Transcript Atomic Spectra •If you “engergize” an atom in a variety of ways, it will give off light with characteristic wavelengths •What wavelengths you.

Atomic Spectra
•If you “engergize” an atom in a variety of ways, it will give off light
with characteristic wavelengths
•What wavelengths you get are a characteristic of the atom
1
1
1
•There were patterns in the wavelengths,


but no one understood why
A B C
•Triplets were not uncommon
Diffraction
•Hydrogen had the simplest pattern
Grating
2
n
   364.7 nm  2
n 4
1
1 1 
   91.17 nm    2 
4
n


•Later, more lines were discovered
1
 1
   91.17 nm   2  2 
m n 
1
m, n  1, 2,3,
nm
Sample Problem:
1
 1
   91.17 nm   2  2 
m n 
1
For what values of m and n will you get red light,
with wavelength between 625 and 740 nm?
1
1
 1
625 nm   91.17 nm  2  2   740 nm
m n 
1
1
 1
6.86   2  2   8.12
m n 
1
1
0.123  2  2  0.146
m n
•If m is 3 or bigger, you can’t get bigger than 1/9 =0.1111
•If m is 1, then n is two or bigger, and the difference is no
smaller than 0.75
1
0.123  0.2500  2  0.146
•m = 2
n
•You need to subtract at least 0.1  n is not too large
•n = 3
0.123  0.2500  0.1111  0.146
Rutherford Scattering
•Radioactive decay produces alpha particles
•Charge +2e
•Mass = Helium atom mass
•7000+ times heavier than electron
•Expected it would deflect – slightly – when it went through
an atom

•1909: Geiger and Marsden perform this
experiment under the direction of Ernest Rutherford
What they found:
•Alpha particles would deflect, occasionally, by large angles
•Even backwards
•Rutherford concluded that the positive charge was concentrated in a tiny region called
kqQ
the nucleus
b = impact parameter
Q = Ze F  2
r
+
b
r

q = 2e

   m v b
cot   
 2  kqQ
2
If you hit Gold (Au, Z = 79) with Ekin = 3.0 MeV -particles,
1. How close do you have to get to be deflected by 90?
2. What is the area of the target to be deflected by 90?
kqQ
   2kZe2
kZe2
b
cot   
cot  45  
2
2
m v
Ekin
 2  m v
Continued . . .
Rutherford Scattering (cont.)
If you hit Gold (Au, Z = 79) with Ekin = 3.0 MeV -particles,
1. How close do you have to get to be deflected by 90?
2. What is the area of the target to be deflected by 90?
8.99 10 N  m / C   79  1.602 10 C 
kZe

b

6
19
Ekin
3.0 10 eV 1.602 10 J/eV 
 3.79 1014 m  37.9 fm
9
37.9 fm
2
2
2
19
2
deflects < 90
deflects > 90
+
Target area:
 = b2 = 4512 fm2
The fact that this formula
worked told Rutherford the
nucleus was smaller than this,
smaller than 38 fm in radius . . .
How small is the nucleus?
How small a nucleus can one see by this method?
•You can’t see the nucleus until you get close enough
•Closest you can come happens when you are headed straight for the nucleus

Ekin
+
2
kqQ
2
Zke


,
Rmin
Rmin
•Conservation of energy gives you the closest approach
•Rutherford realized he needed higher energies and smaller Z
•He used Aluminim (Z = 13) with Ekin = 7.7 MeV
Rmin
2Zke

Ekin
2 13 8.99 10 N  m / C
9
2

6
7.7

10

1.602 10
eV 1.602 10 J/eV 
2
19
2
C
2
19
 4.86 1015 m  4.86 fm
He saw that the scattering changed, and concluded that the nucleus was a few fm
across
The atom and the Solar System
Atoms are about a = 0.1 nm in radius
Nucleus is about R = 10 fm in radius
The Solar system
•Sun’s mass is 1048 times Jupiter’s
mass
•Radius of Neptune’s orbit is 6500
times radius of Sun
•Sun stays put and all planets orbit it
The hydrogen atom
•Nuclear mass is 1836 times
electron’s mass
•Radius of atom is about 10,000
times radius of nucleus
Could the atom be like
a mini-solar system?
The Rutherford model of the atom
1911: Rutherford develops his model
•Nucleus at the center, nearly stationary
•Electrons orbit nucleus in ellipses, like planets
Problems with the model:
•Accelerating electrons should radiate, and
edrop into nucleus
•What about electron-electron interactions?
•No predictions of spectra of atoms
e-
+
e-
e-
Equations you need
E  hf  
 f  c  3.00 108 m/s
eVmax  hf  
Values of h, h-bar,
e, k, kB, me
U
 kBT 
3
15  c 

2
a 1010 m  0.1 nm
 h 2
h
   
1  cos  
me c
R 5 1015 m  5 fm
h
 2.426 1012 m
me c
4
2d cos   m
T  2.8978 10 m  K
3
 
m v 2b
cot   
 2  kqQ
2Zke 2
Rmin 
Ekin
1
 1
   91.17 nm   2  2 
m n 
1
The Bohr Model of the atom
1913: Extension of Rutherford model, with three new assumptions
1. Electrons only go in circular orbits due to attraction of nucleus
2. When the electron changes its configuration, energy is
E  hf  
emitted in the form of a single photon of energy
3. The electrons must have angular momentum that is an integer multiple of
2
me  n 
ke 2

  2
r  me r 
r
n
v
me r
2
n
ke
 2
3
me r
r
Electrostatic
ke 2
F 2
Attraction:
r
e-
Subs
2 2
L  n  me vr
2
2
me v
ke
 2
r
r
Centripetal Force: F 
me v
2
+
2
2
n
rn 
me ke 2
The Bohr Model of the atom (part 2)
n
v
me r
2
2
n
rn 
me ke 2
Define:
Bohr Radius
•Diameter of H-atom for n = 1 is 0.106 nm!
n  me ke 2  ke2
v
 2 2  
me  n 
n
a0 
2
me ke
2
 0.05297 nm
rn  a0 n2
v c  n
e-
Define: Fine Structure Constant
ke 2  1

c 137
•Electrons are non-relativistic in hydrogen
But wait, there’s more!
+
The Bohr Model of the atom (part 3)
2
ke
v
n
2
2
n
rn 
me ke 2
2
ke

c
Energy has two components:
•Kinetic energy:
Ekin
•Potential energy*
Epot
2
2 4
m
k
e


ke
2
e
1
1
 2 mev  2 me 
 
2 2
2 n
n

2
2
ke
ke
  Fdr   2 dr  
er
r
2 4
2
m
k
e
e
2  me ke 
  ke  2 2    2 2
n
 n 
E  Ekin  Epot
2
13.6 eV
me k 2e4
 2 me c 2

E 2 2 
2
2
n
2n
2 n
* Note: the sign of this equation assumes the force F is directed inwards.
+
The Bohr Model of the atom (part 4)
What happens when the electron “jumps” from one level to another?
2. When the electron changes its configuration, energy is emitted in
the form of a single photon of energy
2 c
mk e
1
1
 hf  E  e 2  2  2 

2
m n 
2 4
1 me k e  1
1 

 2
3 
2
 4 c  m n 
2 4
Define:
Rydberg Constant
me k 2e4
R 
3
4 c
Compare:
1
 1
R  2  2
m n 
R1  91.13 nm
level n
me k 2e4
E 2 2
2 n
e-
1
1

1
 1
   91.17 nm   2  2 
m n 
level m
+
1
E
 2 me c 2
2n 2
Reduced Mass
1

137
mec2  0.5110 MeV / c2
13.6 eV
E
n2
•Calculated treating nucleus as stationary
•Proton weighs 1836 times more than electron
•In fact, proton moves a very little bit as electron orbits it
•Can be corrected for by using “reduced mass”
•me  
1
1 1
Mm


M

m

•The real formula for energy levels of Hydrogen:
•Spectroscopy can easily tell the difference
•Slight corrections for isotopes
•0.027% for 1H vs. 2H
•This is why wavelengths were slightly off
•Huge correction for “unusual” atoms
e-
M m
E
 c
2
2n
2
2
+
Sample Problem
Positronium consists of a bound state of an electron (m = 0.511
MeV/c2) and an anti-electron, or positron (same mass). What is
the binding energy of positronium in the ground state?
Mm
me2 1


 2 me
M  m 2me
E
 c
2
2n
2
2

 2 me c 2
4n
2
1

137

5.1110 eV
5
4 137  1
2
2
 6.81 eV
e-
e+
Hydrogen-like atoms
•Atoms with one electron in them
•H, He+, Li+2, Be+3, . . . , U+91
•Bohr model could be easily modified to make them fit
2
•e2  Ze2
Electrostatic
ke
F 2
Attraction:
r
Exact
E
 2 Z 2 c2
2n 2
Mm

M m
v Z

c
n
+
 Ze
kZe
F 2
r
Approx.
13.6 eV  Z 2

E
n
2
r  n2a0 Z
a0  0.05297 nm
Note: This
ignores
relativity
2
Sample Problem
What is the energy and wavelength of light emitted when a
Ca+19 (Z = 20 ) electron jumps from level 7 down to level 6?
 13.6 eV  202

 111.0 eV

2
2
13.6 eV  Z 

7

E

2
2
13.6
eV
20
n



 151.1 eV
2

6
E  40.1 eV
c hc  4.136 1015 eV  s  3.00 108 m/s 
8
 

3.09

10
m

f
E
40.1 eV
  30.9 nm
Multiple Electron Atoms: X-rays
Could it be used to explain other atoms, with multiple electrons?
•Generally failed
•Succeeded approximately for innermost electrons
•Fast moving electron hits an inner electron and knocks it free
•Outer electron falls to lowest level and fills in the void
•Measure wavelength/energy of emitted photon
e-
Why it might work:
•Innermost electron is near nucleus with large charge
•Other electrons are outside of it
e-
1
2
hf  E  13.6 eV  Z 1  2 
 n 
e-
e-
1/2
13.6 eV 
1 
f 
1  2  
 h  n 
Z
ee-
?
e-
+
ee-
e-
ee-
X-rays from inner electrons
1/2
13.6 eV 
1 
f 
1  2  
 h  n 
Z
?
•1913 Henry Moseley measures X-ray spectra
•Finds this relationship almost works for inner electrons
1/2
13.6 eV 
1 
f 
1  2  
 h  n 
 Z  1
e-
•Why Z – 1?
•Innermost electron slightly screened by
other electron
•Also found a relationship for next level out
1/2
13.6 eV  1 1  
f 
  2 
 h  4 n 
ee-
e-
 Z  7.4
•There are more electrons screening the nucleus
+
e-
e-
e-
ee-
e-
e-
e-
Testing one of Bohr’s Hypotheses
2. When the electron changes its configuration, energy
is emitted in the form of a single photon of energy E  hf 

•If the atom has only discrete energy levels, then there should be a minimum amount
of energy that can be added to it
•Anything that collides with it with less energy must collide elastically
•When it does collide inelastically, the energy will be transmitted to the atom
•The atom then has to release a photon to lose a level
e-
Predictions:
•When you hit the threshold of energy, scattering (energy
loss) will suddenly increase
•The target atoms will start to glow at threshold energy
+
e-
e-
The Franck-Hertz Experiment
heater
thin gas
wire screen
e+
V
+
–
measure
current
–
Experiment performed, 1914
•At moderate voltage, current is high
•Drops suddenly as you pass threshold
•Drops periodically at higher voltage
•Gas glows at predicted wavelength above first threshold
Assessing the Bohr Model
The Good:
•Accurately predicted hydrogen line spectrum
•Predicted isotopic and other effects
•Did pretty well at predicting X-ray lines for heavier elements
•Clear evidence that atoms had “levels” of energy
•Evidence that quantum effects would help describe the atom
The Bad:
•No idea how to proceed for heavier elements
•“Strength” of spectral lines was not predicted
The Ugly:
•Arbitrary and unmotivated assumptions