Helium - University of California, Berkeley
Download
Report
Transcript Helium - University of California, Berkeley
Helium
Nikki Meshkat
Physics 138
Outline
•
•
•
•
•
•
Interesting facts about Helium
Solving the Schrodinger Equation for He
Ground State
Excited States
Spin
Solve integrals for various cases
Fun Facts about Helium
• Helium is named after the Greek god of the sun “Helios”
• Discovered on the sun before it was discovered on
Earth. Pierre Janssen, a French astronomer, noticed a
yellow line in the sun’s spectrum while he was studying a
solar eclipse in 1868. Then Sir Norman Lockyer, an
English astronomer, realized that this yellow line could
not be produced by any known element at the time. He
named it Helium.
• Helium is produced in stars by fusion of hydrogen. This
process creates alpha particles, which is Helium without
its electrons.
Fun Facts
• Helium is the second most abundant element in the
universe after Hydrogen
• Thermal conductivity is higher than all gases except
Hydrogen
• Lowest boiling (4.22 K) and melting (0.95 K) points
among all elements
• Liquid Helium is an important cryogenic material, used to
study superconductivity and to cool superconducting
magnets.
• DOE’s Jefferson Lab uses large amounts of liquid helium
to operate its superconductive electron accelerator.
More Fun Facts
• Helium created on earth by alpha decay of radioactive
elements. The alpha particles trap electrons to become
He, which is then trapped in natural gas.
• There are no known compounds that contain Helium,
although attempts are being made to produce Helium
Diflouride.
• When Helium is inhaled, it excites the higher harmonics
of the vocal tract, since the speed of sound in He is 3
times that in air. One’s voice is a linear combination of
the fundamental frequency plus higher harmonics, so
Helium causes there to be more contribution from the
higher harmonics to your voice and less contribution
from the lower frequencies.
Physics of Helium
Let’s look at the Schrodinger Equation with the
Hamiltonian for Helium:
2
2
2
2
2
2e
2e
e
2
2
{
1
2
} E
2m
2m
r1
r2
r12
How can we solve it?
Ignore mutual repulsion term!!!
First solve an easier problem
• Approximate as two Hydrogenic Ions and
perform seperation of variables
( H1 H 2 ) E
( 0)
E ( 0) E1 E2
2 2 Ze2
H1
1
2m
r1
2 2 Ze 2
H2
2
2m
r2
(1) (2)
H1 (1) E1 (1) H 2 (2) E2 (2)
For ground state:
E1 E2 54.4eV
E ( 0) 109eV
What about the repulsion term?
Think of it as a perturbation:
( H 0 H ' ) E
H 0 H1 H 2
E E ( 0) E
H ' E
H 0 E ( 0)
For example, the ground state has energy:
1
2
2
dr
4
r
2
2 4r
1
1
2 dr2
1s
0 0
r12
1
1s 2 R1Zs 2 ( r1 ) R1Zs 2 ( r2 )
4
E e 2
1*s
How about the excited states?
Let’s look at the case where one electron is in the 1s state
and the other electron is in any state (call it nl).
1
4
unl (2) Rnl (r2 )Ylm (2 , 2 )
u1s (1) R1s ( r1 )
But wait, there’s a degeneracy, so we need degenerate perturbation theory!
u1s (1)unl (2)
u1s (2)unl (1)
Then the wave function is just a linear combination of these basis functions
au1s (1)unl (2) bu1s (2)unl (1)
Plug and chug away!
u1*s (1)unl* (2){H ' (au1s (1)unl (2) bu1s (2)unl (1)) E(au1s (1)unl (2) bu1s (2)unl (1))}
u1*s (2)unl* (1){H ' (au1s (1)unl (2) bu1s (2)unl (1)) E(au1s (1)unl (2) bu1s (2)unl (1))}
J
{
K
K a
a
} E
J b
b
2
e
J | u1s (1) |2
| unl (2) |2 dr13dr23
r12
J 1s ( r1 ) nl ( r2 )
1s (r1 ) e | u1s (1) |2
e2
K u (1)u (2) u1s (2)unl (1)dr13dr23
r12
*
1s
*
nl
1 3 3
dr1 dr2
r12
nl (r2 ) e | unl (2) |2
So what does this all mean?
E J K
Plugging this back in, we get b=a or b= -a
E E ( 0) J K
S
space
1
{u1s (1)unl (2) u1s ( 2)unl (1)}
2
E E ( 0) J K
A
space
1
{u1s (1)unl (2) u1s (2)unl (1)}
2
Don’t forget spin!
Since electrons are fermions, the total wavefunction must be
antisymmetric under exchange of particles:
S
space
A
spin
A
S
space
spin
S
spin
S
spin
1
{ }
2
S
spin
A
spin
1
{ }
2
Evaluation of Direct Integrals
Ground state: Coulomb repulsion between electron clouds
The electrostatic potential at r2 produced by electron 1 is:
r2
V12 ( r2 )
0
( r1 )
r12
dr13
But spherical symmetry means that the charge in r1 r2
acts like a point charge at the origin:
V12 ( r2 )
r2
Q( r2 )
r2
where
Q( r2 ) ( r1 )4r12 dr1
0
So electrostatic energy from repulsion is
E12 V12 ( r2 ) ( r2 )4r22 dr2
The total energy of the repulsion between electrons is twice this since there
is an equal contribution to the energy from V21 ( r1 )
r2
J 1s 2
1
2 | R1s ( r1 ) |
| R1s ( r2 ) |2 r12 dr1r22 dr2 34eV
r1
0 0
2
Is this a good answer?
• Recall that total energy without mutual
repulsion term was -109 eV
• Thus total energy is -109+34=-75eV
• To go from singly ionized He to doubly
ionized He takes 54.4 eV, thus first
ionization energy is 75-54=21eV
• Measured ionization energy is 24.6eV
• Why wrong answer?
Variational Method
• Notice that expectation value of
perturbation 34eV is not small compared
to binding energy 75eV, thus we aren’t
completely justified in using perturbation
theory.
• Better approximation to wave function can
be found by using variational method.
Evaluation of Direct Integrals
Excited state: 1snl
Assume the excited electron lies outside the 1s wave function.
Then the inner electron screens the outer electron from the full
nuclear charge, so that the outer electron feels only 1 proton:
2
2 1
H0
(12 22 ) e 2 ( )
2m
r1 r2
H ' e2 (
1 1
)
r12 r2
Expanding 1/r_12 in terms of spherical harmonics…
J 1snl e
2
(
0 r2
1 1
) | R1s ( r1 ) |2 | Rnl ( r2 ) |2 r12 dr1r22 dr2
r1 r2
J1s 2 p 2.8 102 eV
E ( 0) 57.8eV
Evaluation of Exchange Integral
• Similar trick with expansion of 1/r_12 term
K1snl
r1
2
2
e 2 R1s ( r1 ) Rnl ( r1 ) R1s ( r2 ) Rnl ( r2 )r1 dr1r2 dr2
3r2
2
2K1s 2 p 0.21eV
Close to measured value of 0.25eV
Conclusion
• Helium can be solved by first looking at
Hydrogenic case and then adding
perturbation term
• This method does not give perfect results
for the ground state, so the variational
method works best in this case
• Methods used for Helium can be extended
to solving many-electron atoms