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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

)

J. F. Harrison Michigan State University 1

*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

J. F. Harrison Michigan State University 2

*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

J. F. Harrison Michigan State University 3