Second
Midterm
2203,
Nov.
3.
2011
Name:
Solutions
Show
all
of
your
work
on
the
pages
contained
in
the
examination.
Problem
1)
(23
pts)
This
is
a
problem
about
using
Hund’s
Rule
and
the
Pauli
Principle
in
a
many
electron
atom,
Nitrogen
(Z=7)
a) (4
pts)
Write
down
the
ground
state
configuration
of
N
using
the
Pauli
Principle.
Electronic
configuration
1s22s22p3
b) (5
pts)
Use
Hund’s
rule
and
the
Pauli
Principle
to
describe
the
ground
state
using
the
boxes
below.
You
need
to
use
an
up
and
down
arrow
to
indicate
the
spin
state
of
each
electron.
c) (5
pts)
What
is
the
total
z
component
of
spin
and
angular
momentum
for
this
state?
Lz=
0
Sz=
3/2
d) (
4
pts)
The
ionization
energy
of
a
1s
electron
is
399
eV.
What
is
Zeff.
Zeff=
5.42
Ionization = 13.6 Z 2eff eV
399eV
= Z 2eff
13.6eV
Z eff = 5.42
e) (4
pts)
Why
doesn’t
Zeff=Z?
The other electrons screen the Z charge from the nucleus.
The othere 1s electron and the two 2s electrons
1
f) (5
pts)
Describe
what
is
wrong
with
each
of
the
following
electronic
configurations
(Pauli
Principle).
If
the
configuration
satisfies
the
Pauli
Principle,
is
it
the
lowest
energy
configuration
(Hund’s
rules).
#1)
#2)
#3)
#4)
Pauli principle satisfied, but Hund's first rule is not satisfied to maximize spin.
Three electrons in the 2s shell violated the Pauli Principle
Pauli Principle is satisfied but this violates both the first Hund"s rule
Two electrons in the same state in the 2s shell violates Pauli's principle
#5) There are two electrons in the same m l = 1 state, Pauli Principle.
2
Problem
2)
(12
pts)
In
this
problem
you
will
calculate
the
length
of
the
J
vector
for
two
different
quantum
states
on
a
H
atom.
Assume
that
the
z
direction
is
pointing
towards
the
top
of
the
page.
For an electron in the n=2, l=1 state--m l = 1, 0, −1.

For part a) I meant by up to use m l = 1 and Sz = , but since I did not specify this any m l = 0 was correct.
2
For (b) l = 0, m l = 0
(6
pts)
The
electron
in
the
H
atom
is
in
the
n=2,
l=1
state.
Find
the
following
for
the
J
vector
for
a
parallel
spin
(up
spin)
The
length
of
the
|J|
vector.
|j|=
1.94

3
3⎛ 3 ⎞
15
For l = 1, j= , and |J|= ⎜ + 1⎟  =

2
2⎝2 ⎠
2
Jz =
3
 See the figure.
2
3/2
The
z
component
of
the
J
vector.
Jz=

a) (6
pts)
The
electron
in
the
H
atom
is
in
the
n=2,
l=0
state.
Find
the
following
for
the
J
vector
for
a
parallel
spin
(up
spin)
The
length
of
the
|J|
vector.
|j|=
3
2

Look at L, its length is l ( l + 1)=0
Look at spin: length is s ( s + 1) =
3
| J |=
 See Figure
2
1
Jz = 
2
The
z
component
of
the
J
vector.
Jz=
3

2
1/2
3

Problem 3 (15 pts) The
figure
below
shows
three
different
1D
potentials
all
supporting
bound
states:
The
probability
densities
as
a
function
of
x
for
the
n
=
8
state
(the
7th
excited
state)
for
each
of
the
potentials
are
shown
below.
Match
each
curve
to
the
potential
where
it
belongs.
Explain
your
answers.
(Pay
attention
to
symmetry,
and
how
wavelength
and
probability
density
changes
with
x).
(a) (5
pts)
Which
Potential
produces
wave
function
(a)?
Why?
This belongs to the Hydrogen like potential. It is symmetric about the center
and the intensities of the max. are not uniform.
(b) (5
pts)
Which
Potential
produces
wave
function
(b)?
Why?
This belongs to the square well potential. It is symmetric about the center
and the intensities of the max. are uniform.
(c) (5
pts)
Which
Potential
produces
wave
function
(c)?
Why?
This belongs to the Morse potential. It is not symmetric about the center
and the intensities of the max. are at the right. The wavelenth is longer on the right.
4
Problem
4)
(17
pts)
The
figures
show
three
different
radial
wave
functions
( n ≤ 3 )
for
the
H
atom
(Rn.l).
You
may
think
there
are
more
than
one
options,
if
so
state
why.
(a) (4
pts)
Determine
the
n
and
l
values
for
the
wavefunction
to
the
right?
Give
your
reasons.
There are n-l -1 nodes.
(a) Two nodes so n=3, l=0 3s state
(b) (4
pts)
Determine
the
n
and
l
values
for
the
second
wavefunction?
Give
your
reasons.
There are n-l -1 nodes.
(b) One node, not an s state so n=3, l=1 3p
(c)
(5
pts)
Determine
the
n
and
l
values
for
the
third
wavefunction?
Give
your
reasons.
There are n-l -1 nodes.
(c) no nodes, not an s state
could be 3d, or 2p: Maximium indicates 3d
(d)
(4
pts)
The
bottom
figure
is
a
plot
of
|Rn,l|2.
Which
wave
function
does
it
represent?
There are no nodes so it is a 2p or 3d
This is not a picture of a p function so it has to be 3d This should have told you that the answer to (c) was 3d!
5
Problem
#5)
(6
pts)
What
is
the
degeneracy
of
the
n=4
energy
level
in
a
Hydrogen
atom
including
spin?
Show
your
work?
2n2=32
For the n=4 state we have
n=4, l=3, m l = 3, m s = ±1 / 2
n=4, l=3, m l = 2, m s = ±1 / 2
n=4, l=3, m l = 1, m s = ±1 / 2
n=4, l=3, m l = 0, m s = ±1 / 2
n=4, l=3, m l = −1, m s = ±1 / 2
n=4, l=3, m l = −2, m s = ±1 / 2
n=4, l=3, m l = −3, m s = ±1 / 2
n=4, l=2, m l = 2, m s = ±1 / 2
n=4, l=2, m l = 1, m s = ±1 / 2
n=4, l=2, m l = 0, m s = ±1 / 2
n=4, l=2, m l = −1, m s = ±1 / 2
n=4, l=2, m l = −2, m s = ±1 / 2
n=4, l=1, m l = 1, m s = ±1 / 2
n=4, l=1, m l = 1, m s = ±1 / 2
n=4, l=1, m l = 0, m s = ±1 / 2
n=4, l=1, m l = 0, m s = ±1 / 2
n=4, l=0, m l = 0, m s = ±1 / 2
6
Problem
7.
(6
points)
The
picture
below
is
from
a
website
aimed
at
illustrating
the
ordinary
Zeeman
effect
for
undergraduates
like
you.
It
is
suppose
to
show
the
transitions
for
a
7s
(top)
to
a
6p
state
(bottom)
of
H
when
the
atom
is
in
a
10T
magnetic
field.
a) (2pts)
First
calculate
the
energy
from
7s
to
6p
with
B=0.
Is
there
range
correct?
See
“real
spectrum.”
1⎤
⎡1
ΔE = 0.1 eV ΔE = 13.6 ⎢ 2 − 2 ⎥ eV = 0.1eV 7 ⎦
⎣6
b)
(4pts)
Write
down
(very
specifically)
everything
that
is
wrong
with
this
picture.
1) The energy of the transition is wrong. 2) A 7s state with l=0 has no Zeeman splitting. 3) The 6p level has l=2. This means that there are 5 lines in a magnetic field. 4) The energy separation is really wrong. 8