Quantization of energy.pptx

Quantization of Energy: Theory of atomic model.
Rutherford Atomic Model:
From the results of his Alpha particle scattering experiment, Rutherford gave a atomic model. According
to this model, an atom consists of a massive, positively charged tiny core called nucleus with the much
lighter electrons orbiting around it in circular paths. The necessary centripetal force is provided by
electrostatic force of attraction of nucleus on the electrons. Moreover, the total number of positive
charges at the nucleus is equal to the total number of negative charge carried by the electrons orbiting
around it so that the atom is electrically neutral.
Drawbacks of Rutherford model:
i) Rutherford model could not explain the stability of atoms. According to classical theory of
electromagnetism, the electron orbiting the nucleus of Rutherford atom is in accelerated state and it
should emit radiation. As such, its energy should go on decreasing, resulting into a gradually
decreasing radius of the orbit until the electron collapses into the nucleus. However no such situation
has been observed.
ii) This theory could not also explain the line spectra observed in elements like hydrogen.
+
-
12/17/2022 Aroma English Secondary School

Quantization of Energy: Theory of atomic model.
Bohr’s Atomic Model:
Neil Bohr, in 1913, added the following postulates to Rutherford model in order to overcome its drawbacks.
1. An electron is allowed to revolve around the nucleus only in certain permitted orbits, known as the stationary
orbits.
2. The permitted orbits are those in which the angular momentum (L = mvr) of the electron is an integral multiple of
h/2π. That means-
mvr = n
h
2π
, where n = 1, 2,3,…….
This is called Bohr’s quantization condition.
3. An electron radiates energy only when it jumps from a higher energy level (Ef) to a lower energy level (Ei). The
energy of the photon emitted: hν = Ef – Ei
This is called Bohr’s frequency condition.

Quantization of Energy
Bohr’s Theory of Hydrogen (H) and Hydrogen like (He+, Li++) atoms.
The electrostatic force of attraction of nucleus on the electron is given by:
Fe =
Ze e
4π𝜀or2 =
Ze2
4π𝜀or2
And, the centripetal force on the electron is-
Fc =
mv2
r
For the circular orbit,
Fc = Fe
Ze
r
v
e

Fc = Fe
or
mv2
r
=
Ze2
4π𝜀or2
or mv2 =
Ze2
4π𝜀or
-------------------(1)
Now,
a) Radius of Bohr’s orbit:
According to Bohr’s quantization condition,
mvr = n
h
2π
or v =
nh
2πmr
-------------------------- (2)
Substituting this value in equation (1),
m
nh
2πmr
2
=
Ze2
4π𝜀or
Ze
r
v
e

or m
n2h2
4π2m2r2 =
Ze2
4π𝜀or
or r =
𝜀on2h2
πZe2m
where n = 1, 2, 3,
Thus for nth Bohr orbit, we write-
rn =
𝛆𝐨𝐧𝟐𝐡𝟐
𝛑𝐙𝐞𝟐𝐦
-------------------------- (3)
Calculation of Radii for H-atom
For H atom Z = 1. So, rn =
εoh2
πe2m
n2
=
8.85×10−12× 6.62×10−34 2
π 1.6×10−19 2×9.1×10−31 n2
= (0.53 × 10−10) n2
= (0.53)n2 A
Ze
r
v
e

rn = 0.53 × n2 A
Thus, for first orbit, n = 1. So, r1= 0.53 A
For the second orbit, n = 2. and, r2= 0.53 × 4 A = 2.12 A, and so on.
b) Velocity of Electron: Substituting, r =
𝜀on2h2
πZe2m
in equation (2), we obtain-
v =
nh
2πm
×
πZe2m
𝜀on2h2
or v =
Ze2
2εonh
For hydrogen atom Z = 1. So velocity of electron in nth Bohr orbit is written as-
vn =
𝐞𝟐
𝟐𝛆𝐨𝐧𝐡
Calculation of velocity:
Ze
r
v
e

vn =
𝐞𝟐
𝟐𝛆𝐨𝐧𝐡
Calculation of velocity: vn =
(1.6×10−19)2
2×8.85×10−12×6.62×10−34 ×
1
n
= 2.19 × 106 ×
1
n
=
c
3×108
2.19×106
×
1
n
= (
c
137
)
1
n
where c is the speed of light in vaccum.
## You may find the frequency of electron in an orbit of hydrogen by using
f =
v
2πr
Ze
r
v
e

c) Energy of the electron in nth orbit:
Total energy in nth orbit can be written as
En = K. E. + P. E. --------------------- (1)
For the electron in nth orbit of hydrogen atom,
mvn
2
rn
=
e2
4π𝜀orn
2 --------------------- (2)
And from Bohr’s quantization condition:
mvnrn =
nh
2π
--------------------- (3)
Dividing equation (2) by (3)
mvn
2
rn×mvnrn
=
e2
4π𝜀orn
2 ×
2π
nh
or vn =
e2
2𝜀onh
--------------------- (4)
Using this value in equation (3)
m
e2
2𝜀onh
rn =
nh
2π
rn =
𝜀on2h2
πZe2m
--------------------- (5)
+e
r
v
-e

c) Energy of the electron in nth orbit: vn =
e2
2𝜀onh
---------- (4), rn =
𝜀on2h2
πZe2m
-------------- (5)
Therefore, its kinetic energy, K. E. =
1
2
mvn
2
=
1
2
m
e2
2εonh
2
(Using eqn. 4)
or K. E. =
me4
8𝜀𝑜
2n2h2 --------------------- (6)
And, the electrostatic potential energy,
P. E. =
e(−e)
4𝜋εorn
= −
e2
4𝜋εo
πe2m
εon2h2 (Using eqn. 5)
or P. E. = −
me4
4𝜀𝑜
2n2h2 --------------------- (7)
Using equations 6 and 7 in 1, the total energy in the nth orbit of H atom becomes-
En =
me4
8𝜀𝑜
2n2h2 + −
me4
4𝜀𝑜
2n2h2
En = −
𝐦𝐞𝟒
𝟖𝜺𝒐
𝟐𝐧𝟐𝐡𝟐 --------------------- (8)
The negative sign means that the electron is bound to the nucleus by an attractive electrostatic field.

c) Calculation of the energy levels:
The total energy in the nth state can be written as- En = −
me4
8𝜀𝑜
2h2
1
n2
This gives En = −
9.1×10−31× 1.6×10−19 4
8× 8.85×10−12 2× 6.62×10−34 2
1
n2 = − 21.76 × 10−19 1
n2 Joules
= −
21.76×10−19
1.6×10−19
1
n2 eV
= − 13.6 ×
1
n2 eV
Now for the first orbit, n = 1. So, E1 = -13.6 eV (First energy level)
For the second orbit, n = 2. So, E2 = -13.6/4 eV = -3.4 eV (Second energy level)
For the third orbit, n = 3. So, E3 = -13.6/9 eV = -1.51 eV (Third energy level)
For the fourth orbit, n = 4. So, E4 = -13.6/16 eV = -0.85 eV and so on.
Similarly, for n = ∞, E∞= -13.6/∞ = 0 eV
The energy levels E1, E2, E3 … are also known as the ground state, first excited state, second excited state,…. respectively. The
energy level E∞ = 0 is known as the Free State, where the electron no longer remains bound to the nucleus.

d) Energy Level Diagram of Hydrogen:
A diagram showing different energy levels of hydrogen with their relative spacing is called its energy level diagram.
The energy of different states of hydrogen are given by: En = −
me4
8𝜀𝑜
2h2
1
n2 = − 13.6 ×
1
n2 eV.
This gives: E1 = -13.6 eV, E2 = -3.4 eV, E3 = -1.51 eV, E4 = -0.85 eV, E5 = -0.54 eV,……………..and E∞= 0 eV.
So the energy level diagram of hydrogen will be as shown in the figure below:
-13.6 n = 1
-3.4 n = 2
-1.51 n = 3
-0.85 n = 4
-0.54 n = 5
n = ∞
0
Ground Level
Energy
(eV)

e) Excitation Energy and Potentials:
Excitation Energy:
The minimum energy required to excite an electron from the ground state to any of the excited states
is known as the excitation energy.
Therefore,
The first excitation energy , E2 – E1 = -3.4 – (-13.6) = 10.2 eV,
Second excitation energy, E3 – E1 = -1.51 – (-13.6) = 12.09 eV.
Third excitation energy, E4 – E1 = -0.85 – (-13.6) = 12.75 eV and so on.
-13.6 n = 1 (ground state)
-3.4 n = 2 (1st excited state)
-1.51 n = 3 (2nd excited state)
-0.85 n = 4 (3rd excited state)
-0.54 n = 5
n = ∞ (freestate)
0
Ground Level
Energy
(eV)

e) Excitation Energy and Potentials:
Excitation Potential:
1 eV is the energy gained by an electron when it is accelerated through a potential difference of 1
volt. Therefore
The excitation potential of an energy level of an atom is defined as the electric potential through
which an electron has to be accelerated so that when it collides with the atom, the atom is excited to
that energy level.
It means that, to excite a hydrogen atom in ground state to the first excited state, the colliding
electron has to be accelerated under a p. d. of 10.2 volts, so that its energy will be equal to the first
excitation energy, i.e. 10.2 eV.
Therefore, the first excitation potential of hydrogen is 10.2 volts. Similarly, second excitation potential
is equal to 12.09 volts, third excitation potential is 12.75 volts, etc.

f) Ionization energy and Ionization potential.
The minimum energy required to ionize an atom, that is, to excite the atom from ground state (n =1) to
the Free State (n =∞) is known as ionization energy.
Again, the ionization potential of an atom is defined as the potential through which an extra electron is to
be accelerated so that when it collides with the atom, the atom is ionized.
For hydrogen, ionization energy is equal to E∞- E1 = 0 – (-13.6) = 13.6 eV.
In order to gain this much energy, colliding electron needs to be accelerated under a p. d. of 13.6 volts.
Therefore, the ionization potential of hydrogen is 13.6 volts.
-13.6 n = 1 (ground state)
-3.4 n = 2 (1st excited state)
-1.51 n = 3 (2nd excited state)
-0.85 n = 4 (3rd excited state)
-0.54 n = 5
n = ∞ (freestate)
0
Ground Level
Energy
(eV)

g) Spectral Series of Hydrogen:
When an electron jumps from a higher orbit with n = n2 to a lower orbit with n = n1, it emits energy in the form of radiation. The
energy of radiation emitted is equal to the difference between the energy of two levels.
Thus, radiated energy- hν = En2
− En1
--------------------- (1)
Where ν (pronounced as ‘new’) is the frequency of radiation emitted.
From Bohr’s theory of hydrogen atom, we have-
En1
= −
me4
8𝜀𝑜
2h2
1
n1
2
And, En2
= −
me4
8𝜀𝑜
2h2
1
n2
2
Therefore, from equation (1)
hν = −
me4
8𝜀𝑜
2h2
1
n2
2 +
me4
8𝜀𝑜
2h2
1
n1
2
or hν =
me4
8𝜀𝑜
2h2
1
n1
2 −
1
n2
2
But speed of light, c = νλ, where λ is the wavelength of radiation emitted.
So hν = h
c
𝜆
=
me4
8𝜀𝑜
2h2
1
n1
2 −
1
n2
2
giving-
1
λ
=
me4
8𝜀𝑜
2ch3
1
n1
2 −
1
n2
2

g) Spectral Series of Hydrogen:
1
λ
=
me4
8𝜀𝑜
2ch3
1
n1
2 −
1
n2
2
This is written as
1
λ
= Ry
1
n1
2 −
1
n2
2 --------------------- (2)
Here, Ry =
me4
8𝜀𝑜
2ch3 is a constant quantity having value equal to 1.097×107 m-1 and is known as Rydberg’s constant.
With different values of n1 and n2 , equation (2) gives rise to a series of spectral lines, known as spectral series of
hydrogen. They are:
i) Lyman series: This series corresponds to the transitions of electron from n2 = 2, 3, 4, ….. to n1 = 1. Thus for Lyman series,
1
𝜆
= 𝑅𝑦
1
12 −
1
𝑛2
2 with n2 = 2, 3, 4, …..
The spectral lines in this series lie in the ultra violet (UV) region.
ii) Balmer series: This series corresponds to the transitions of electron from n2 = 3, 4, 5,… to n1 = 2. Thus for Balmer series,
1
𝜆
= 𝑅𝑦
1
22 −
1
𝑛2
2 with n2 = 3, 4, 5,…..
The spectral lines in this series lie in the visible region.
iii) Paschen series: This series corresponds to the transitions of electron from n2 = 4, 5, 6, to n1 = 3. Thus for Paschen series,
1
𝜆
= 𝑅𝑦
1
32 −
1
𝑛2
2 with n2 = 4, 5, 6,….., and it corresponds to Infrared region.

g) Spectral Series of Hydrogen
iv) Brackett series: This series corresponds to the transitions of electron from n2 = 5, 6, 7,….. to n1 = 4.
Thus for Brackett series,
1
𝜆
= 𝑅𝑦
1
42 −
1
𝑛2
2 with n2 = 5, 6, 7,…..
This series corresponds to the infrared (IR) region.
v) Pfund series: This series corresponds to the transitions of electron from n2 = 6, 7,….. to n1 = 5.
Thus for Pfund series,
1
𝜆
= 𝑅𝑦
1
52 −
1
𝑛2
2 with n2 = 6, 7,….., which lies in far infrared region.
These transitions are shown in following diagram:
n =1
n =2
n =3
n =4
n =5
n =6
n = ∞
Lyman Series
Balmer Series
Paschen Series
Brackett Series
Pfund Series

Emission spectra and Absorption Spectra
The spectra which is obtained due to the transition of electrons from higher energy states to the lower energy states
is called emission spectra. When an electron jumps down from a higher energy states to lower energy states, the
wavelength of the emitted line is given by:
hc
λ
= En2
− En1
. Such spectra is obtained when electric discharge is
passed through a discharge tube containing the experimental gas at low pressure or when atoms in the material are
excited by heating it to a high temperature. The emission spectra of Hydrogen is as shown in fig. 1 below.
The absorption spectra is obtained due to the transition of electrons from lower energy states to higher energy states
when light is allowed to pass through a cold, dilute gas. The gas atoms absorb the photons of particular wavelengths
for their excitation and the spectral lines corresponding to these wavelengths are found missing from the resulting
spectrum. These absorbed wavelengths appear as dark lines in the spectrum. If an atom is excited from a lower
energy state to a higher energy state, the wavelength of absorbed photon is given by:
hc
λ
= En2
− En1
.
n =1
n =2
n =3
n =4
n =5
n =6
n = ∞
Fig. 1: Emission spectra of Hydrogen
Lyman
Series
Balmer
Series
Paschen Series
Brackett
Series
Pfund
Series
12/17/2022
Aroma English Secondary School,
n =1
n =2
n =3
n =4
n =5
n =6
n = ∞
Fig. 2: Absorption spectra of Hydrogen
Lyman Series
Balmer Series
Paschen Series
Brackett Series
Pfund Series

Limitations of Bohr’s theory
1. It does not explain about the observed fine structure in Hydrogen spectrum, where an apparently single
spectral line is actually composed of two or more closely spaced lines.
2. It fails to explain the spectra of complex atoms containing more than one electrons.
3. It does not predict about the relative intensities of spectral lines.
To account for these limitations Bohr-Sommerfeld atomic model was introduced, which uses the
concept of elliptical orbits. The circular orbits are only the special case of these elliptical orbits.
De Broglie Hypothesis: Wave-Particle Duality of Matter
Light (or Radiation) behaves as wave in the phenomenon like Interference, Diffraction, Polarization etc, and as
particle in the phenomenon like Photoelectric Effect, Compton Effect etc. It means that light behaves both as wave
and as particle.
In 1924, de Broglie suggested that not only light, matter should also possess dual nature on the basis of following
facts:
i. Total energy density of universe exists both as radiation and matter. As radiation has dual nature, matter should
as well.
ii. Nature loves symmetry. So just like radiation, matter should also have a dual nature.
12/17/2022

De Broglie Hypothesis: Wave-Particle Duality of Matter
According to him, a particle having a linear momentum equal to ‘p’ has an associated wavelength ‘λ’ given by:
λ =
h
p
Where h is Planck’s constant. The wavelength given by this equation is known as de Broglie wavelength.
Examples: de Broglie wavelength of electron or proton:
Let, mass of particle = m
Charge of the particle = e
If the particle is accelerated by a p. d. equal to V,
Gain in kinetic energy,
1
2
m𝑣2 = eV
or, its speed, 𝑣 =
2eV
m
Now, de Broglie wavelength, λ =
h
p
=
h
m𝑣
or, λ =
h
m
m
2eV
or, λ =
h
2meV
From Planck’s quantum theory, E = hν
And from Einstein theory, E = mc2
, so that
hν = mc2
or h
c
λ
= mc2
or
h
λ
= mc, which is equivalent to momentum (p) of photon.
Then for particle of momentum p = mv,
h
λ
= p = mv
or λ =
h
p
=
h
mv
12/17/2022
For electron, m = me = 9.1 × 10−31 kg
For proton, m = mp = 1.67 × 10−27 kg
So, λe =
h
2meeV
for electron. And,
λp =
h
2mpeV
for proton.
Q(1). Calculate de Broglie wavelength for (i) proton and (ii)
electron, accelerated through the p. d. of 1 KV
Q(2). If matter has wave nature, Why don’t we observe it when a
ball is falling?

Heisenberg’s Uncertainty Principle
In 1927, German Physicist Werner Heisenberg developed a principle which states that- “It is impossible to determine
precisely and simultaneously the values of the canonically conjugate variables”.
This is known as Heisenberg’s Uncertainty principle. The canonically conjugate variables refers to the pair of physical
quantities which describe the motion of an atomic system. Position (𝑥) and momentum (p), energy (E) and time (t)
etc are examples of such variables.
According to this principle, the position(𝑥) and momentum(P) of a particle can’t be determined simultaneously with
a perfect accuracy. If ∆𝑥 is the uncertainty in position and ∆p is the uncertainty in momentum then these
uncertainties are limited by the relation,
∆𝑥 × ∆p ≥
h
2π
Likewise for the measurement of energy(E) and time(t), uncertainty principle will be
∆E × ∆t ≥
h
2π
Applications:
This principle can be used to show how an electron can’t exist inside the nucleus, to estimate the ground state
energy of a system, etc.
12/17/2022
Q. By using Heisenberg’s uncertainty principle, show that an electron can’t
exist in nucleus.
The size of nucleus is ~10−14 m. If electron exists inside nucleus it can
stay anywhere within it. So, maximum uncertainty of its position in
nucleus would be ∆𝑥 = 10−14
m
According to Heisenberg’s uncertainty principle, the uncertainty in
momentum of the electron is given by (∆𝑥)(∆p) ≥
h
2π
or ∆p ≥
h
2π∆𝑥
or ∆p ≥
6.626×10−34
2π×10−14 = 1.055 × 10−20
kgm/s
It means, the momentum of electron should at least be equal to 1.055 ×
10−20
kg m/s. That is-
p > 1.055 × 10−20 kg m/s,
or mev = p > 1.055 × 10−20
Where me is mass and v is the speed of electron.
Therefore, speed of electron, v >
1.055×10−20
me
=
1.055×10−20
9.1×10−31 = 1.6 ×
1010m/s which is greater than the speed of light c, and is impossible.
Thus, it is clear that an electron can never exist inside the nucleus

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