today’s outline - november 30, 2015segre/phys405/15f/lecture_24.pdf · c. segre (iit) phys 405 -...
TRANSCRIPT
Today’s Outline - November 30, 2015
• Rolling dice example
• Ideal gas
• Comparison of statistical models
• Blackbody radiation
• Chapter 5 problem requests
Please fill out course evaluation!
Final Exam:Monday, December 7, 14:00 – 16:00Room 204 SB
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 1 / 16
Today’s Outline - November 30, 2015
• Rolling dice example
• Ideal gas
• Comparison of statistical models
• Blackbody radiation
• Chapter 5 problem requests
Please fill out course evaluation!
Final Exam:Monday, December 7, 14:00 – 16:00Room 204 SB
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 1 / 16
Today’s Outline - November 30, 2015
• Rolling dice example
• Ideal gas
• Comparison of statistical models
• Blackbody radiation
• Chapter 5 problem requests
Please fill out course evaluation!
Final Exam:Monday, December 7, 14:00 – 16:00Room 204 SB
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 1 / 16
Today’s Outline - November 30, 2015
• Rolling dice example
• Ideal gas
• Comparison of statistical models
• Blackbody radiation
• Chapter 5 problem requests
Please fill out course evaluation!
Final Exam:Monday, December 7, 14:00 – 16:00Room 204 SB
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 1 / 16
Today’s Outline - November 30, 2015
• Rolling dice example
• Ideal gas
• Comparison of statistical models
• Blackbody radiation
• Chapter 5 problem requests
Please fill out course evaluation!
Final Exam:Monday, December 7, 14:00 – 16:00Room 204 SB
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 1 / 16
Today’s Outline - November 30, 2015
• Rolling dice example
• Ideal gas
• Comparison of statistical models
• Blackbody radiation
• Chapter 5 problem requests
Please fill out course evaluation!
Final Exam:Monday, December 7, 14:00 – 16:00Room 204 SB
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 1 / 16
Today’s Outline - November 30, 2015
• Rolling dice example
• Ideal gas
• Comparison of statistical models
• Blackbody radiation
• Chapter 5 problem requests
Please fill out course evaluation!
Final Exam:Monday, December 7, 14:00 – 16:00Room 204 SB
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 1 / 16
Today’s Outline - November 30, 2015
• Rolling dice example
• Ideal gas
• Comparison of statistical models
• Blackbody radiation
• Chapter 5 problem requests
Please fill out course evaluation!
Final Exam:Monday, December 7, 14:00 – 16:00Room 204 SB
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 1 / 16
Example: Rolling dice
To show how the degeneraciesbecome weighted to the cen-ter of the distribution, take thecase of rolling multiple 6-sideddice
Take, for example, 2 dice. Thepossible sums range from 2 to12 with multiplicities:
The total number of possibili-ties is 36 = 62.
For 3 dice, it becomes 216 = 63
Sum Mult.2 13 24 35 46 57 68 59 4
10 311 212 1
Sum Mult.3 14 35 66 107 158 219 25
10 2711 2712 2513 2114 1515 1016 617 318 1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 2 / 16
Example: Rolling dice
To show how the degeneraciesbecome weighted to the cen-ter of the distribution, take thecase of rolling multiple 6-sideddice
Take, for example, 2 dice. Thepossible sums range from 2 to12 with multiplicities:
The total number of possibili-ties is 36 = 62.
For 3 dice, it becomes 216 = 63
Sum Mult.2 13 24 35 46 57 68 59 4
10 311 212 1
Sum Mult.3 14 35 66 107 158 219 25
10 2711 2712 2513 2114 1515 1016 617 318 1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 2 / 16
Example: Rolling dice
To show how the degeneraciesbecome weighted to the cen-ter of the distribution, take thecase of rolling multiple 6-sideddice
Take, for example, 2 dice. Thepossible sums range from 2 to12 with multiplicities:
The total number of possibili-ties is 36 = 62.
For 3 dice, it becomes 216 = 63
Sum Mult.2 13 24 35 46 57 68 59 4
10 311 212 1
Sum Mult.3 14 35 66 107 158 219 25
10 2711 2712 2513 2114 1515 1016 617 318 1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 2 / 16
Example: Rolling dice
To show how the degeneraciesbecome weighted to the cen-ter of the distribution, take thecase of rolling multiple 6-sideddice
Take, for example, 2 dice. Thepossible sums range from 2 to12 with multiplicities:
The total number of possibili-ties is 36 = 62.
For 3 dice, it becomes 216 = 63
Sum Mult.2 13 24 35 46 57 68 59 4
10 311 212 1
Sum Mult.3 14 35 66 107 158 219 25
10 2711 2712 2513 2114 1515 1016 617 318 1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 2 / 16
Example: Rolling dice
To show how the degeneraciesbecome weighted to the cen-ter of the distribution, take thecase of rolling multiple 6-sideddice
Take, for example, 2 dice. Thepossible sums range from 2 to12 with multiplicities:
The total number of possibili-ties is 36 = 62.
For 3 dice, it becomes 216 = 63
Sum Mult.2 13 24 35 46 57 68 59 4
10 311 212 1
Sum Mult.3 14 35 66 107 158 219 25
10 2711 2712 2513 2114 1515 1016 617 318 1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 2 / 16
Example: Rolling dice
To show how the degeneraciesbecome weighted to the cen-ter of the distribution, take thecase of rolling multiple 6-sideddice
Take, for example, 2 dice. Thepossible sums range from 2 to12 with multiplicities:
The total number of possibili-ties is 36 = 62.
For 3 dice, it becomes 216 = 63
Sum Mult.2 13 24 35 46 57 68 59 4
10 311 212 1
Sum Mult.3 14 35 66 107 158 219 25
10 2711 2712 2513 2114 1515 1016 617 318 1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 2 / 16
Example: Rolling dice
and for 4 dice, 1296 = 64
Sum Mult.4 15 46 107 208 359 56
10 8011 10412 12513 140
Sum Mult.14 14615 14016 12517 10418 8019 5620 3521 2022 1023 424 1
0
2
4
6
8
10
12
14
16
18
0 20 40 60 80 100
Pro
ba
bili
ty
Normalized Count
2 dice
3 dice
4 dice
as N →∞ the distributionapproaches a Gaussisn
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 3 / 16
Example: Rolling dice
and for 4 dice, 1296 = 64
Sum Mult.4 15 46 107 208 359 56
10 8011 10412 12513 140
Sum Mult.14 14615 14016 12517 10418 8019 5620 3521 2022 1023 424 1
0
2
4
6
8
10
12
14
16
18
0 20 40 60 80 100
Pro
ba
bili
ty
Normalized Count
2 dice
3 dice
4 dice
as N →∞ the distributionapproaches a Gaussisn
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 3 / 16
Example: Rolling dice
and for 4 dice, 1296 = 64
Sum Mult.4 15 46 107 208 359 56
10 8011 10412 12513 140
Sum Mult.14 14615 14016 12517 10418 8019 5620 3521 2022 1023 424 1
0
2
4
6
8
10
12
14
16
18
0 20 40 60 80 100
Pro
ba
bili
ty
Normalized Count
2 dice
3 dice
4 dice
as N →∞ the distributionapproaches a Gaussisn
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 3 / 16
Example: Rolling dice
and for 4 dice, 1296 = 64
Sum Mult.4 15 46 107 208 359 56
10 8011 10412 12513 140
Sum Mult.14 14615 14016 12517 10418 8019 5620 3521 2022 1023 424 1
0
2
4
6
8
10
12
14
16
18
0 20 40 60 80 100
Pro
ba
bili
ty
Normalized Count
2 dice
3 dice
4 dice
as N →∞ the distributionapproaches a Gaussisn
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 3 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square well
assuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)dk =
1
8
4πk2dk
(π3/V )
=V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk
= Ve−α(
m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)dk =
1
8
4πk2dk
(π3/V )
=V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk
= Ve−α(
m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)dk =
1
8
4πk2dk
(π3/V )
=V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk
= Ve−α(
m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)
dk =1
8
4πk2dk
(π3/V )
=V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk
= Ve−α(
m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)
dk =1
8
4πk2dk
(π3/V )
=V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk
= Ve−α(
m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)dk =
1
8
4πk2dk
(π3/V )
=V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk
= Ve−α(
m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)dk =
1
8
4πk2dk
(π3/V )=
V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk
= Ve−α(
m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)dk =
1
8
4πk2dk
(π3/V )=
V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk
= Ve−α(
m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)dk =
1
8
4πk2dk
(π3/V )=
V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk
= Ve−α(
m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)dk =
1
8
4πk2dk
(π3/V )=
V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk
= Ve−α(
m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)dk =
1
8
4πk2dk
(π3/V )=
V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk
= Ve−α(
m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)dk =
1
8
4πk2dk
(π3/V )=
V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk = Ve−α
(m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Ideal gas example
Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is
as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek
for distinguishable particles
Ek =~2
2mk2
~k =
(πnxlx,πnyly,πnzlz
)dk =
1
8
4πk2dk
(π3/V )=
V
2π2k2dk
Nn= dne−(α+βEn)
and the constraint that N =∑
Nn becomes
N =V
2π2e−α
∫ ∞0
k2e−β~2k2/2mdk = Ve−α
(m
2πβ~2
)3/2
e−α =N
V
(2πβ~2
m
)3/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 4 / 16
Significance of α and β
The constraint on the total energy, E =∑
NnEn becomes, in this context
E =V
2π2e−α
~2
2m
∫ ∞0
k4e−β~2k2/2mdk
=3V
2βe−α
(m
2πβ~2
)3/2
substituting for e−α obtained pre-viously
this is similar to the classical ther-modynamic expression for energyper molecule
we postulate that β is related to thetemperature
while α is related to the chemicalpotential
e−α =N
V
(2πβ~2
m
)3/2
E =3N
2βE
N=
3
2kBT
β =1
kBT
µ(T ) ≡ −αkBT
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 5 / 16
Significance of α and β
The constraint on the total energy, E =∑
NnEn becomes, in this context
E =V
2π2e−α
~2
2m
∫ ∞0
k4e−β~2k2/2mdk
=3V
2βe−α
(m
2πβ~2
)3/2
substituting for e−α obtained pre-viously
this is similar to the classical ther-modynamic expression for energyper molecule
we postulate that β is related to thetemperature
while α is related to the chemicalpotential
e−α =N
V
(2πβ~2
m
)3/2
E =3N
2βE
N=
3
2kBT
β =1
kBT
µ(T ) ≡ −αkBT
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 5 / 16
Significance of α and β
The constraint on the total energy, E =∑
NnEn becomes, in this context
E =V
2π2e−α
~2
2m
∫ ∞0
k4e−β~2k2/2mdk =
3V
2βe−α
(m
2πβ~2
)3/2
substituting for e−α obtained pre-viously
this is similar to the classical ther-modynamic expression for energyper molecule
we postulate that β is related to thetemperature
while α is related to the chemicalpotential
e−α =N
V
(2πβ~2
m
)3/2
E =3N
2βE
N=
3
2kBT
β =1
kBT
µ(T ) ≡ −αkBT
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 5 / 16
Significance of α and β
The constraint on the total energy, E =∑
NnEn becomes, in this context
E =V
2π2e−α
~2
2m
∫ ∞0
k4e−β~2k2/2mdk =
3V
2βe−α
(m
2πβ~2
)3/2
substituting for e−α obtained pre-viously
this is similar to the classical ther-modynamic expression for energyper molecule
we postulate that β is related to thetemperature
while α is related to the chemicalpotential
e−α =N
V
(2πβ~2
m
)3/2
E =3N
2βE
N=
3
2kBT
β =1
kBT
µ(T ) ≡ −αkBT
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 5 / 16
Significance of α and β
The constraint on the total energy, E =∑
NnEn becomes, in this context
E =V
2π2e−α
~2
2m
∫ ∞0
k4e−β~2k2/2mdk =
3V
2βe−α
(m
2πβ~2
)3/2
substituting for e−α obtained pre-viously
this is similar to the classical ther-modynamic expression for energyper molecule
we postulate that β is related to thetemperature
while α is related to the chemicalpotential
e−α =N
V
(2πβ~2
m
)3/2
E =3N
2β
E
N=
3
2kBT
β =1
kBT
µ(T ) ≡ −αkBT
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 5 / 16
Significance of α and β
The constraint on the total energy, E =∑
NnEn becomes, in this context
E =V
2π2e−α
~2
2m
∫ ∞0
k4e−β~2k2/2mdk =
3V
2βe−α
(m
2πβ~2
)3/2
substituting for e−α obtained pre-viously
this is similar to the classical ther-modynamic expression for energyper molecule
we postulate that β is related to thetemperature
while α is related to the chemicalpotential
e−α =N
V
(2πβ~2
m
)3/2
E =3N
2β
E
N=
3
2kBT
β =1
kBT
µ(T ) ≡ −αkBT
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 5 / 16
Significance of α and β
The constraint on the total energy, E =∑
NnEn becomes, in this context
E =V
2π2e−α
~2
2m
∫ ∞0
k4e−β~2k2/2mdk =
3V
2βe−α
(m
2πβ~2
)3/2
substituting for e−α obtained pre-viously
this is similar to the classical ther-modynamic expression for energyper molecule
we postulate that β is related to thetemperature
while α is related to the chemicalpotential
e−α =N
V
(2πβ~2
m
)3/2
E =3N
2βE
N=
3
2kBT
β =1
kBT
µ(T ) ≡ −αkBT
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 5 / 16
Significance of α and β
The constraint on the total energy, E =∑
NnEn becomes, in this context
E =V
2π2e−α
~2
2m
∫ ∞0
k4e−β~2k2/2mdk =
3V
2βe−α
(m
2πβ~2
)3/2
substituting for e−α obtained pre-viously
this is similar to the classical ther-modynamic expression for energyper molecule
we postulate that β is related to thetemperature
while α is related to the chemicalpotential
e−α =N
V
(2πβ~2
m
)3/2
E =3N
2βE
N=
3
2kBT
β =1
kBT
µ(T ) ≡ −αkBT
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 5 / 16
Significance of α and β
The constraint on the total energy, E =∑
NnEn becomes, in this context
E =V
2π2e−α
~2
2m
∫ ∞0
k4e−β~2k2/2mdk =
3V
2βe−α
(m
2πβ~2
)3/2
substituting for e−α obtained pre-viously
this is similar to the classical ther-modynamic expression for energyper molecule
we postulate that β is related to thetemperature
while α is related to the chemicalpotential
e−α =N
V
(2πβ~2
m
)3/2
E =3N
2βE
N=
3
2kBT
β =1
kBT
µ(T ) ≡ −αkBT
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 5 / 16
Significance of α and β
The constraint on the total energy, E =∑
NnEn becomes, in this context
E =V
2π2e−α
~2
2m
∫ ∞0
k4e−β~2k2/2mdk =
3V
2βe−α
(m
2πβ~2
)3/2
substituting for e−α obtained pre-viously
this is similar to the classical ther-modynamic expression for energyper molecule
we postulate that β is related to thetemperature
while α is related to the chemicalpotential
e−α =N
V
(2πβ~2
m
)3/2
E =3N
2βE
N=
3
2kBT
β =1
kBT
µ(T ) ≡ −αkBT
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 5 / 16
Significance of α and β
The constraint on the total energy, E =∑
NnEn becomes, in this context
E =V
2π2e−α
~2
2m
∫ ∞0
k4e−β~2k2/2mdk =
3V
2βe−α
(m
2πβ~2
)3/2
substituting for e−α obtained pre-viously
this is similar to the classical ther-modynamic expression for energyper molecule
we postulate that β is related to thetemperature
while α is related to the chemicalpotential
e−α =N
V
(2πβ~2
m
)3/2
E =3N
2βE
N=
3
2kBT
β =1
kBT
µ(T ) ≡ −αkBT
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 5 / 16
Particle distributions
It is conventional to divide the expressions for number of particles in astate by the degeneracy of the state to get an occupation fraction for eachstate and call the energy per state ε = En
Nn
dn= n(ε) =
[e(ε−µ)/kBT ]−1, Maxwell− Boltzmann
[e(ε−µ)/kBT + 1]−1, Fermi− Dirac
[e(ε−µ)/kBT − 1]−1, Bose− Einstein
if we apply the same constraints as before to identical fermions or bosonsin the 3-D infinite well, we obtain
N =V
2π2
∫ ∞0
k2
e [(~2k2/2m)−µ]/kBT ± 1dk
E =V
2π2~2
2m
∫ ∞0
k4
e [(~2k2/2m)−µ]/kBT ± 1dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 6 / 16
Particle distributions
It is conventional to divide the expressions for number of particles in astate by the degeneracy of the state to get an occupation fraction for eachstate and call the energy per state ε = En
Nn
dn= n(ε) =
[e(ε−µ)/kBT ]−1, Maxwell− Boltzmann
[e(ε−µ)/kBT + 1]−1, Fermi− Dirac
[e(ε−µ)/kBT − 1]−1, Bose− Einstein
if we apply the same constraints as before to identical fermions or bosonsin the 3-D infinite well, we obtain
N =V
2π2
∫ ∞0
k2
e [(~2k2/2m)−µ]/kBT ± 1dk
E =V
2π2~2
2m
∫ ∞0
k4
e [(~2k2/2m)−µ]/kBT ± 1dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 6 / 16
Particle distributions
It is conventional to divide the expressions for number of particles in astate by the degeneracy of the state to get an occupation fraction for eachstate and call the energy per state ε = En
Nn
dn= n(ε) =
[e(ε−µ)/kBT ]−1, Maxwell− Boltzmann
[e(ε−µ)/kBT + 1]−1, Fermi− Dirac
[e(ε−µ)/kBT − 1]−1, Bose− Einstein
if we apply the same constraints as before to identical fermions or bosonsin the 3-D infinite well, we obtain
N =V
2π2
∫ ∞0
k2
e [(~2k2/2m)−µ]/kBT ± 1dk
E =V
2π2~2
2m
∫ ∞0
k4
e [(~2k2/2m)−µ]/kBT ± 1dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 6 / 16
Particle distributions
It is conventional to divide the expressions for number of particles in astate by the degeneracy of the state to get an occupation fraction for eachstate and call the energy per state ε = En
Nn
dn= n(ε) =
[e(ε−µ)/kBT ]−1, Maxwell− Boltzmann
[e(ε−µ)/kBT + 1]−1, Fermi− Dirac
[e(ε−µ)/kBT − 1]−1, Bose− Einstein
if we apply the same constraints as before to identical fermions or bosonsin the 3-D infinite well, we obtain
N =V
2π2
∫ ∞0
k2
e [(~2k2/2m)−µ]/kBT ± 1dk
E =V
2π2~2
2m
∫ ∞0
k4
e [(~2k2/2m)−µ]/kBT ± 1dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 6 / 16
Particle distributions
It is conventional to divide the expressions for number of particles in astate by the degeneracy of the state to get an occupation fraction for eachstate and call the energy per state ε = En
Nn
dn= n(ε) =
[e(ε−µ)/kBT ]−1, Maxwell− Boltzmann
[e(ε−µ)/kBT + 1]−1, Fermi− Dirac
[e(ε−µ)/kBT − 1]−1, Bose− Einstein
if we apply the same constraints as before to identical fermions or bosonsin the 3-D infinite well, we obtain
N =V
2π2
∫ ∞0
k2
e [(~2k2/2m)−µ]/kBT ± 1dk
E =V
2π2~2
2m
∫ ∞0
k4
e [(~2k2/2m)−µ]/kBT ± 1dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 6 / 16
Particle distributions
It is conventional to divide the expressions for number of particles in astate by the degeneracy of the state to get an occupation fraction for eachstate and call the energy per state ε = En
Nn
dn= n(ε) =
[e(ε−µ)/kBT ]−1, Maxwell− Boltzmann
[e(ε−µ)/kBT + 1]−1, Fermi− Dirac
[e(ε−µ)/kBT − 1]−1, Bose− Einstein
if we apply the same constraints as before to identical fermions or bosonsin the 3-D infinite well, we obtain
N =V
2π2
∫ ∞0
k2
e [(~2k2/2m)−µ]/kBT ± 1dk
E =V
2π2~2
2m
∫ ∞0
k4
e [(~2k2/2m)−µ]/kBT ± 1dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 6 / 16
Particle distributions
It is conventional to divide the expressions for number of particles in astate by the degeneracy of the state to get an occupation fraction for eachstate and call the energy per state ε = En
Nn
dn= n(ε) =
[e(ε−µ)/kBT ]−1, Maxwell− Boltzmann
[e(ε−µ)/kBT + 1]−1, Fermi− Dirac
[e(ε−µ)/kBT − 1]−1, Bose− Einstein
if we apply the same constraints as before to identical fermions or bosonsin the 3-D infinite well, we obtain
N =V
2π2
∫ ∞0
k2
e [(~2k2/2m)−µ]/kBT ± 1dk
E =V
2π2~2
2m
∫ ∞0
k4
e [(~2k2/2m)−µ]/kBT ± 1dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 6 / 16
Particle distributions
It is conventional to divide the expressions for number of particles in astate by the degeneracy of the state to get an occupation fraction for eachstate and call the energy per state ε = En
Nn
dn= n(ε) =
[e(ε−µ)/kBT ]−1, Maxwell− Boltzmann
[e(ε−µ)/kBT + 1]−1, Fermi− Dirac
[e(ε−µ)/kBT − 1]−1, Bose− Einstein
if we apply the same constraints as before to identical fermions or bosonsin the 3-D infinite well, we obtain
N =V
2π2
∫ ∞0
k2
e [(~2k2/2m)−µ]/kBT ± 1dk
E =V
2π2~2
2m
∫ ∞0
k4
e [(~2k2/2m)−µ]/kBT ± 1dk
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 6 / 16
Comparison of statistics
The three statistics lead to very different behavior in occupation number
0
2
4
6
8
10
0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3
1 K50 K300 K
n(E
)
ε/µ
Maxwell-Boltzmann
0
2
4
6
8
10
0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3
1 K
50 K
300 K
n(E
)
ε/µ
Bose-Einstein
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2 1.4
1 K50 K
300 K
n(E
)
ε/µ
Fermi-Dirac
Maxwell-Boltzmann: n(ε)→∞ as ε < µ and exponential behavior withtemperature
Bose-Einstein: occupation number is always infinite for ε < µ
Fermi-Dirac: n(ε) ≡ 1 as T → 0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 7 / 16
Comparison of statistics
The three statistics lead to very different behavior in occupation number
0
2
4
6
8
10
0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3
1 K50 K300 K
n(E
)
ε/µ
Maxwell-Boltzmann
0
2
4
6
8
10
0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3
1 K
50 K
300 K
n(E
)
ε/µ
Bose-Einstein
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2 1.4
1 K50 K
300 K
n(E
)
ε/µ
Fermi-Dirac
Maxwell-Boltzmann: n(ε)→∞ as ε < µ and exponential behavior withtemperature
Bose-Einstein: occupation number is always infinite for ε < µ
Fermi-Dirac: n(ε) ≡ 1 as T → 0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 7 / 16
Comparison of statistics
The three statistics lead to very different behavior in occupation number
0
2
4
6
8
10
0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3
1 K50 K300 K
n(E
)
ε/µ
Maxwell-Boltzmann
0
2
4
6
8
10
0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3
1 K
50 K
300 K
n(E
)
ε/µ
Bose-Einstein
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2 1.4
1 K50 K
300 K
n(E
)
ε/µ
Fermi-Dirac
Maxwell-Boltzmann: n(ε)→∞ as ε < µ and exponential behavior withtemperature
Bose-Einstein: occupation number is always infinite for ε < µ
Fermi-Dirac: n(ε) ≡ 1 as T → 0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 7 / 16
Comparison of statistics
The three statistics lead to very different behavior in occupation number
0
2
4
6
8
10
0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3
1 K50 K300 K
n(E
)
ε/µ
Maxwell-Boltzmann
0
2
4
6
8
10
0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3
1 K
50 K
300 K
n(E
)
ε/µ
Bose-Einstein
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2 1.4
1 K50 K
300 K
n(E
)
ε/µ
Fermi-Dirac
Maxwell-Boltzmann: n(ε)→∞ as ε < µ and exponential behavior withtemperature
Bose-Einstein: occupation number is always infinite for ε < µ
Fermi-Dirac: n(ε) ≡ 1 as T → 0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 7 / 16
Blackbody spectrum
Photons are bosons and so now we can use our understanding of quantumstatistics to compute the distribution of phonons in a rectangular box withthe following assumptions
1 The energy of a photon is ε = hν = ~ω2 The wave number is k = 2π/λ = ω/c
3 Only two spin states occur, s = 1 but m = ±1
4 The number of photons is NOT conserved
since the number of photons is notconserved, we can set µ → 0 andwe have
the degeneracy in terms of ω is
the energy density, Nω~ω/V in theinterval dω is
Nn=dn
e(ε−µ)/kBT − 1
Nω =dk
e~ω/kBT − 1
dk =V
π2c3ω2dω
ρ(ω)dω =~ω3dω
π2c3(e~ω/kBT − 1)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 8 / 16
Blackbody spectrum
Photons are bosons and so now we can use our understanding of quantumstatistics to compute the distribution of phonons in a rectangular box withthe following assumptions
1 The energy of a photon is ε = hν = ~ω
2 The wave number is k = 2π/λ = ω/c
3 Only two spin states occur, s = 1 but m = ±1
4 The number of photons is NOT conserved
since the number of photons is notconserved, we can set µ → 0 andwe have
the degeneracy in terms of ω is
the energy density, Nω~ω/V in theinterval dω is
Nn=dn
e(ε−µ)/kBT − 1
Nω =dk
e~ω/kBT − 1
dk =V
π2c3ω2dω
ρ(ω)dω =~ω3dω
π2c3(e~ω/kBT − 1)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 8 / 16
Blackbody spectrum
Photons are bosons and so now we can use our understanding of quantumstatistics to compute the distribution of phonons in a rectangular box withthe following assumptions
1 The energy of a photon is ε = hν = ~ω2 The wave number is k = 2π/λ = ω/c
3 Only two spin states occur, s = 1 but m = ±1
4 The number of photons is NOT conserved
since the number of photons is notconserved, we can set µ → 0 andwe have
the degeneracy in terms of ω is
the energy density, Nω~ω/V in theinterval dω is
Nn=dn
e(ε−µ)/kBT − 1
Nω =dk
e~ω/kBT − 1
dk =V
π2c3ω2dω
ρ(ω)dω =~ω3dω
π2c3(e~ω/kBT − 1)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 8 / 16
Blackbody spectrum
Photons are bosons and so now we can use our understanding of quantumstatistics to compute the distribution of phonons in a rectangular box withthe following assumptions
1 The energy of a photon is ε = hν = ~ω2 The wave number is k = 2π/λ = ω/c
3 Only two spin states occur, s = 1 but m = ±1
4 The number of photons is NOT conserved
since the number of photons is notconserved, we can set µ → 0 andwe have
the degeneracy in terms of ω is
the energy density, Nω~ω/V in theinterval dω is
Nn=dn
e(ε−µ)/kBT − 1
Nω =dk
e~ω/kBT − 1
dk =V
π2c3ω2dω
ρ(ω)dω =~ω3dω
π2c3(e~ω/kBT − 1)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 8 / 16
Blackbody spectrum
Photons are bosons and so now we can use our understanding of quantumstatistics to compute the distribution of phonons in a rectangular box withthe following assumptions
1 The energy of a photon is ε = hν = ~ω2 The wave number is k = 2π/λ = ω/c
3 Only two spin states occur, s = 1 but m = ±1
4 The number of photons is NOT conserved
since the number of photons is notconserved, we can set µ → 0 andwe have
the degeneracy in terms of ω is
the energy density, Nω~ω/V in theinterval dω is
Nn=dn
e(ε−µ)/kBT − 1
Nω =dk
e~ω/kBT − 1
dk =V
π2c3ω2dω
ρ(ω)dω =~ω3dω
π2c3(e~ω/kBT − 1)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 8 / 16
Blackbody spectrum
Photons are bosons and so now we can use our understanding of quantumstatistics to compute the distribution of phonons in a rectangular box withthe following assumptions
1 The energy of a photon is ε = hν = ~ω2 The wave number is k = 2π/λ = ω/c
3 Only two spin states occur, s = 1 but m = ±1
4 The number of photons is NOT conserved
since the number of photons is notconserved, we can set µ → 0 andwe have
the degeneracy in terms of ω is
the energy density, Nω~ω/V in theinterval dω is
Nn=dn
e(ε−µ)/kBT − 1
Nω =dk
e~ω/kBT − 1
dk =V
π2c3ω2dω
ρ(ω)dω =~ω3dω
π2c3(e~ω/kBT − 1)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 8 / 16
Blackbody spectrum
Photons are bosons and so now we can use our understanding of quantumstatistics to compute the distribution of phonons in a rectangular box withthe following assumptions
1 The energy of a photon is ε = hν = ~ω2 The wave number is k = 2π/λ = ω/c
3 Only two spin states occur, s = 1 but m = ±1
4 The number of photons is NOT conserved
since the number of photons is notconserved, we can set µ → 0 andwe have
the degeneracy in terms of ω is
the energy density, Nω~ω/V in theinterval dω is
Nn=dn
e(ε−µ)/kBT − 1
Nω =dk
e~ω/kBT − 1
dk =V
π2c3ω2dω
ρ(ω)dω =~ω3dω
π2c3(e~ω/kBT − 1)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 8 / 16
Blackbody spectrum
Photons are bosons and so now we can use our understanding of quantumstatistics to compute the distribution of phonons in a rectangular box withthe following assumptions
1 The energy of a photon is ε = hν = ~ω2 The wave number is k = 2π/λ = ω/c
3 Only two spin states occur, s = 1 but m = ±1
4 The number of photons is NOT conserved
since the number of photons is notconserved, we can set µ → 0 andwe have
the degeneracy in terms of ω is
the energy density, Nω~ω/V in theinterval dω is
Nn=dn
e(ε−µ)/kBT − 1
Nω =dk
e~ω/kBT − 1
dk =V
π2c3ω2dω
ρ(ω)dω =~ω3dω
π2c3(e~ω/kBT − 1)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 8 / 16
Blackbody spectrum
Photons are bosons and so now we can use our understanding of quantumstatistics to compute the distribution of phonons in a rectangular box withthe following assumptions
1 The energy of a photon is ε = hν = ~ω2 The wave number is k = 2π/λ = ω/c
3 Only two spin states occur, s = 1 but m = ±1
4 The number of photons is NOT conserved
since the number of photons is notconserved, we can set µ → 0 andwe have
the degeneracy in terms of ω is
the energy density, Nω~ω/V in theinterval dω is
Nn=dn
e(ε−µ)/kBT − 1
Nω =dk
e~ω/kBT − 1
dk =V
π2c3ω2dω
ρ(ω)dω =~ω3dω
π2c3(e~ω/kBT − 1)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 8 / 16
Blackbody spectrum
Photons are bosons and so now we can use our understanding of quantumstatistics to compute the distribution of phonons in a rectangular box withthe following assumptions
1 The energy of a photon is ε = hν = ~ω2 The wave number is k = 2π/λ = ω/c
3 Only two spin states occur, s = 1 but m = ±1
4 The number of photons is NOT conserved
since the number of photons is notconserved, we can set µ → 0 andwe have
the degeneracy in terms of ω is
the energy density, Nω~ω/V in theinterval dω is
Nn=dn
e(ε−µ)/kBT − 1
Nω =dk
e~ω/kBT − 1
dk =V
π2c3ω2dω
ρ(ω)dω =~ω3dω
π2c3(e~ω/kBT − 1)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 8 / 16
Blackbody spectrum
Photons are bosons and so now we can use our understanding of quantumstatistics to compute the distribution of phonons in a rectangular box withthe following assumptions
1 The energy of a photon is ε = hν = ~ω2 The wave number is k = 2π/λ = ω/c
3 Only two spin states occur, s = 1 but m = ±1
4 The number of photons is NOT conserved
since the number of photons is notconserved, we can set µ → 0 andwe have
the degeneracy in terms of ω is
the energy density, Nω~ω/V in theinterval dω is
Nn=dn
e(ε−µ)/kBT − 1
Nω =dk
e~ω/kBT − 1
dk =V
π2c3ω2dω
ρ(ω)dω =~ω3dω
π2c3(e~ω/kBT − 1)C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 8 / 16
Blackbody spectrum
0
2
4
6
8
10
12
14
0 0.5 1 1.5 2 2.5 3
Inte
nsity (
arb
.)
Wavelength (µm)
5000 K
4000 K
3000 K
ρ(ω) =~ω3
π2c3(e~ω/kBT − 1)
Universal curves with reducedtemperature
As ω → 0 (λ → ∞) dropsoff as λ−3 avoiding ultravioletcatastrophe
As ω → ∞ (λ → 0) dropsto zero exponentially, movingpeak position up in λ
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 9 / 16
Blackbody spectrum
0
2
4
6
8
10
12
14
0 0.5 1 1.5 2 2.5 3
Inte
nsity (
arb
.)
Wavelength (µm)
5000 K
4000 K
3000 K
ρ(ω) =~ω3
π2c3(e~ω/kBT − 1)
Universal curves with reducedtemperature
As ω → 0 (λ → ∞) dropsoff as λ−3 avoiding ultravioletcatastrophe
As ω → ∞ (λ → 0) dropsto zero exponentially, movingpeak position up in λ
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 9 / 16
Blackbody spectrum
0
2
4
6
8
10
12
14
0 0.5 1 1.5 2 2.5 3
Inte
nsity (
arb
.)
Wavelength (µm)
5000 K
4000 K
3000 K
ρ(ω) =~ω3
π2c3(e~ω/kBT − 1)
Universal curves with reducedtemperature
As ω → 0 (λ → ∞) dropsoff as λ−3 avoiding ultravioletcatastrophe
As ω → ∞ (λ → 0) dropsto zero exponentially, movingpeak position up in λ
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 9 / 16
Blackbody spectrum
0
2
4
6
8
10
12
14
0 0.5 1 1.5 2 2.5 3
Inte
nsity (
arb
.)
Wavelength (µm)
5000 K
4000 K
3000 K
ρ(ω) =~ω3
π2c3(e~ω/kBT − 1)
Universal curves with reducedtemperature
As ω → 0 (λ → ∞) dropsoff as λ−3 avoiding ultravioletcatastrophe
As ω → ∞ (λ → 0) dropsto zero exponentially, movingpeak position up in λ
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 9 / 16
Problem 5.9
(a) Suppose you put both electrons in a helium atom into the n = 2 statefrom which state the atom is ionized to He+. What would the energyof the emitted electron be?
(b) Describe, quantitatively, the spectrum of the helium ion, He+
(a) The energy of each electron in thehelium atom is given by
initially the total energy is
after the ionization the energy of theion is
the energy of the emitted electron is
(b) The He+ is a hydrogenic atom withZ = 2 so the spectrum is the sameas for the Hydrogen atom except for aprefactor of 4
E =Z 2E1
n2, Z = 2
Etot = 2 · 4E1
4= −27.2 eV
Eion =4E1
12= −54.4 eV
Ee = Etot − Eion = 27.2 eV
1
λ= 4R
(1
n2f− 1
n2i
)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 10 / 16
Problem 5.9
(a) Suppose you put both electrons in a helium atom into the n = 2 statefrom which state the atom is ionized to He+. What would the energyof the emitted electron be?
(b) Describe, quantitatively, the spectrum of the helium ion, He+
(a) The energy of each electron in thehelium atom is given by
initially the total energy is
after the ionization the energy of theion is
the energy of the emitted electron is
(b) The He+ is a hydrogenic atom withZ = 2 so the spectrum is the sameas for the Hydrogen atom except for aprefactor of 4
E =Z 2E1
n2, Z = 2
Etot = 2 · 4E1
4= −27.2 eV
Eion =4E1
12= −54.4 eV
Ee = Etot − Eion = 27.2 eV
1
λ= 4R
(1
n2f− 1
n2i
)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 10 / 16
Problem 5.9
(a) Suppose you put both electrons in a helium atom into the n = 2 statefrom which state the atom is ionized to He+. What would the energyof the emitted electron be?
(b) Describe, quantitatively, the spectrum of the helium ion, He+
(a) The energy of each electron in thehelium atom is given by
initially the total energy is
after the ionization the energy of theion is
the energy of the emitted electron is
(b) The He+ is a hydrogenic atom withZ = 2 so the spectrum is the sameas for the Hydrogen atom except for aprefactor of 4
E =Z 2E1
n2, Z = 2
Etot = 2 · 4E1
4= −27.2 eV
Eion =4E1
12= −54.4 eV
Ee = Etot − Eion = 27.2 eV
1
λ= 4R
(1
n2f− 1
n2i
)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 10 / 16
Problem 5.9
(a) Suppose you put both electrons in a helium atom into the n = 2 statefrom which state the atom is ionized to He+. What would the energyof the emitted electron be?
(b) Describe, quantitatively, the spectrum of the helium ion, He+
(a) The energy of each electron in thehelium atom is given by
initially the total energy is
after the ionization the energy of theion is
the energy of the emitted electron is
(b) The He+ is a hydrogenic atom withZ = 2 so the spectrum is the sameas for the Hydrogen atom except for aprefactor of 4
E =Z 2E1
n2, Z = 2
Etot = 2 · 4E1
4= −27.2 eV
Eion =4E1
12= −54.4 eV
Ee = Etot − Eion = 27.2 eV
1
λ= 4R
(1
n2f− 1
n2i
)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 10 / 16
Problem 5.9
(a) Suppose you put both electrons in a helium atom into the n = 2 statefrom which state the atom is ionized to He+. What would the energyof the emitted electron be?
(b) Describe, quantitatively, the spectrum of the helium ion, He+
(a) The energy of each electron in thehelium atom is given by
initially the total energy is
after the ionization the energy of theion is
the energy of the emitted electron is
(b) The He+ is a hydrogenic atom withZ = 2 so the spectrum is the sameas for the Hydrogen atom except for aprefactor of 4
E =Z 2E1
n2, Z = 2
Etot = 2 · 4E1
4= −27.2 eV
Eion =4E1
12= −54.4 eV
Ee = Etot − Eion = 27.2 eV
1
λ= 4R
(1
n2f− 1
n2i
)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 10 / 16
Problem 5.9
(a) Suppose you put both electrons in a helium atom into the n = 2 statefrom which state the atom is ionized to He+. What would the energyof the emitted electron be?
(b) Describe, quantitatively, the spectrum of the helium ion, He+
(a) The energy of each electron in thehelium atom is given by
initially the total energy is
after the ionization the energy of theion is
the energy of the emitted electron is
(b) The He+ is a hydrogenic atom withZ = 2 so the spectrum is the sameas for the Hydrogen atom except for aprefactor of 4
E =Z 2E1
n2, Z = 2
Etot = 2 · 4E1
4= −27.2 eV
Eion =4E1
12= −54.4 eV
Ee = Etot − Eion = 27.2 eV
1
λ= 4R
(1
n2f− 1
n2i
)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 10 / 16
Problem 5.9
(a) Suppose you put both electrons in a helium atom into the n = 2 statefrom which state the atom is ionized to He+. What would the energyof the emitted electron be?
(b) Describe, quantitatively, the spectrum of the helium ion, He+
(a) The energy of each electron in thehelium atom is given by
initially the total energy is
after the ionization the energy of theion is
the energy of the emitted electron is
(b) The He+ is a hydrogenic atom withZ = 2 so the spectrum is the sameas for the Hydrogen atom except for aprefactor of 4
E =Z 2E1
n2, Z = 2
Etot = 2 · 4E1
4= −27.2 eV
Eion =4E1
12= −54.4 eV
Ee = Etot − Eion = 27.2 eV
1
λ= 4R
(1
n2f− 1
n2i
)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 10 / 16
Problem 5.9
(a) Suppose you put both electrons in a helium atom into the n = 2 statefrom which state the atom is ionized to He+. What would the energyof the emitted electron be?
(b) Describe, quantitatively, the spectrum of the helium ion, He+
(a) The energy of each electron in thehelium atom is given by
initially the total energy is
after the ionization the energy of theion is
the energy of the emitted electron is
(b) The He+ is a hydrogenic atom withZ = 2 so the spectrum is the sameas for the Hydrogen atom except for aprefactor of 4
E =Z 2E1
n2, Z = 2
Etot = 2 · 4E1
4= −27.2 eV
Eion =4E1
12= −54.4 eV
Ee = Etot − Eion = 27.2 eV
1
λ= 4R
(1
n2f− 1
n2i
)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 10 / 16
Problem 5.9
(a) Suppose you put both electrons in a helium atom into the n = 2 statefrom which state the atom is ionized to He+. What would the energyof the emitted electron be?
(b) Describe, quantitatively, the spectrum of the helium ion, He+
(a) The energy of each electron in thehelium atom is given by
initially the total energy is
after the ionization the energy of theion is
the energy of the emitted electron is
(b) The He+ is a hydrogenic atom withZ = 2 so the spectrum is the sameas for the Hydrogen atom except for aprefactor of 4
E =Z 2E1
n2, Z = 2
Etot = 2 · 4E1
4= −27.2 eV
Eion =4E1
12= −54.4 eV
Ee = Etot − Eion = 27.2 eV
1
λ= 4R
(1
n2f− 1
n2i
)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 10 / 16
Problem 5.9
(a) Suppose you put both electrons in a helium atom into the n = 2 statefrom which state the atom is ionized to He+. What would the energyof the emitted electron be?
(b) Describe, quantitatively, the spectrum of the helium ion, He+
(a) The energy of each electron in thehelium atom is given by
initially the total energy is
after the ionization the energy of theion is
the energy of the emitted electron is
(b) The He+ is a hydrogenic atom withZ = 2 so the spectrum is the sameas for the Hydrogen atom except for aprefactor of 4
E =Z 2E1
n2, Z = 2
Etot = 2 · 4E1
4= −27.2 eV
Eion =4E1
12= −54.4 eV
Ee = Etot − Eion = 27.2 eV
1
λ= 4R
(1
n2f− 1
n2i
)
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 10 / 16
Problem 5.9
(a) Suppose you put both electrons in a helium atom into the n = 2 statefrom which state the atom is ionized to He+. What would the energyof the emitted electron be?
(b) Describe, quantitatively, the spectrum of the helium ion, He+
(a) The energy of each electron in thehelium atom is given by
initially the total energy is
after the ionization the energy of theion is
the energy of the emitted electron is
(b) The He+ is a hydrogenic atom withZ = 2 so the spectrum is the sameas for the Hydrogen atom except for aprefactor of 4
E =Z 2E1
n2, Z = 2
Etot = 2 · 4E1
4= −27.2 eV
Eion =4E1
12= −54.4 eV
Ee = Etot − Eion = 27.2 eV
1
λ= 4R
(1
n2f− 1
n2i
)C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 10 / 16
Problem 5.12
(a) Figure out the electron configurations for the firsttwo rows of the Periodic Table (up to neon).
(b) Figure out the corresponding total angular momenta,for the first four elements. List all the possibilitiesfor boron, carbon, and nitrogen
(a) hydrogen: (1s) helium: (1s)2
lithium: (1s)2(2s) beryllium: (1s)2(2s)2
boron: (1s)2(2s)2(2p) carbon: (1s)2(2s)2(2p)2
nitrogen: (1s)2(2s)2(2p)3 oxygen: (1s)2(2s)2(2p)4
fluorine: (1s)2(2s)2(2p)5 neon: (1s)2(2s)2(2p)6
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 11 / 16
Problem 5.12
(a) Figure out the electron configurations for the firsttwo rows of the Periodic Table (up to neon).
(b) Figure out the corresponding total angular momenta,for the first four elements. List all the possibilitiesfor boron, carbon, and nitrogen
(a) hydrogen: (1s)
helium: (1s)2
lithium: (1s)2(2s) beryllium: (1s)2(2s)2
boron: (1s)2(2s)2(2p) carbon: (1s)2(2s)2(2p)2
nitrogen: (1s)2(2s)2(2p)3 oxygen: (1s)2(2s)2(2p)4
fluorine: (1s)2(2s)2(2p)5 neon: (1s)2(2s)2(2p)6
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 11 / 16
Problem 5.12
(a) Figure out the electron configurations for the firsttwo rows of the Periodic Table (up to neon).
(b) Figure out the corresponding total angular momenta,for the first four elements. List all the possibilitiesfor boron, carbon, and nitrogen
(a) hydrogen: (1s) helium: (1s)2
lithium: (1s)2(2s) beryllium: (1s)2(2s)2
boron: (1s)2(2s)2(2p) carbon: (1s)2(2s)2(2p)2
nitrogen: (1s)2(2s)2(2p)3 oxygen: (1s)2(2s)2(2p)4
fluorine: (1s)2(2s)2(2p)5 neon: (1s)2(2s)2(2p)6
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 11 / 16
Problem 5.12
(a) Figure out the electron configurations for the firsttwo rows of the Periodic Table (up to neon).
(b) Figure out the corresponding total angular momenta,for the first four elements. List all the possibilitiesfor boron, carbon, and nitrogen
(a) hydrogen: (1s) helium: (1s)2
lithium: (1s)2(2s)
beryllium: (1s)2(2s)2
boron: (1s)2(2s)2(2p) carbon: (1s)2(2s)2(2p)2
nitrogen: (1s)2(2s)2(2p)3 oxygen: (1s)2(2s)2(2p)4
fluorine: (1s)2(2s)2(2p)5 neon: (1s)2(2s)2(2p)6
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 11 / 16
Problem 5.12
(a) Figure out the electron configurations for the firsttwo rows of the Periodic Table (up to neon).
(b) Figure out the corresponding total angular momenta,for the first four elements. List all the possibilitiesfor boron, carbon, and nitrogen
(a) hydrogen: (1s) helium: (1s)2
lithium: (1s)2(2s) beryllium: (1s)2(2s)2
boron: (1s)2(2s)2(2p) carbon: (1s)2(2s)2(2p)2
nitrogen: (1s)2(2s)2(2p)3 oxygen: (1s)2(2s)2(2p)4
fluorine: (1s)2(2s)2(2p)5 neon: (1s)2(2s)2(2p)6
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 11 / 16
Problem 5.12
(a) Figure out the electron configurations for the firsttwo rows of the Periodic Table (up to neon).
(b) Figure out the corresponding total angular momenta,for the first four elements. List all the possibilitiesfor boron, carbon, and nitrogen
(a) hydrogen: (1s) helium: (1s)2
lithium: (1s)2(2s) beryllium: (1s)2(2s)2
boron: (1s)2(2s)2(2p)
carbon: (1s)2(2s)2(2p)2
nitrogen: (1s)2(2s)2(2p)3 oxygen: (1s)2(2s)2(2p)4
fluorine: (1s)2(2s)2(2p)5 neon: (1s)2(2s)2(2p)6
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 11 / 16
Problem 5.12
(a) Figure out the electron configurations for the firsttwo rows of the Periodic Table (up to neon).
(b) Figure out the corresponding total angular momenta,for the first four elements. List all the possibilitiesfor boron, carbon, and nitrogen
(a) hydrogen: (1s) helium: (1s)2
lithium: (1s)2(2s) beryllium: (1s)2(2s)2
boron: (1s)2(2s)2(2p) carbon: (1s)2(2s)2(2p)2
nitrogen: (1s)2(2s)2(2p)3 oxygen: (1s)2(2s)2(2p)4
fluorine: (1s)2(2s)2(2p)5 neon: (1s)2(2s)2(2p)6
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 11 / 16
Problem 5.12
(a) Figure out the electron configurations for the firsttwo rows of the Periodic Table (up to neon).
(b) Figure out the corresponding total angular momenta,for the first four elements. List all the possibilitiesfor boron, carbon, and nitrogen
(a) hydrogen: (1s) helium: (1s)2
lithium: (1s)2(2s) beryllium: (1s)2(2s)2
boron: (1s)2(2s)2(2p) carbon: (1s)2(2s)2(2p)2
nitrogen: (1s)2(2s)2(2p)3
oxygen: (1s)2(2s)2(2p)4
fluorine: (1s)2(2s)2(2p)5 neon: (1s)2(2s)2(2p)6
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 11 / 16
Problem 5.12
(a) Figure out the electron configurations for the firsttwo rows of the Periodic Table (up to neon).
(b) Figure out the corresponding total angular momenta,for the first four elements. List all the possibilitiesfor boron, carbon, and nitrogen
(a) hydrogen: (1s) helium: (1s)2
lithium: (1s)2(2s) beryllium: (1s)2(2s)2
boron: (1s)2(2s)2(2p) carbon: (1s)2(2s)2(2p)2
nitrogen: (1s)2(2s)2(2p)3 oxygen: (1s)2(2s)2(2p)4
fluorine: (1s)2(2s)2(2p)5 neon: (1s)2(2s)2(2p)6
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 11 / 16
Problem 5.12
(a) Figure out the electron configurations for the firsttwo rows of the Periodic Table (up to neon).
(b) Figure out the corresponding total angular momenta,for the first four elements. List all the possibilitiesfor boron, carbon, and nitrogen
(a) hydrogen: (1s) helium: (1s)2
lithium: (1s)2(2s) beryllium: (1s)2(2s)2
boron: (1s)2(2s)2(2p) carbon: (1s)2(2s)2(2p)2
nitrogen: (1s)2(2s)2(2p)3 oxygen: (1s)2(2s)2(2p)4
fluorine: (1s)2(2s)2(2p)5
neon: (1s)2(2s)2(2p)6
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 11 / 16
Problem 5.12
(a) Figure out the electron configurations for the firsttwo rows of the Periodic Table (up to neon).
(b) Figure out the corresponding total angular momenta,for the first four elements. List all the possibilitiesfor boron, carbon, and nitrogen
(a) hydrogen: (1s) helium: (1s)2
lithium: (1s)2(2s) beryllium: (1s)2(2s)2
boron: (1s)2(2s)2(2p) carbon: (1s)2(2s)2(2p)2
nitrogen: (1s)2(2s)2(2p)3 oxygen: (1s)2(2s)2(2p)4
fluorine: (1s)2(2s)2(2p)5 neon: (1s)2(2s)2(2p)6
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 11 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 ,
J = 12 −→
2S1/2
Li: (1s)2(2s)
L = 0, S = 12 ,
J = 12 −→
2S1/2
He: (1s)2
L = 0,S = 0,
J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0,
J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2
−→ 2S1/2
Li: (1s)2(2s)
L = 0, S = 12 ,
J = 12 −→
2S1/2
He: (1s)2
L = 0,S = 0,
J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0,
J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 ,
J = 12 −→
2S1/2
He: (1s)2
L = 0,S = 0,
J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0,
J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 ,
J = 12 −→
2S1/2
He: (1s)2
L = 0, S = 0,
J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0,S = 0,
J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 ,
J = 12 −→
2S1/2
He: (1s)2
L = 0, S = 0, J = 0
−→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0,
J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 ,
J = 12 −→
2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0,
J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 ,
J = 12 −→
2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0,
J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2
−→ 2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0,
J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0,
J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0,
J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0
−→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 ,
J = 12 ,
32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32
−→ 2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0;
J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1,
2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2,
2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0,
1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1,
1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1,
0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→
3D3,3D2,
3D1,1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,
3P2,3P1,
3P0,1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,
3S1,1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
H: (1s)
L = 0, S = 12 , J = 1
2 −→2S1/2
Li: (1s)2(2s)
L = 0, S = 12 , J = 1
2 −→2S1/2
He: (1s)2
L = 0, S = 0, J = 0 −→ 1S0
Be: (1s)2(2s)2
L = 0, S = 0, J = 0 −→ 1S0
B: (1s)2(2s)2(2p)
L = 1, S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 12 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,
2P3/2,2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,
2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.12(b)
N: (1s)2(2s)2(2p)3
L = 3, 2, 1, 0; S = 32 ,
12 ;
J = 92 ,
72 ,
52 ,
32 ,
72 ,
52 ,
72 ,
52 ,
32 ,
12 ,
52 ,
32 ,
52 ,
32 ,
12 ,
32 ,
12 ,
32 ,
12 −→
4F9/2,4F7/2,
4F5/2,4F3/2,
2F7/2,2F5/2,
4D7/2,4D5/2,
4D3/2,4D1/2,
2D5/2,2D3/2,
4P5/2,4P3/2,
4P1/2,2P3/2,
2P1/2,
4S3/2,2S1/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 13 / 16
Problem 5.13
(a) Hund’s first rule says that, consistent with the Pauli principle, thestate with the highest spin (S) will have the lowest energy. Whatwould this predict in the case of the excited states of helium?
(b) Hund’s second rule says that, for a given spin, the state with thehighest total orbital angular momentum (L), consistent with overallantisymmetrization, will have the lowest energy. Why doesn’t carbonhave L = 2?
(c) Hund’s third rule says that if a subshell (n,l) is no more than halffilled, then J = |L− S | has the lowest energy. use this to resolve theboron ambiguity of problem 5.12(b).
(d) Use Hund’s rules, together with the fact that a symmetric spin statemust go with an antisymmetric position state (and vice versa) toresolve the carbon and nitrogen ambiguities of problem 5.12(b).
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 14 / 16
Problem 5.13 - solution
(a) The configuration of the excited state of helium is (1s)1(2s)1
This state can either be the singlet (S = 0) parahelium or the triplet(S = 1) orthohelium.
Hund’s first rule identifies, correctly, that orthohelium will have a lowerenergy than parahelium.
(b) From problem 5.12, we had
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
The first rule forces the highest total spin state, which would be S = 1,which is symmetric. Thus, the spatial function must be antisymmetric andwhile L = 2 is the largest value, it is symmetric and so the ground statemust have L = 1, leaving 3P2,
3P1,3P0 as possibilities
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 15 / 16
Problem 5.13 - solution
(a) The configuration of the excited state of helium is (1s)1(2s)1
This state can either be the singlet (S = 0) parahelium or the triplet(S = 1) orthohelium.
Hund’s first rule identifies, correctly, that orthohelium will have a lowerenergy than parahelium.
(b) From problem 5.12, we had
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
The first rule forces the highest total spin state, which would be S = 1,which is symmetric. Thus, the spatial function must be antisymmetric andwhile L = 2 is the largest value, it is symmetric and so the ground statemust have L = 1, leaving 3P2,
3P1,3P0 as possibilities
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 15 / 16
Problem 5.13 - solution
(a) The configuration of the excited state of helium is (1s)1(2s)1
This state can either be the singlet (S = 0) parahelium or the triplet(S = 1) orthohelium.
Hund’s first rule identifies, correctly, that orthohelium will have a lowerenergy than parahelium.
(b) From problem 5.12, we had
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
The first rule forces the highest total spin state, which would be S = 1,which is symmetric. Thus, the spatial function must be antisymmetric andwhile L = 2 is the largest value, it is symmetric and so the ground statemust have L = 1, leaving 3P2,
3P1,3P0 as possibilities
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 15 / 16
Problem 5.13 - solution
(a) The configuration of the excited state of helium is (1s)1(2s)1
This state can either be the singlet (S = 0) parahelium or the triplet(S = 1) orthohelium.
Hund’s first rule identifies, correctly, that orthohelium will have a lowerenergy than parahelium.
(b) From problem 5.12, we had
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
The first rule forces the highest total spin state, which would be S = 1,which is symmetric. Thus, the spatial function must be antisymmetric andwhile L = 2 is the largest value, it is symmetric and so the ground statemust have L = 1, leaving 3P2,
3P1,3P0 as possibilities
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 15 / 16
Problem 5.13 - solution
(a) The configuration of the excited state of helium is (1s)1(2s)1
This state can either be the singlet (S = 0) parahelium or the triplet(S = 1) orthohelium.
Hund’s first rule identifies, correctly, that orthohelium will have a lowerenergy than parahelium.
(b) From problem 5.12, we had
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
The first rule forces the highest total spin state, which would be S = 1,which is symmetric. Thus, the spatial function must be antisymmetric andwhile L = 2 is the largest value, it is symmetric and so the ground statemust have L = 1, leaving 3P2,
3P1,3P0 as possibilities
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 15 / 16
Problem 5.13 - solution
(a) The configuration of the excited state of helium is (1s)1(2s)1
This state can either be the singlet (S = 0) parahelium or the triplet(S = 1) orthohelium.
Hund’s first rule identifies, correctly, that orthohelium will have a lowerenergy than parahelium.
(b) From problem 5.12, we had
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
The first rule forces the highest total spin state, which would be S = 1,which is symmetric. Thus, the spatial function must be antisymmetric andwhile L = 2 is the largest value, it is symmetric and so the ground statemust have L = 1, leaving 3P2,
3P1,3P0 as possibilities
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 15 / 16
Problem 5.13 - solution
(a) The configuration of the excited state of helium is (1s)1(2s)1
This state can either be the singlet (S = 0) parahelium or the triplet(S = 1) orthohelium.
Hund’s first rule identifies, correctly, that orthohelium will have a lowerenergy than parahelium.
(b) From problem 5.12, we had
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
The first rule forces the highest total spin state, which would be S = 1,which is symmetric.
Thus, the spatial function must be antisymmetric andwhile L = 2 is the largest value, it is symmetric and so the ground statemust have L = 1, leaving 3P2,
3P1,3P0 as possibilities
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 15 / 16
Problem 5.13 - solution
(a) The configuration of the excited state of helium is (1s)1(2s)1
This state can either be the singlet (S = 0) parahelium or the triplet(S = 1) orthohelium.
Hund’s first rule identifies, correctly, that orthohelium will have a lowerenergy than parahelium.
(b) From problem 5.12, we had
C: (1s)2(2s)2(2p)2
L = 2, 1, 0; S = 1, 0; J = 3, 2, 1, 2, 2, 1, 0, 1, 1, 0 −→3D3,
3D2,3D1,
1D2,3P2,
3P1,3P0,
1P1,3S1,
1S0
The first rule forces the highest total spin state, which would be S = 1,which is symmetric. Thus, the spatial function must be antisymmetric andwhile L = 2 is the largest value, it is symmetric and so the ground statemust have L = 1, leaving 3P2,
3P1,3P0 as possibilities
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 15 / 16
Problem 5.13 - solution
(c) For boron, we had the configuration
B: (1s)2(2s)2(2p)
L = 1,S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
Since the outermost shell is less than half filled, J = |L− S | = 12 and the
ground state must be 2P1/2
(d) For carbon, we know that S = 1 and L = 1 and sincethe outer shell is less than half-filled, J = |L− S | = 0 sothe ground state must be 3P0.
For nitrogen, let’s use my graphical trick, which main-tains the correct symmetries for spin and spatial states.The configuration of nitrogen is (1s)2(2s)2(2p)3
S = 32 L = −1 + 0 + 1
= 0
J = L + S = 32 −→ 4S3/2
+1
0
2p-1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 16 / 16
Problem 5.13 - solution
(c) For boron, we had the configuration
B: (1s)2(2s)2(2p)
L = 1,S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
Since the outermost shell is less than half filled, J = |L− S | = 12 and the
ground state must be 2P1/2
(d) For carbon, we know that S = 1 and L = 1 and sincethe outer shell is less than half-filled, J = |L− S | = 0 sothe ground state must be 3P0.
For nitrogen, let’s use my graphical trick, which main-tains the correct symmetries for spin and spatial states.The configuration of nitrogen is (1s)2(2s)2(2p)3
S = 32 L = −1 + 0 + 1
= 0
J = L + S = 32 −→ 4S3/2
+1
0
2p-1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 16 / 16
Problem 5.13 - solution
(c) For boron, we had the configuration
B: (1s)2(2s)2(2p)
L = 1,S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
Since the outermost shell is less than half filled, J = |L− S | = 12 and the
ground state must be 2P1/2
(d) For carbon, we know that S = 1 and L = 1 and sincethe outer shell is less than half-filled, J = |L− S | = 0 sothe ground state must be 3P0.
For nitrogen, let’s use my graphical trick, which main-tains the correct symmetries for spin and spatial states.The configuration of nitrogen is (1s)2(2s)2(2p)3
S = 32 L = −1 + 0 + 1
= 0
J = L + S = 32 −→ 4S3/2
+1
0
2p-1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 16 / 16
Problem 5.13 - solution
(c) For boron, we had the configuration
B: (1s)2(2s)2(2p)
L = 1,S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
Since the outermost shell is less than half filled, J = |L− S | = 12 and the
ground state must be 2P1/2
(d) For carbon, we know that S = 1 and L = 1 and sincethe outer shell is less than half-filled, J = |L− S | = 0 sothe ground state must be 3P0.
For nitrogen, let’s use my graphical trick, which main-tains the correct symmetries for spin and spatial states.The configuration of nitrogen is (1s)2(2s)2(2p)3
S = 32 L = −1 + 0 + 1
= 0
J = L + S = 32 −→ 4S3/2
+1
0
2p-1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 16 / 16
Problem 5.13 - solution
(c) For boron, we had the configuration
B: (1s)2(2s)2(2p)
L = 1,S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
Since the outermost shell is less than half filled, J = |L− S | = 12 and the
ground state must be 2P1/2
(d) For carbon, we know that S = 1 and L = 1 and sincethe outer shell is less than half-filled, J = |L− S | = 0 sothe ground state must be 3P0.
For nitrogen, let’s use my graphical trick, which main-tains the correct symmetries for spin and spatial states.
The configuration of nitrogen is (1s)2(2s)2(2p)3
S = 32 L = −1 + 0 + 1
= 0
J = L + S = 32 −→ 4S3/2
+1
0
2p-1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 16 / 16
Problem 5.13 - solution
(c) For boron, we had the configuration
B: (1s)2(2s)2(2p)
L = 1,S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
Since the outermost shell is less than half filled, J = |L− S | = 12 and the
ground state must be 2P1/2
(d) For carbon, we know that S = 1 and L = 1 and sincethe outer shell is less than half-filled, J = |L− S | = 0 sothe ground state must be 3P0.
For nitrogen, let’s use my graphical trick, which main-tains the correct symmetries for spin and spatial states.The configuration of nitrogen is (1s)2(2s)2(2p)3
S = 32 L = −1 + 0 + 1
= 0
J = L + S = 32 −→ 4S3/2
+1
0
2p-1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 16 / 16
Problem 5.13 - solution
(c) For boron, we had the configuration
B: (1s)2(2s)2(2p)
L = 1,S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
Since the outermost shell is less than half filled, J = |L− S | = 12 and the
ground state must be 2P1/2
(d) For carbon, we know that S = 1 and L = 1 and sincethe outer shell is less than half-filled, J = |L− S | = 0 sothe ground state must be 3P0.
For nitrogen, let’s use my graphical trick, which main-tains the correct symmetries for spin and spatial states.The configuration of nitrogen is (1s)2(2s)2(2p)3
S = 32 L = −1 + 0 + 1
= 0
J = L + S = 32 −→ 4S3/2
+1
0
2p-1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 16 / 16
Problem 5.13 - solution
(c) For boron, we had the configuration
B: (1s)2(2s)2(2p)
L = 1,S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
Since the outermost shell is less than half filled, J = |L− S | = 12 and the
ground state must be 2P1/2
(d) For carbon, we know that S = 1 and L = 1 and sincethe outer shell is less than half-filled, J = |L− S | = 0 sothe ground state must be 3P0.
For nitrogen, let’s use my graphical trick, which main-tains the correct symmetries for spin and spatial states.The configuration of nitrogen is (1s)2(2s)2(2p)3
S = 32
L = −1 + 0 + 1
= 0
J = L + S = 32 −→ 4S3/2
+1
0
2p-1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 16 / 16
Problem 5.13 - solution
(c) For boron, we had the configuration
B: (1s)2(2s)2(2p)
L = 1,S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
Since the outermost shell is less than half filled, J = |L− S | = 12 and the
ground state must be 2P1/2
(d) For carbon, we know that S = 1 and L = 1 and sincethe outer shell is less than half-filled, J = |L− S | = 0 sothe ground state must be 3P0.
For nitrogen, let’s use my graphical trick, which main-tains the correct symmetries for spin and spatial states.The configuration of nitrogen is (1s)2(2s)2(2p)3
S = 32 L = −1 + 0 + 1
= 0
J = L + S = 32 −→ 4S3/2
+1
0
2p-1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 16 / 16
Problem 5.13 - solution
(c) For boron, we had the configuration
B: (1s)2(2s)2(2p)
L = 1,S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
Since the outermost shell is less than half filled, J = |L− S | = 12 and the
ground state must be 2P1/2
(d) For carbon, we know that S = 1 and L = 1 and sincethe outer shell is less than half-filled, J = |L− S | = 0 sothe ground state must be 3P0.
For nitrogen, let’s use my graphical trick, which main-tains the correct symmetries for spin and spatial states.The configuration of nitrogen is (1s)2(2s)2(2p)3
S = 32 L = −1 + 0 + 1 = 0
J = L + S = 32 −→ 4S3/2
+1
0
2p-1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 16 / 16
Problem 5.13 - solution
(c) For boron, we had the configuration
B: (1s)2(2s)2(2p)
L = 1,S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
Since the outermost shell is less than half filled, J = |L− S | = 12 and the
ground state must be 2P1/2
(d) For carbon, we know that S = 1 and L = 1 and sincethe outer shell is less than half-filled, J = |L− S | = 0 sothe ground state must be 3P0.
For nitrogen, let’s use my graphical trick, which main-tains the correct symmetries for spin and spatial states.The configuration of nitrogen is (1s)2(2s)2(2p)3
S = 32 L = −1 + 0 + 1 = 0
J = L + S = 32
−→ 4S3/2
+1
0
2p-1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 16 / 16
Problem 5.13 - solution
(c) For boron, we had the configuration
B: (1s)2(2s)2(2p)
L = 1,S = 12 , J = 1
2 ,32 −→
2P1/2,2P3/2
Since the outermost shell is less than half filled, J = |L− S | = 12 and the
ground state must be 2P1/2
(d) For carbon, we know that S = 1 and L = 1 and sincethe outer shell is less than half-filled, J = |L− S | = 0 sothe ground state must be 3P0.
For nitrogen, let’s use my graphical trick, which main-tains the correct symmetries for spin and spatial states.The configuration of nitrogen is (1s)2(2s)2(2p)3
S = 32 L = −1 + 0 + 1 = 0
J = L + S = 32 −→ 4S3/2
+1
0
2p-1
C. Segre (IIT) PHYS 405 - Fall 2015 November 30, 2015 16 / 16