Transcript Material since exam 3 - UW-Madison Department of Physics

Material since exam 3
• De Broglie wavelength, wavefunctions, probabilities
• Uncertainty principle
• Particle in a box
– Wavefunctions, energy, uncertainty relation
– 1D, 2D, and 3D box, wavefunctions, energy
• 3D hydrogen atom
– Quantum #’s, physical meaning of quantum #’s
– Energies and wavefunctions
– Orbital magnetic dipole moment, electron spin
• Multielectron atoms
– State energies, electron configuration, periodic table
– Lasers
• Nuclear physics
– Isotopes, nuclear binding energy
– Radioactive decay
• Decay rates, activity, radiation damage
• Types of decay, half-life, radioactive dating.
Matter waves
• If light waves have particle-like properties,
maybe matter has wave properties?
• de Broglie postulated that the
wavelength of matter
is related to momentum as

h
p
• This is called
the de Broglie wavelength.
Nobel prize, 1929
Matter Waves
• deBroglie postulated that matter has wavelike
properties.
• deBroglie wavelength 
 h/p
Example:
Wavelength of electron with 10 eV of energy:

Kinetic energy
E KE 

p
2
 p
2m
h
2 mE

KE
2 mE
KE
hc
2
2 mc E KE

1240 eV  nm
2 0.511  10 eV
6
10 eV 
 0.39 nm
Heisenberg Uncertainty
Principle
• Using
– x = position uncertainty
– p = momentum uncertainty
• Heisenberg showed that the product
( x )  ( p ) is always greater than ( h / 4 )
Often write this as
p ~ / 2
 x ‘h-bar’
is pronounced
where


h
2
Planck’s
constant
The wavefunction
• Particle has a wavefunction (x)
2
0.2

0.1 5
2(x)
0.1
0.0 5
0
x
-0.05
-4
-3
-2
-1
0
1
2
dx
Very small x-range

2
 x dx
x2
Larger x-range


x1
Entire x-range

2



3
4
= probability to find particle in
infinitesimal range dx about x
 x dx

x
2
= probability to find particle
between x1 and x2
 x dx
 1 particle must be somewhere
Question
2
0.5nm-1
0.2
0.1 5
0.1
What is probability that particle is found
in 0.01nm wide region about -0.2nm?
0.0 5
0
A.
0.001
B.
0.005
C.
0.01
D.
0.05
E.
0.1
-0.05
-4
About what is probability that particle is
in the region -1.0nm<x<0.0nm?
A.
0.1
B.
0.4
C.
0.5
D.
1.5
E.
3.0
-0.8nm
-3
-2
-0.2nm
-1
0
0
1
2
x
3
Particle in 1D box
n
n=3
p
h

h
E 
3
2
2L /3

n=1

h
2
8 mL
h

2L /2
2
2m

 n=2
p
2
2
h
2
2
8 mL
2
h
h
2L
8 mL
2
2
L
Wavefunction
Probability
Particle in box energy levels
• Quantized momentum
p
h

h

2L /n
 np o
• Energy = kinetic

E 
2m

np o 
2
2m
 n Eo
• Or Quantized Energy
2
Energy

p
2
n=5
n=4
En  n Eo
n=3
n=quantum number
n=2
n=1
2
3-D particle in box: summary
• Three quantum numbers (nx,ny,nz) label each state
– nx,y,z=1, 2, 3 … (integers starting at 1)
• Each state has different motion in x, y, z
• Quantum numbers determine
px 
– Momentum in each direction: e.g.
– Energy: E 
p
2
x
2m
2

py
2

pz
2 m 2 m
h
n
 nx
x
h
2L
 E o n x  n y  n z 
2
2
• Some quantum states have same energy

2
Question
How many 3-D particle in box spatial quantum
states have energy E=18Eo?
A. 1
B. 2
C. 3
D. 5
E. 6
E  E o n x  n y  n z 
2
n

x
2
2
, n y , n z   4,1,1, 1, 4,1, (1,1, 4 )
Q: ask what state is this?
(121)
(211)
All these states have the same
energy, but different probabilities
(112)
3D hydrogen atom
mS
Spin magnetic quantum
number
+1/2 or -1/2
2
For hydrogen atom:
•
•
•
•
n describes energy of orbit : E  13 .6 / n 2 eV
ℓ describes the magnitude of orbital angular momentum
m ℓ describes the angle of the orbital angular momentum
 of the spin angular moment
ms describes the angle
 n  2 


0


 m  0 

1 
m s   

2 

Other elements: Li has 3
electrons
 n  2 


0


 m  0 

1 
m s   

2 
 n  2 


1


 m  0 

1 
m s   

2 
 n  2 


1


 m  0 

1 
m s   

2 
 n  2 


1


 m  1 

1 
m s   

2 




 n  2 


1


 m  1 

1 
m s   

2 
 n  1 


 0 

 m  0 


m s   1 / 2 
 n  2 


1


 m   1 

1 
m s   

2 
n=2 states, 
8 total, 1 occupied
n=1 states,
2 total, 2 occupied
 n  1 


 0 

 m  0 


m s   1 / 2 
 n  2 


1


 m   1 

1 
m s   

2 
one spin up, one spin down
Question
Inert gas atoms are ones that have just enough electrons to finish filling
a p-shell (except for He). How many electrons do next two inert gas
atoms after helium ( neon (Ne) and argon (Ar) ) have.
In this range of atomic number the subshells fill in order of increasing
angular momentum.
A. 10 & 18
B. 4 & 8
C. 8 & 16
D. 12 & 20
E. 6 & 10
Multi-electron atoms
• Electrons interact with
nucleus (like hydrogen)
• Also with other electrons
• Causes energy to
depend on ℓ
States fill in order of energy:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d
Energy depends
on n and
ℓ
Energy
depends only
on n
The periodic table
• Atoms in same column
have ‘similar’ chemical properties.
• Quantum mechanical explanation:
similar ‘outer’ electron configurations.
H
1s1
Li
2s1
Na
3s1
K
4s1
Be
2s2
Mg
3s2
Ca
4s2
Sc
3d1
Y
3d2
8 more
transition
metals
B
2p1
Al
3p1
Ga
4p1
C
2p2
Si
3p2
Ge
4p2
N
2p3
P
3p3
As
4p3
O
2p4
S
3p4
Se
4p4
F
2p5
Cl
3p5
Br
4p5
He
1s2
Ne
2p6
Ar
3p6
Kr
4p6
Electron Configurations
Atom
Configuration
H
1s1
He
1s2
Li
1s22s1
Be
1s22s2
B
1s22s22p1
Ne
etc
1s shell filled
1s22s22p6
(n=1 shell filled noble gas)
2s shell filled
2p shell filled
(n=2 shell filled noble gas)
Ruby laser operation
Relaxation to
metastable state
(no photon
emission)
3
e
V
2
Metastable
state
e
V
PUMP
1
e
V
Transition by
stimulated emission of
photon
Ground state
Isotopes
# protons
• Carbon has
6 protons, 6 electrons (Z=6):
this is what makes it carbon.
Total # nucleons
12
C
6
• Most common form of carbon has 6 neutrons
in the nucleus. Called 12C
Another form of carbon has
6 protons, 8 neutrons in the nucleus. This is 14C.
This is a different ‘isotope’ of carbon
Isotopes: same # protons, different # neutrons
Nuclear matter


Any particle in nucleus, neutron or proton, is
called a nucleon.
“A” is atomic mass number


A=total number of nucleons in nucleus.
Experimental result

All nuclei have ~ same (incredibly high!) density of
2.3x1017kg/m3
Volume A = number of nucleons
Radius  A1/3

r  ro A


1/ 3
,
ro  1.2 fm  1.2  10
 15
m
Binding energy
• Calculate binding energy from masses
E binding  Zm
p
 Zm
e
 Nm N  m nucleus c
 Zm
e
E binding  Zm H  Nm N  m atom c

 Mass of
Hydrogen atom
(1.0078 u)

2
2
Mass of atom with
Z protons, N neutrons

Atomic masses well-known-> easier to use
Biological effects of radiation
• Radiation damage depends on
– Energy deposited / tissue mass (1 Gy (gray) = 1J/kg)
– Damaging effect of particle (RBE, relative biological effectiveness)
Radiation type
RBE
X-rays
Gamma rays
Beta particles
Alpha particles
1
1
1-2
10-20
• Dose equivalent = (Energy deposited / tissue mass) x RBE
– Units of Sv (sieverts) [older unit = rem, 1 rem=0.01 Sv]
– Common units mSv (10-3Sv), mrem (10-3rem)
– Common ‘safe’ limit = 500 mrem/yr
(5 mSv/yr)
Exposure from 60Co source
4
• 60Co source has an activity of 1 µCurie 3 .7  10 decays / s 
• Each decay: 1.3 MeV photon emitted
• Hold in your fist for one hour
– all particles absorbed by a 1 kg section
of your body for 1 hour

• Energy absorbed in 1 kg =
1 .3  10
6
eV
1 .6  10
19
J / eV
3 .7  10
4
decays / s 1hr 3600 s / hr   2 .8  10
5
J
What dose do you receive? A. 0.5 rem
2.8  10
5
J / kg 1rad /0.01 J / kg 
 0.003 rad = 0.003 rem
B. 0.3 rem
C. 0.1 rem
D. 0.05 rem
E. 0.003 rem
Quantifying radioactivity

Decay rate r



Prob( nucleus decays in time t ) = r t
(Units of becquerel (1 Bq=1 s-1) or
curie (1 Ci=3.7x1010 s-1)
Activity R

Mean # decays / s = rN,
Half-life t1/2

(Units of s-1)
N=# nuclei in sample
(Units of s)
time for half of nuclei to decay = t1/2 
N  N oe
rt
ln 2
r

0 .693
r
Activity of Radon
• 222Rn has a half-life of 3.83 days.
• Suppose your basement has 4.0 x 108
such nuclei in the air. What is the activity?
We are trying to find number of decays/sec.
So we have to know decay constant to get R=rN
r
0.693
t1 / 2
R
dN

0.693
3.83 days  86 , 400 s / day
 rN  2.09  10
6
 2.09  10
6
s
s  4.0  10 nuclei  836 decays / s
8
dt
R  836 decays / s 
1Ci
2.7  10 decays / s
10
 0.023  Ci
Decay summary
• Alpha decay
– Nucleus emits He nucleus (2 protons, 2 neutrons)
– Nucleus loses 2 protons, 2 neutrons
• Beta- decay
– Nucleus emits electron
– Neutron changes to proton in nucleus
• Beta+ decay
– Nucleus emits positron
– Proton changes to neutron in nucleus
• Gamma decay
– Nucleus emits photon
as it drops from excited state