BOHR’S ATOM MODEL

Bohr proposed an atom model in the year 1913 in order to explain

the spectra emitted by hydrogen atom.
Bohr took Rutherford’s conclusion as the basis for his atom model.
That is an atom consists of a positively charged nucleus at its centre
and is surrounded by electrons in the extra nuclear space.
The drawback of the Rutherford atom model is the stability of the atom.
If an electron is at rest, it is attracted by the nucleus. If it is revolving around the nucleus,
it radiates energy. Due to the continuous emission of energy by the electron
while revolving around the nucleus, the energy of the electron decreases
and it finally spiral down into the nucleus. But an atom is found to be a stable one.
He rectified the drawback of the Rutherford atom model by the following postulates.
The basic postulates of Bohr’s theory are a combination of the Classical physics and
Planck’s quantum theory of radiation.

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BASIC POSTULATES:

1. An electron cannot revolve around the nucleus in all orbits

arbitrarily as suggested by the classical theory. The electrons can revolve around the nucleus
only in those allowed orbits (or permitted orbits) for which
ℎ
the angular momentum of the electron is an integral multiple of ,
2𝜋

where ‘h’ is the Planck’s constant 6.625 𝑋 10−34 𝐽𝑠 .

These orbits are called Stationary orbits and an electron revolving in these orbits
does not radiate energy.
Explanation: Let us consider an electron of mass ′𝑚′ moving with a speed ′𝑣′ in an orbit
of radius′𝑟′. Then, the angular momentum 𝐿 of the electron,
𝑳= 𝒎𝒗𝒓 −−−−− 𝒂
According to Bohr first postulate, the angular momentum 𝐿 of an electron is
𝒉
𝑳=𝒏 −−−−− 𝒃
𝟐𝝅

𝒉
𝒎𝒗𝒓 =𝒏
𝟐𝝅
Where 𝑛 = 1, 2, 3, 4, 𝑒𝑡𝑐 for the first, second, third and fourth orbits respectively.
It is called the Principal quantum number.
2. An atom radiates energy, only when an electron jumps from a stationary orbit of higher energy
to another orbit of lower energy. If the electron jumps from an initial orbit of energy 𝐸𝑖
to a final orbit of energy 𝐸𝑓 ,
𝐄𝐢 − 𝐄𝐟
a photon of frequency 𝛎 = is emitted.
𝐡
The relation between the energy released and frequency of radiation emitted is,
𝑬
𝑬𝟐 − 𝑬𝟏 = 𝒉𝝂 𝑎𝑛𝑑 𝝂 =
𝒉

BOHR FORMULAE:
Based on the above postulates, Bohr derived the formulae for
(i) Radii of the stationary orbits 𝑟𝑛 and
(ii) The total energy of the electron in the orbit 𝐸𝑛 .
 Consider an atom whose nucleus has a charge ′𝑍𝑒′ and mass ′𝑀′ .
Let an electron of charge ′ − 𝑒′ and mass ′𝑚′ move around the nucleus in an orbit of radius ′𝑟′
as shown in the figure. ∵ 𝑀 >> 𝑚, the nucleus is stationary. Hence the mass of the nucleus
does not come into calculations.

The electrostatic force of attraction between the nucleus and the electron 𝐅𝐞 is,
𝑍𝑒 2
𝐅𝐞 = −−−−− 1
4𝜋𝜀0 𝑟 2
Where 𝜀0 is the permittivity of free space and has the value 8.854 𝑋 10−12 𝐹𝑚−1 .
mv 2
The centrifugal force experienced by the electron 𝐅𝐜 = −−−−− 2
r
The atom will be stable only if the centrifugal force experienced by the electron 𝐅𝐜
is balanced by the electrostatic force of attraction between the nucleus and
the revolving electrons 𝐅𝐞 .
𝐅𝐜 = 𝐅𝐞
m 𝟐 𝑍𝑒 2
𝐯 = −−−−− 3
r 4𝜋𝜀0 𝑟 2

According to Bohr’s first postulate,

Squaring on both sides, we get,
𝒉
𝒎𝒗𝒓=𝒏 𝑛2 ℎ2
𝟐𝝅 𝒗𝟐 = −−−−− 4
𝑛ℎ 4𝜋 2 𝑚2 𝑟 2
𝒗=
2𝜋 𝑚𝑟
Substituting the value of 𝒗𝟐 from Equation (4) in Equation (3)
𝑚 𝑛2 ℎ2 𝑍𝑒 2
=
𝒓 4𝜋 2 𝑚2 𝑟 2 4𝜋𝜀0 𝑟 2
1 𝑍𝑒 2 4𝜋 2 𝑚2 𝑟 2 1
=
𝒓 4𝜋𝜀0 𝑟 2 𝑛2 ℎ2 𝑚
𝟒𝛑𝛆𝟎 𝐫 𝟐 𝒏𝟐 𝒉𝟐 𝒎
𝒓=
𝒁𝒆𝟐 𝟒𝝅𝟐 𝒎𝟐 𝐫 𝟐 1
𝒏𝟐 𝒉𝟐 𝛆𝟎
𝑟=
𝒁𝒆𝟐 𝝅𝒎
Therefore, the radius of the 𝑛𝑡ℎ permissible orbit of hydrogen can be written as,
𝒏𝟐 𝒉𝟐 𝜺𝟎
𝒓𝒏 = 𝟐 −−−−− 𝟓
𝒁𝒆 𝝅 𝒎
𝐧𝟐 h2 ε0
𝐫𝐧 = ∵ Z = 1 for the Hydrogen atom
π me2
From the above equation, we find that 𝐫𝐧 𝛂 𝐧𝟐
Thus the radii of the orbits are in the ratio of 𝟏 ∶ 𝟒 ∶ 𝟗 ∶ 𝟏𝟔 ∶ 𝟐𝟓, etc.
Let us calculate the radii of the orbits of a hydrogen atom. From Equation (5),
n2 h2 ε0
rn = For hydrogen Z = 1
Ze2 π m
n2 h2 ε0 n2 6.625 X 10−34 2 8.854 X 10−12
rn = =
π me2 3.14 9.11 X 10−31 1.6 X 10−19 2
𝐫𝐧 = 𝟎. 𝟓𝟑 𝐧𝟐 𝐀𝐨 − − − − − 6
For the first orbit 𝑛 = 1 , the radius 𝑟1 is,
2
r1 = 0.53 1 = 0.53 1 = 0.53 Ao
𝐫𝟏 = 𝟎. 𝟓𝟑 𝐀𝐨
Similarly, for the second orbit 𝑛 = 2 , the radius 𝑟2 is,
2
r2 = 0.53 2 = 0.53 4 = 2.12 Ao
𝐫𝟐 = 𝟐. 𝟏𝟐 𝐀𝐨 and so on.

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CALCULATION OF ENERGY:
The total energy of the electron 𝐸𝑛 in an orbit is the sum of its
potential and kinetic energies. The potential energy of the electron is considered to be zero
when it is at an infinite distance from the nucleus. The potential energy of an electron in an orbit
is given by the work done in bringing the electron from infinity to that orbit,
which is obtained by integrating the electrostatic force of attraction
between the nucleus and electron from the limits ∞ 𝑡𝑜 𝑟.
𝑟
𝑍𝑒 2
Potential energy of the electron = 𝑑𝑟
4𝜋𝜀0 𝑟 2
∞

− 𝐙𝐞𝟐
𝐏. 𝐄 𝐨𝐟 𝐭𝐡𝐞 𝐞𝐥𝐞𝐜𝐭𝐫𝐨𝐧 = −−−−− 7
𝟒𝛑𝛆𝟎 𝐫
From Equation 3

mv 2 𝑍𝑒 2 𝑍𝑒 2 mv 2 𝑍𝑒 2 𝑍𝑒 2
= mv 2 = = K. E =
𝐫 4𝜋𝜀0 𝑟 𝟐 4𝜋𝜀0 𝑟 𝟐 𝟐 4𝜋𝜀0 𝑟 8𝜋𝜀0 𝑟

𝐙𝐞𝟐
𝐊. 𝐄 𝐨𝐟 𝐭𝐡𝐞 𝐞𝐥𝐞𝐜𝐭𝐫𝐨𝐧 = −−−−− 8
𝟖𝛑𝛆𝟎 𝐫
Total energy of the electron in the nth orbit, 𝐄𝐧 = 𝐏. 𝐄 + 𝐊. 𝐄
− Ze2 Ze2
En = +
4πε0 r 8πε0 r
− 2 Ze2 + Ze2
En =
8πε0 r
− Ze2
En =
8πε0 r
− Ze2 1
En =
8πε0 𝐫
Substituting the value of ′𝐫′ from Equation (5), we get
− Ze2 1 n2 h2 ε0 𝟏 𝐙𝐞𝟐 𝛑𝐦
En = ∵ rn = ; =
8πε0 𝐫 Ze2 πm 𝐫𝐧 𝐧𝟐 𝐡𝟐 𝛆𝟎
− Ze2 Ze2 𝛑 m
En =
8 𝛑 ε0 n2 h2 ε0
 − 𝐙 𝟐 𝐞𝟒 𝐦
𝐄𝐧 = −−−−− 9
𝟖𝛆𝟎 𝟐 𝐧𝟐 𝐡𝟐
The negative value of the energy implies that the electron is bounded to the nucleus.
As the value of ′𝒏′ increases, ′𝑬𝒏 ′ also increases.
𝟏
∴ 𝐄𝐧 𝜶 −
𝐧𝟐
Hence, the outer orbits have greater energies than the inner orbits.

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BOHR’S INTERPRETATION TO HYDROGEN SPECTRUM:
If an electron jumps from an outer orbit 𝑛2 of higher energy
to an inner orbit 𝑛1 of lower energy, the frequency of the radiation emitted is,
𝐸𝑛 2 − 𝐸𝑛 1
𝜈=
ℎ
1
𝜈= 𝑬𝒏𝟐 − 𝑬𝒏𝟏 − − − − − 10
ℎ

− Z 2 e4 m − Z 2 e4 m
E𝑛 2 = From Eqn. 9 E𝑛 1 = From Eqn. 9
8ε0 2 𝑛2 2 h2 8ε0 2 𝑛1 2 h2

− 𝐦 𝐞𝟒 − 𝐦 𝐞𝟒
𝐄𝒏𝟐 = ∵ Z=1 𝐄𝒏𝟏 = ∵ Z=1
𝟖 𝛆𝟎 𝟐 𝒏𝟐 𝟐 𝐡𝟐 𝟖 𝛆𝟎 𝟐 𝒏𝟏 𝟐 𝐡𝟐

∴ The frequency 𝜈 of the spectral line emitted is,

1 − m e4 − m e4
𝜈= −
ℎ 8 ε0 2 𝑛2 2 h2 8 ε0 2 𝑛1 2 h2

1 − m e4 m e4
𝜈= +
ℎ 8 ε0 2 𝒏𝟐 𝟐 h2 8 ε0 2 𝒏𝟏 𝟐 h2

m e4 −1 1
𝜈= + 2
8 ε0 2 h3 𝑛2 2 𝑛1

𝐦 𝐞𝟒 𝟏 𝟏
𝝂= − − − − − − 11
𝟖 𝛆𝟎 𝟐 𝐡𝟑 𝒏𝟏 𝟐 𝒏𝟐 𝟐

𝜈 m e4 1 1
= 2 3 2
− 2 − − − − − 12
𝐜 8 ε0 𝐜 h 𝑛1 𝑛2
The number of waves containing in unit length in vacuum,
𝛎 1 𝛎 1
= =𝛎 ∵ c = 𝛎λ or =
𝐜 λ 𝐜 λ
Where ′𝜆′ is the wavelength and ′𝑐′ is the velocity of light.
me4 1 1
𝜈= 2 3 2
− 2
8 ε0 𝐜 h 𝑛1 𝑛2
𝟏 𝟏 me4
𝝂=𝑹 𝟐
− ∵ R=
𝒏𝟏 𝒏𝟐 𝟐 8 ε0 2 𝐜 h3
Where ‘R’ is a constant known as Rydberg constant. The above equation is the general expression
for the wave number of the radiation emitted by an electron as it jumps from higher orbit 𝑛2
to the lower orbit 𝑛1 .
Substituting the values of m, e, ε0 and h in the following equation, we get,

me4
R=
8 ε0 2 𝐜 h3

9.1 X 10−31 1.6 X 10−19 4

R=
8 8.854 X 10−12 2 3 X 108 6.625 X 10−34 3

R = 10.97 X 106 m−1 .

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SPECTRAL SERIES OF HYDROGEN ATOM:
(a) Lyman series: When an electron jumps from the second, third, fourth…etc., orbits
to the first orbit, the spectral lines are in the ultraviolet regions, Hence
n1 = 1
n2 = 2, 3, 4, etc.
1 1
𝜈=𝑅 2
− 2
𝑛1 𝑛2
i When n1 = 1 and n2 = 2
1 1
𝜈1 = 𝑅 2
− 2
1 2
1 1 4−1 3
𝜈1 =𝑅 − =𝑅 =𝑅
1 4 4 4
3
𝜈1 = 𝑅
4
1 3𝑅 1
= ∵ 𝜈=
λ1 4 𝜆
4 4 4 𝑋 10−6 4 𝑋 10−6
λ1 = = 6
= = = 0.1215 𝑋 10−6
3𝑅 3 10.97 X 10 3 10.97 32.91
𝛌𝟏 = 𝟏𝟐𝟏𝟓 𝐗 𝟏𝟎−𝟏𝟎 𝐦 𝐨𝐫 𝛌𝟏 = 𝟏𝟐𝟏𝟓 𝐀𝐨

ii When n1 = 1 and n2 = 3
1 1
𝜈2 = 𝑅 2
− 2
1 3
1 1 9−1 8
𝜈2 =𝑅 − =𝑅 =𝑅
1 9 9 9
8
𝜈2 = 𝑅
9
1 8𝑅 1
= ∵ 𝜈=
λ2 9 𝜆
9 9 9 𝑋 10−6 9 𝑋 10−6
λ2 = = = = = 0.1026 𝑋 10−6
8𝑅 8 10.97 X 106 8 10.97 87.76
𝛌𝟐 = 𝟏𝟎𝟐𝟔 𝐗 𝟏𝟎−𝟏𝟎 𝐦 𝐨𝐫 𝛌𝟐 = 𝟏𝟎𝟐𝟔 𝐀𝐨 and so on.

This series is known as Lyman series. The shortest wavelength of this series
is obtained by Substituting 𝑛2 = ∞.
(b) Balmer series: When an electron jumps from outer to second orbit,
i.e., 𝑛1 = 2, 𝑛2 = 3, 4, 5, 𝑒𝑡𝑐.
1 1
ν=R −
n1 2 n2 2
1 1
ν=R − where n2 = 3, 4, 5 etc.
22 n2 2
This series is known as Balmer series and it lies in the visible region of the spectrum.
The first line in this series 𝑛=3 is called 𝐻𝛼 line, the second line 𝑛=4
is called 𝐻𝛽 line and so on.
(c) Paschen series: If an electron jumps from higher orbits, i.e., 𝑛2 = 4, 5, 6, . . 𝑒𝑡𝑐
and it ends at 𝑛1 = 3 , the spectral line emitted is said to be Paschen series.
It falls on the Infrared region and are given by 𝑛1 = 3 and 𝑛2 = 4, 5, 6, . . 𝑒𝑡𝑐.
1 1
ν=R −
n1 2 n2 2
(d)

1 1
ν=R 2
− 2 where n2 = 4, 5, 6 etc.
3 n2
(d) Bracket series: If 𝑛1 = 4 and 𝑛2 = 5, 6, 7, . . . 𝑒𝑡𝑐, we get, the Bracket series.
1 1
ν=R 2
− 2
n1 n2
(e)

1 1
ν=R − where n2 = 5, 6, 7 etc.
42 n2 2
(e) Pfund series: If 𝑛1 = 5 and 𝑛2 = 6, 7, 8, . . . . 𝑒𝑡𝑐, we get, the Pfund series.
1 1
ν=R 2
− 2
n1 n2
(f)

1 1
ν=R 2
− 2 where n2 = 6, 7, 8 etc.
5 n2
The Bracket and Pfund series lie in the far infrared region of the hydrogen spectrum.
The spectral series of hydrogen atom are shown in the figure.

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ENERGY LEVEL DIAGRAM:
The diagrammatic representation of the energy ′𝐸′ is known as energy level diagram.
Substituting the values of m, e, Z and 𝜀0 in the Equation (9), we get,

− Z 2 e4 m
En =
8ε0 2 n2 h2

− 1 2 1.6 X 10−19 4
9.1 X 10−31
En =
8 8.854 X 10−12 2 n 2 6.625 X 10−34 2

𝟏𝟑. 𝟔
𝐄𝐧 = − 𝐞𝐕
𝐧𝟐
13.6
For n = 1, E1 = − = − 13.6 eV
12
13.6
For n = 2, E2 = − = − 3.40 eV
22
13.6
For n = 3, E1 = − = − 1.51 eV
32

The discrete energy levels and the possible spectral lines emitted are shown in figure.
The energy level corresponding to n = 1 is known as ground state 𝐸1 and
the other energy levels 𝐸2 , 𝐸3 , 𝐸4 etc. are known as excited states.
DRAWBACK’S OF BOHR’S THEORY:
Although Bohr’s theory could successfully explain the spectrum of hydrogen,
yet it had the following shortcomings.
1. This theory is applicable only to hydrogen-like single electron atoms and
fails in the case of atoms with two or more electrons.
2. In the spectrum of hydrogen, certain spectral lines are not single lines
but a group of closed lines with slightly different frequencies.
Bohr’s theory could not explain these fine features of the hydrogen spectrum.
3. It does not explain why only circular orbits should be chosen
when elliptical orbits are also possible.
4. As electrons exhibits wave properties also, so orbits of electrons
cannot be exactly defined in Bohr’s theory.
5. It does not tell anything about the relative intensities of the various spectral lines.
Bohr’s theory predicts only the frequencies of these lines.
6. It does not explain the further splitting of spectral lines in a magnetic field (Zeeman effect)
or in an electric field (Stark effect).

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