1. Complex atoms Structure of nucleus
PHY123
11/24/2018
Lecture XVIII

2. Concepts Quantum numbers, quantum state Pauli principle Periodic table
Isotopes
Binding energy
Fission and fusion
11/24/2018
Lecture XVIII

3. Pauli principle Electron has spin 1/2 Electron is a fermion:
Not more than one electron can be in each quantum state (Pauli principle)
Pauli principle is responsible for periodic table (Mendeleev)
NB. If particle has an integer spin (0 or 1) it is a boson - all particles tend to fall in the same state.
Example – photons (lasers use this principle)
11/24/2018
Lecture XVIII

4. Electron quantum state
Principle quantum number n=1,2,3,4,…
determines energy level, higher E for higher n
Orbital quantum number l
For each n l can be 0,1,2,3, …(n-1)
l states are leveled by letters
s: l=0; p: l=1; d: l=2; f: l=3; g:l=4
E.g. n=5, then l can be 0, 1, 2, 3, 4
Possible l states are s,p,d,f,g
n=1, only l=0 s-state is possible
11/24/2018
Lecture XVIII

5. Electron quantum statez
Orbital quantum number is a vector length l
Its projection on z axis is another q.n. – magnetic quantum number ml
ml can be only integer
11/24/2018
Lecture XVIII

6. Electron quantum statez
All electrons have spin=1/2
It is a vector
Its projection on z axis is another q.n. – spin ms
ms can be only
11/24/2018
Lecture XVIII

7. Possible number of states
For each given n,l,ml  2 possible states:
ms=+1/2; ms=-1/2
For each given n,l  2(2l+1) states:
ml=+l, +(l-1), …+2,+1, 0, -1, -2, …-(l-1), -l  (2l+1)
For each ml: ms=+1/2; ms=-1/2  2
For each given n  2n2 possible states
l=0,1,2,3,4,….(n-1)
For each l: ml=+l, +(l-1), …+2,+1, 0, -1, -2, …-(l-1), -l  (2l+1)
For each (l,ml) ms=+1/2; ms=-1/2
11/24/2018
Lecture XVIII

8. State labeling 2p3 means “3 electrons in p state (l=1) at n=2 level”
4f5 means “5 electrons in f state (l=3) at n=4 level”
States filled up from lower to higher n
For each n from lower to higher l
11/24/2018
Lecture XVIII

9. Periodic table Elements in increasing mass
MA=NpMp+NnMn
But mass does not determine chemical properties
Atomic number Z does. Z=Np- charge of nucleus = number of electrons
H Z=1 1 electron
n=1, l=0 (s), ml=0
1s1
He Z=2 2 electrons
1s2
2 e: spin up, spin down
Li Z=3 3 electrons:
2 e on 1s shell
1 e on n=2, l=0 (s), ml=0
1s22s1
11/24/2018
Lecture XVIII

10. Electron structure 2n2 states for each n 2(2l+1) states for each n,l
2 states for each n,l,ml
s(l=0), p(l=1), d(l=2), f(l=3)
2 states in each s subshell
6 states in each p subshell
10 states in each d subshell
14 states in each f subshell
n=1,2,3,4….
l=0,1,2,…(n-1)
ml=-l,…-1,0,1,…+l
ms=+1/2; -1/2
3p4: 4 electrons on
n=3, l=1 subshell
11/24/2018
Lecture XVIII

11. Test problem #1 The following configuration is allowed 1s22s22p63s3
True
False
11/24/2018
Lecture XVIII

12. Test problem #2
The following configuration is allowed 1s22s22p63s23p54s2
True
False
11/24/2018
Lecture XVIII

13. Test problem #3 The following configuration is allowed 1s22s22p62d1
True
False
11/24/2018
Lecture XVIII

14. Atomic spectra
Forbidden (not completely though), allowed transitions (Dl=±1), because photon has spin 1 and angular momentum is conserved
Inner energy shells (called K-shells) “see” almost full potential of the nucleus proportional to (Z-1)2
Observe atomic spectra in X-ray scattering
11/24/2018
Lecture XVIII

15. Nucleons and nuclei Proton and neutron are called nucleons
1u= x10-27kg=931.5MeV/c2
Protons have positive charge , neutrons have zero charge
A new kind of force – “strong” force – holds nucleons inside nucleus. Unlike EM force strong force is “short distance”  nuclei are very tightly packed
11/24/2018
Lecture XVIII

16. Nucleus radius Volume proportional to the number of nucleons:
11/24/2018
Lecture XVIII

17. Isotopes
Mass of the nucleus does not determine chemical properties Atomic number Z does.
Same chemical element (Z=Np) can have different number of neutrons –isotopes:
Carbon Z=66 p, but it can have
5n: 11C, 6n: 12C, 7n: 13C, n: 14C, 9n: 15C, 10n: 16C
Only 12C is stable (98.9% of C on Earth)
11/24/2018
Lecture XVIII

18. Masses of nuclei and nucleons
Let’s compare nucleus mass with the sum of nucleon masses (from appendix D in Giancoli):
Al: Z=13  13 p, A=27 27-13=14 n
M(13p+14n)=13x u+14x = u
M(Al)= u
Where did ( = u) go?
Recall that mass is a form of energy
To extract a nucleon from the nucleus you must supply energy W:
M(Al)+W=M(13p)+M(14n)
Hence M(Al)<M(13p)+M(14n)
W is called binding energy
11/24/2018
Lecture XVIII

19. Binding energy per nucleon
Fusion
Fission
Most tightly woven nuclei
in the middle of the Mendeleev’s table
Energetically most favorable place
Too many protons
Electrostatic repulsion
Not enough nucleons to
build up strong force
11/24/2018
Lecture XVIII

20. Nuclear fission
Discovered by German scientists Otto Hahn and Fritz Strassmann in 1938
235U splits into two nuclei when bombarded by neutrons
More neutrons are produced in the reaction:
n+ 235U141Ba+92Kr+3n
This means a chain reaction is possible when the minimum mass of uranium reaches critical mass
Tremendous amount of energy is released
11/24/2018
Lecture XVIII

21. Chain reaction Uncontrollable chain reaction – fission bomb
Controllable chain reaction – nuclear reactors
11/24/2018
Lecture XVIII

22. Nuclear reactor
11/24/2018
Lecture XVIII

23. Fusion reactions in the Sun
The key is to bring two protons together close enough for the strong force to overcome electrostatic repulsion – need high T:
1H+1H2H+e++n
1H+2H3He+g
3He+3He4He+1H+1H
This is how heavier (Z>1) elements were produced in the first place (only upto iron or nickel)
Heavier elements are produced in the middle of the starts or supernova explosions by absorbing extra energy
Fission bomb is used to ignite fusion bomb – hydrogen bomb
11/24/2018
Lecture XVIII