cathode rays.

Characteristics of Protons:

• Protons are positively charged.
• Protons are located in the nucleus.
• Protons have mass 0f 1.0078 a.m.u. (1.672 ×  10^-27  Kg.). This mass of a proton is considered as unit mass ( 1 a.m.u. ).
• Mass of one proton is almost equal to the mass of one hydrogen atom.
• The proton carries a positive charge of  1.6 ×  10^-19  C. This charge carried by the roton is considered to be the unit positive charge.
• Proton is denoted by   1 H 1  or 1 P 1. I.e. it has unit positive charge and unit mass.
• All atoms contain proton.

Discovery of Neutron:

In 1920 Rutherford proposed existence of the third neutral particle in an atom. Neutron was discovered by Sir James Chadwick in 1932. He observed that when Beryllium (Be) is bombarded by α -particles; particles having no charge and mass equal to that at proton were produced (ejected).  He called them as neutrons.

Characteristics of Neutrons:

• Neutrons have no charge i.e. they are electrically neutral.
• They are located in the nucleus of an atom.
• Mass of neutron is of 1.008665 a.m.u  (1.675 ×  10^-27  Kg ). For practical purpose, this mass is assumed as unit mass.
• Mass of neutron is nearly as that of the proton.
• Neutron is denoted as 0 n 1.

Other Subatomic Particles:

Mesons, positrons, neutrinos, antiprotons are other subatomic particles. Other subatomic particles discovered recently are quarks, antiquarks, pions, gluons, boson and god particle.

Discovery of X-rays:

X-Rays were discovered by Wilhelm Roentgen in 1895. He discovered that when cathode rays strike metal with a high atomic number, radiation of very short wavelength was emitted. He was unable to explain the nature of the emitted rays hence he called these rays as X-rays.

Characteristics of X-Rays:

• These rays are capable of penetrating through wood, paper, and flesh but are stopped by bones and metallic substance.
• X-Rays are electromagnetic waves.
• They are chargeless.
• X-rays produce fluorescence in many substances. The fluorescence in different substances has different characteristics.
• X-ray kill some form of animal tissues.
• X-ray affect photographic plates.
• X-rays travel by velocity of light (3 x 10⁸ m/s) in air or vacuum.
• They are not deflected by electric or magnetic fields.
• They ionize air or gas through which they pass.

Moseley’s Contribution:

Moseley in 1913 found that each element when bombarded by high-velocity electrons, emit X-rays of different frequency.  Thus the atomic number of many elements was determined accurately by the following relationship.

Atomic number α square root of the frequency of X-rays. He gave relation Where a and k are constant. Z is an atomic number and υ is the frequency of X-ray. Thus he stated that frequency of radiation from elements depends upon the number and arrangement of unit particles in their atoms. Hence atomic number (z) is a fundamental property of all elements.

Important Terms and Concepts:

Atomic number (Z) :

The number of protons (positive charge) present in the nucleus of an atom of a particular element is called the atomic number of that element. It is denoted by letter ‘Z’.

Neutron number (N): 

The number of neutrons present in the nucleus of an atom is known as neutron number. It is denoted by ‘N’

Mass number (A): 

The total number of protons and neutrons present in the nucleus of an atom of the element is called mass number. The mass number is denoted as ‘A’.

A  =   Z + N

Representation of Atom in Symbolic Form: 

Generally, every atom X is represented as

Concept of Isotopes, Isobars, and Isotones:

Isotopes: 

Different atoms of the same element having the same atomic number but having different mass numbers are known as isotopes.

Examples :

Characteristics of Isotopes:

• Different Atoms of the same element having the same atomic number but having different mass numbers are known as isotopes.
• Isotopes are the atoms of the same element.
• They have the same atomic number but different mass numbers.
• They have the same number of protons but the different number of neutrons.
• Since they have the same atomic number they show the same chemical properties.
• They occupy the same positions in the periodic table.

Isobars: 

Atoms of the different elements having a different atomic number but having same mass numbers are known as isobars.

Examples :

Characteristics of Isobars:

• Atoms of the different elements having the different atomic numbers but having the same mass numbers are known as isobars.
• Isobars are the atoms of different elements.
• They have the same mass number but different atomic numbers.
• They have a different number of protons and neutrons.
• Since they have different atomic number they show different chemical properties.
• They occupy different positions in the periodic table.

Isotones: 

Atoms of the different elements having the different atomic number, different mass number but having same neutron number are known as isotones.

Examples:

Characteristics of Isotones:

• Atoms of the different elements having the different atomic number, different mass number but having same neutron number are known as isotones.
• Isotones are the atoms of different elements.
• They have the different mass numbers and different atomic numbers.
• They have the same number of neutrons.
• Since they have a different atomic number they show different chemical properties.
• They occupy different positions in the periodic table.

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In this article, we shall study earlier concepts of an atom, discovery of electron, and its characteristics.

600 B.C. Indian saint and philosopher Maharshi Kanad proposed that matter is made up of the smallest individual particles. He called these particles ‘paramanu’. Around 400 B.C. The Greek Philosopher Democritus suggested that all the matter is composed of tiny, discrete, indivisible particles. He called these particles ‘atoms’. Both the ideas are of conceptual nature and didn’t have any experimental evidence.

Assumptions of Dalton’s atomic theory:

The first concept was given by John Dalton. The postulates of his theory are:

• Every element is made up of extremely small particles called an atom.
• The atoms are indivisible and they can neither be created nor be destroyed. Atoms of the same element resemble each other in all respects but differ from the atoms of other elements.
• When chemical compounds are formed they do so by the combination of atoms of different elements in a simple proportion of whole numbers.
• Atoms of different elements may combine in more than one proportion to form different compounds.

Evidence of Electrical Nature of Matter:

When a glass is rubbed with silk or ebonite is rubbed with fur, electricity is generated. Electricity gets transferred from one point to another point through certain substances. This phenomenon indicated that the matter has electrical nature.

Michael Faraday in 1832 passed electricity through the solution and he called the phenomenon as electrolysis. He observed that charged particles migrate towards oppositely charged electrodes. During this process, they accumulate on the electrode or escape out as a gas at the electrode. On the basis of his experiments, he proposed the laws of electrolysis. These laws of electrolysis given by Michael Faraday provide a relation between matter and electricity. These laws assumed the discrete nature of electricity. These discrete particles are called electrons by Loney.

Fundamental particles of an atom:

Protons, neutrons, and electrons that make up an atom are known as the fundamental subatomic particles. Protons and neutrons are present in the nucleus of an atom, they are called intranuclear particles or nucleons. Electrons revolve around the nucleus in a circular orbit. They are called extra-nuclear particles.

Discovery of Electrons:

Electrons were the first of sub-atomic particles to be discovered, by J.J. Thomson in 1859. J.J. Thomson made a detailed study of the discharge of electricity through gases under very low pressure. This experiment was performed using a cathode ray tube (Crooke’s tube).  It consists of a glass tube connected to two metal electrodes at two ends.  There is a side tube connected to a vacuum pump to reduce pressure.

When a gas is subjected to a high potential (5000 to 10000 V) at low pressure, the glass wall of tube glows with fluorescent light.  This glow is because of the bombardment of glass by rays emitted from the cathode.  As the rays arise from the cathode, they are called cathode rays. These rays were found to consist of negatively charged particles with a negligible but definite mass.

Cathode rays produce heat energy when they collide with the matter. It is due to the kinetic energy possessed by the cathode rays. Cathode rays also rotate the paddle wheel in their path. They move from the negative electrode to the positive electrode. Which clearly indicate that the cathode ray consists of a negatively charged particle. Thomson called these particles as negatrons. Stoney changed this name to electrons.

Thomson also showed that the electrons are present in atoms of all elements. Thus electrons are the fundamental particles of an atom.

Charge to Mass Ratio of Electron:

J. J. Thomson measured the ratio of electrical charge (e) to mass (m) of cathode ray particles using specially designed cathode ray tube.

Construction of Apparatus:

The apparatus consists of a discharge tube containing gas at a very low pressure about 0.01 mm of mercury. The discharge tube has cathode C at one end and fluorescent screen S at the other. Anode A consists of a cylinder with a fine bore.

The electric field can be applied between the plates P1 and P2. Plate P1 is positive and plate P2 is negative

The magnetic field can be applied perpendicular to the electric field and perpendicular to the plane of the diagram and into the plane.

Working:

Cathode emits electrons and they are collimated by cylindrical fine bore anode. The velocity of electrons depends on the potential difference between the cathode and the anode. When no field is applied. The electrons move in a straight line and forms spot at O at the centre of the screen.

When cathode rays are passed through an electric field created by applying a potential across the plates P₁ and P₂ only. It is found that the cathode rays particles get deflected towards the positive plate.

When cathode rays are passed through the magnetic field created by applying a strong magnetic field only, it is found that the cathode rays particles get deflected in a circular path.

Thomson applied both the fields simultaneously and adjusted its value such that the spot remains at the centre of the screen at O. Knowing values of V, B and d and using following formula the e/m ratio can be determined.

Where E = V/x. The value of e/m is 1.758820 x 10¹¹ C kg^-16.

Thomson repeated the experiment for different materials of the cathode and found that the e/m ratio is always the same. From this, he concluded that the particles present in cathode rays (electrons) are fundamental particles of any atom of all matter.

Equation of Path of Parabolic path of Electron:

The path of an electron in the electric field is parabolic whose equation is given by

Where y = deflection of the path of the electron in the y-direction

x = Distance between the parallel plates

e = Charge on electron

E = Intensity of the electric field

m = Mass of electron

v = velocity of electron

Radius of Circular Orbit of Electron:

The radius of the circular orbit of an electron in the magnetic field is given by

Where r = radius of the circular path of an electron in magnetic fields

e = Charge on electron

B = Magnetic induction

m = Mass of electron

v = velocity of electron

Alternate Equation for e/m Ratio:

Millikan’s Oil Drop Experiment:

In the first step, the oil drops are allowed to fall between the plates in the absence of an electric field. Due to gravity, they accelerate first, but gradually velocity decreases up to certain minimum value due to air resistance. Then the drops start moving downward with a constant velocity called terminal velocity (v₁). The terminal velocity of the drop is measured. As the velocity is constant, the air resistance and weight of the drop are equal in magnitude. Thus the net force acting on the drop is zero.

In the second step, an electric field is produced in the chamber. A likely looking drop is selected and kept in the middle of the field of view by adjusting the voltage. So that the drop is moving down with constant velocity called terminal velocity (v₂). Thus net force acting on it is zero. Thus the electric force is balancing the downward forces.

The charge on an electron is given by

Where η = Coefficient of the viscosity of gaseous medium (air)

v₁ = Terminal velocity of the oil drop when an electric field is not applied

v₂ = Terminal velocity of the oil drop when an electric field is applied

E = Intensity of the electric field between plates

f = Density of oil

σ = density of the gas (air)

g = acceleration due to gravity

Millikan repeated the experiment by varying the strength of x-rays used for ionization of air no. of times. Due to which the no. of electrons attaching to the oil drop varied. Then he obtained various values for the charge and is found to be an integral multiple of 1.6 x 10^-19 C.

In 1923, Millikan won the Nobel Prize in Physics in part because of this experiment.

The charge on an Electron:

Scientist R. A. Millikan in his oil-drop experiment determined the charge on the electron and he found that the charge on an electron is 1.6022 x 10^-19 C.

Mass of an Electron:

Using e/m ratio and charge on the electron, the mass of an electron is found to be 9.1094 x 10^-31 kg.

Characteristics of Electrons:

• Electrons are negatively charged.
• They are revolving in circular orbits around the nucleus.
• They have mass 0f 0.00055 a.m.u.  (9.11 × 10 ^-31 Kg.). This mass of an electron is negligible compared to other particles. Hence it is taken as zero.
• Its mass is 1/1837 times that of the proton.
• Electrons carry a negative charge of 1.6 × 10 ^-19 C. This charge carried by the electron is considered to be a unit negative charge.

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