Transcript Spectra of Atoms

Spectra of Atoms
When an atom is excited, it emits light. But not in the
continuous spectrum as blackbody radiation! The light is
emitted at discrete wavelengths. The set of wavelengths
emitted by atoms of a given element is absolutely the same for
every atom of that element, no matter how hot or cold it is.
Mystery #4. What is the dynamics that causes line spectra?
Emission spectrum from hydrogen
n0=3
n0=2
n0=1
7
6
n0 = 2
index n
5
4
3
2
1
0
4000
4500
5000
5500
6000
6500
wavelength (Angstroms)
 1
1
1 / n   2  2 
 n0 n 
  Rydberg  1.1105 cm 1
7000
How to explain these numerical series?
Before we tackle that mystery, consider another:
Mystery #5: Why don’t the electrons in an atom spiral
into the nucleus?
In classical electrodynamics, an accelerated charged particle
radiates electromagnetic waves:
When an energetic (free) electron scatters from a nucleus, it emits
Bremmstrahlung radiation. When a charged particle is bent along a
curved trajectory in a magnetic field, it emits synchrotron radiation.
Bohr’s postulate: angular
momentum must be quantized!
Calculate the kinetic energy associated with the orbit of an
electron in a hydrogen atom, using classical mechanics:
mv 2
1 Ze 2

First write the orbit equation:
r
40 r 2
Now extract the angular momentum: L  mvr
40 2
Express orbit radius in terms of L:
r
L
2
Zme
Now quantize L:
L  n  nh / 2
The orbit radius can only have
discrete values:
1   0 2  2
n
r  
2 
Z  me 



Bohr radius
o
a0  0.53 A
Now calculate the quantized energy levels
First the potential energy:
Ze 2
U 
40 r
Now the kinetic energy:
1 Ze 2
1 2
T  mv 
80 r
2
Finally the total energy:
1 Ze 2
En  T  U  
80 r
Now insert the quantized radius:
 me4  1
En   Z  2 2  2
8 0   n




1
2
c / h
 13.6 eV
And recover the series spectra!
1
h
2  1
1 /   En  En0  Z  2  2 
c
 n0 n 


This is a remarkable result!
• The quantum of angular momentum is the
same as the proportionality constant in the
relation of energy and (angular) frequency:
E  
• Angular momentum now plays a very
special role in quantum physics: it can only
take on values that are integer multiples of
this basic unit value.
Let’s work a few problems
4.6 The energy loss per unit time due to an acceleration a
acting on an electron is given by the Larmor formula:
e2a 2
P
60c 3
a) What would be the power loss for an electron in the first
Bohr orbit in a hydrogen atom if it were able to lose energy
by this classical process?
First calculate the (centripetal) acceleration of the electron:
v2
a
r
Use orbit equation to express a in terms of mass, potential
energy, and radius:
v 2 1  1 Ze 2  1

 
r m  40 r  r
1  2h  1
2 13.6 eV
a 
 9 10 22 m / s 2
 
o
m  c  a0 (5 105 eV / c 2 )(0.53 A)
e2a 2
(1.6 10 19 C ) 2 (9 10 22 m / s 2 ) 2 (9 109 Jm / C 2 )
9
P0 


1
.
4

10
Watts
3
8
3
60 c
1.5  (3 10 m / s)
b) Estimate the time it would take the electron to spiral into the
nucleus if it able were to lose energy in this way.
We must realize that, as it spirals in, the radius gets smaller and
smaller. The acceleration a depends upon the radius, and
therefore so does P:
a04
1
a 2

P  P0 4
r
r
We must also realize that the radius depends upon the energy E:
Ze 2
r
80 E
So the change in radius is
 Ze 2  dE  Ze 2  80 r 
a04
 2  

dr  
P0 4 dt
2 
r
 80  E
 80  Ze 
2
We need to solve this as a differential equation:
4


8

P
a
2
0 0 0
dt
r dr  
2
Ze


C
1
d (r 3 )  Cdt
3
r 3  a03  3Ct
The time to collapse to zero radius is then
a03
Ze 2
(9 109 Jm / C 2 )(1.6 1019 C ) 2
10
t



5
.
2

10
s
o
3C 240 P0 a0
6(1.4 109W )(0.53 A)
4.18 a) Calculate the first three energy levels of Li++
Li++ has Z=3 and a single electron (it is a hydrogenic atom).
E   Z 2 (h / c)n 2  32 13.6eV  n 2
n 1
E1  122eV
n2
E 2  30.5eV
n3
E 3  13.5eV
b) What is the ionization potential of Li++?
The ionization potential is the energy difference between
the ground state level and the large-n limit:
  E1  122eV
c) What is the first resonance potential for Li++?
The first resonance potential is the energy needed to
excite the electron from the ground state to the first
excited state:
E  E2  E1  122eV  30.5eV  91.5eV
This is the first (lowest energy) line in the Lyman series of
emission spectral lines.