130_notes.dvi

and that angular momentum is quantizedsuch thatL=n ̄h, forn= 1, 2 , 3 .... This is natural
since ̄hhas units of angular momentum. Bohr correctly postdicted the Hydrogen energies and the
size of the atom.

Balance of forces for circular orbits.
mv^2
r

=

1

4 πǫ 0

e^2

r^2

Angular momentum quantization assumption.

L=mvr=n ̄h

Solve for velocity.

v=

n ̄h

mr

Plug into force equation to get formula forr.

mn^2 ̄h^2 r^2

m^2 r^2 r

=

e^2

4 πǫ 0

n^2 ̄h^2

mr

=

e^2

4 πǫ 0

1

r

=

me^2

4 πǫ 0 ̄h^2

1

n^2

Now we just want to plugvandrinto the energy formula. We write the Hydrogen potential in
terms of the fine structure constant (See section 2.5.2): αSI= 4 πǫ^10 e

2

̄hc≈

1

137.

V(r) =

−α ̄hc

r

1

r

=

αmc

̄h

1

n^2

We now compute the energy levels.

E =

1

2

mv^2 +V(r)

E =

1

2

m

(

n ̄h

mr

) 2

−

α ̄hc

r

E =

1

2

m

(

αn ̄hc

n^2 ̄h

) 2

−

α^2 ̄hmc^2

n^2 ̄h

E =

1

2

m

(αc

n

) 2

−

α^2 mc^2

n^2

E =

1

2

(

α^2 mc^2

n^2

)

−

α^2 mc^2

n^2

=−

α^2 mc^2

2 n^2

The constant α

(^2) mc 2
2 =
511000
2(137)^2 = 13.6 eV. Bohr’s formula gives theright Hydrogen energy
spectrum.