Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture5
Bohr Theory
Bohr assumed that the angular momentum of the electron in a
hydrogen atom is quantized
- We can consider the nucleus to be fixed and the electron to
be revolving about it.
- The force holding the electron in a circular orbit is supplied
by the columbic force of attraction between the proton
and the electron.
 e2 
- If we equate coulomb’s force law  4 r 2  with Eq.(1-21)



then we have


e2
m v2


2 
4

r
r



where

(1-22)
is the permittivity of free space and is equal to
8.85 x10-12 (C2 /N. m2 ).
The factor 4  called coulomb’s law.
- We are assuming here that the electron is revolving around
the fixed nucleus in a circular orbit of radius r.
- Classically , because the electron is constantly being
accelerated accoding to Eq. 1.21 , it should emit
[1]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture5
electromagnetic radiation and loose energy , classical
physics predicts that an electron revolving around the
nucleus will loose energy and spiral into the nucleus and
this is forbidden.
-
It was Bohr’s great contribution to make two non classical
assumption.
- The first of these was to assume the existence of stationary
electron orbit , in defense of classical physics and in this
case he assumed that the angular momentum of the
electron must be quantized according to
L=mvr=nℏ
(1-23)
where n =1,2,3,…… ℏ called (h-bar) is equal to h/2𝛑
- Solving this equation for v and substituting into Eq. 1-22 we
obtain
4  n 2
r
me 2
(1-24)
Thus we see that the radii of the allowed orbits or Bohr
orbit are quantized , the electron can move around the
nucleus only in circular orbits with radii given by Eq. 1.24.
The orbit with the smallest radius is the orbit n=1 (for
hydrogen atom).
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture5
4 (8.85 1012C 2 .N 1.m 2 )(1.055 1034 J .S ) 2
r
(1-25)
(9.110 1031 Kg )(1.602 1019C ) 2
=5.29×10-11 m = 0.529Å
- Total energy= kinetic energy + potential energy
e 2
Potential energy U(r) = 4 r

2
1
e
E  mv2 
2
4  r
(1-26)
(1-27)
Using Eq. 1-22 to eliminate m v2 from (1-27)

1  e2
e2
E 


2  4  r 
4  r
=
e 2
8  r
Substitute r from eq. (1-24)
me 4
E n  2 2 2
8  h n
(1-28)
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture5
- The second assuming for Bohr that the observed spectrum
of the hydrogen atom is due to transition from one allowed
energy state to another.
me 4  1
1 
E  2 2  2  2   h
8  h  n1 n 2 
(1-29)
Where E=h is called the Bohr frequency and is the basic
assumption as the electron falls from one level to another
, the energy evolved is given off as a photon energy
E=h .
We can write Eq. (1-29) in the form of Rydberg formula by
writing h  hc
me 4 1
1
  2 3 ( 2  2)
8 o ch n1 n 2
If we compare Eq. (1-16) and (1-30)
me 4
RH  2 3
8 o ch
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(1-30)
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture5
Example 8
Using the values of the physical constants , Calculate RH
m: 9.10953×10-3 Kg, e=1.602189×10-19 C
𝛆o=8.85419×10-12 C2. N-1.m-2
C: 3×108 m.S-1, h: 6.63×10-34 J.S
And compare it to experimental value 109677.6 Cm-1
Solution:
me 4
RH  2 3
8 o ch
9.10953 103 (1.602 1019 ) 4
RH 
8(8.85 1012 )2  3 108 (6.63 1034 )3
=1.0973×107m-1
=109736 Cm-1
The results surely are makeable agreement with the experimental
value.
[5]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture5
Example 9:
Calculate the ionization energy of the hydrogen atom.
Solution:
The ionization enrgy IE is the energy required to take the
electron from ground state to the first unbond state ; n1=1 ,
n2=∞
1
1
IE  R H ( 2  2 )
1 
IE  R H  109680 Cm 1 
But
  hc
, h=6.63×10-34, c=3×1010 Cm.S-1
ε= 109680×6.63×10-34×3×1010
= 2.179×10-18J
= 2.179×10-18/1.602×10-19= 13.6 e.V
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