Bohr Model of the Atom
The Hydrogen atom consists of a relatively massive positively charged ( e ) nucleus (proton) around which an electron of charge – e moves.
The force attracting the electron to the proton is given by Coulombs law f
G
= e r
ˆ
G
4
πε
0 r 2 r
ε
0
is the magnitude of the position vector from the proton to the electron and
= −
12 C N m
−
2 is the permittivity of free space.
The velocity of the electron is equal to the time derivative of the position vector
G
V= dr dt
and if the orbit is circular r
G
= r cos
θ i ˆ + r sin
θ j ˆ where i
ˆ & ˆ j are unit vectors. The velocity becomes
G
V= d r
G dt
= r d
θ dt
(
− sin
θ i
ˆ + cos
θ j
ˆ
) and the magnitude is
V= r d
θ dt
= r
ω
We need the acceleration which is
G d V d
= dt dt
2
2 r
G
= r d
2
θ dt
2
(
− sin
θ i ˆ + cos
θ ˆ
) j r
⎛ d
θ ⎞ 2
⎝ dt ⎠
( cos
θ i ˆ + sin
θ j ˆ
)
Identifying r
G
ˆ =
( cos
θ i
ˆ + sin
θ j
ˆ
)
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means that the counterbalancing force is rm
⎛ d
θ
⎞
⎝ dt ⎠
2
= rm
ω
2
Equating the two forces gives
4 e
πε
2
0 r 2
= rm
ω
2
= m V 2 r
= m V 2 r
Or
4
πε
0 r
=
( m V
)
2 = p 2 =
= 2 n 2 r 2
Where Bohr used the de Broglie relationship p
= h
λ and assumed that
2
π r
= n
λ n
=
1, 2,3, . This results in the radius of the electrons orbit being r =
=
4
πε
0 = a n 2 n
0
=
1,2,3,
"
Where a
0
=
= 2 4
πε
0 is equal to
−
10 m . Note that it is the smallest radius possible for the electron (
One often refers to n
=
1 n
=
1 ) and is called the Bohr radius.
as the first Bohr orbit, n
=
2 as the second Bohr orbit and so on.
The energy of the electron in the H atom is the sum of its kinetic ( T ) and potential energies ( V )
E T V
= 1
2 m V 2
−
4 e
πε
2
0 r
= −
8
π e
ε
2
0 r
And using the value of the radius found above the energy becomes
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E n
= −
2
( me
4
πε
0
)
2
4
=
2 n
2
=
E
1 n 2 n
=
1, 2,3,
"
Where
E
1
= −
2
4
( me
4
πε
0
)
2 =
2
= − −
18
J
= −
13.606
e V
E
1
is the lowest possible energy for the H atom. As with the radius of the allowed orbits one refers to E
1
as the energy of the electron in the first Bohr orbit, E
2
as the energy of the electron in the second Bohr orbit and so on.
It’s useful to remember the possible energies of the H atom as
E n
= −
13.606
n 2 eV n
=
1,2,3, "
How fast is the electron moving in the allowed orbits? We can estimate this as follows
Since
2
π r
= n
λ = nh p
We have p
=
2 nh
π r
=
2
π nh a n
0
2
= m v
And so
2
π h
=
= a mn a mn
0 0
= v
So the speed (magnitude of the velocity) is v =
= a mn
0
= n x
6
It’s insightful to express this speed relative to the speed of light c. v
= c
137.0
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