Lab 11: Atomic Spectra
Only 1 more lab to go!!
Bohr’s atomic model contains some assumptions about how an electron orbits the nucleus of
an atom.
As an electron orbits the nucleus it’s angular momentum L = I2 (I=mr2) must be quantized.
Only certain angular momentums are allowed. Bohr proposed that the
electron’s angular momentum is an integer, n, times h/2
h
V 
mr    n
2
r
2
Since the electron has quantized angular momentum so to will the
energy levels of the electron. We can find the energy of an electron by:
1 2
q2
E  KE  PE  mv  k
2
r
where k = 9 X 109 Nm2/C2 and q = 1.6 X 10-19 C
The electron experiences an electrostatic force due to the positively charged nucleus. If we
write Newton’s 2nd Law:
Fe  mac
q2
V2
k
m
r
r
Acceleration experienced with
circular motion
We can use these equations to solve for the radius of orbit for the electron as
well as the energy levels of the electron orbits:
2m 2 q 4 1
13.6eV
En   2 2

2
k h n
n2
1eV  1.6 x1016 J
n=1
n=2
n=3
When an electron makes a transition from a higher energy level (nupper)
to a lower energy level (nlower) a photon of light is emitted equal in
energy to the difference in the energy levels made by the electron’s transition. In other words:
E photon  Eupper  Elower
E photon 
hc

hc

 Eupper  Elower
We can rearrange this equation to be:
 1
1 
 R 2  2 

 nlower nupper 
1
R  1.09737 107
1
m
The human eye can see wavelengths between
4000 Ang.- 7000 Ang. Using this information
we can calculate the electron transitions that
will emit photons in the visible spectrum:
n=1
n=2
1

n=3

R
R
 2
4 nupper
n=4
1
n=5

slope = -R
y-int = R/4
1
2
nupper