Lecture 10 - Finite Square Well.pdf-Toda...
Lecture_10_-_Finite_Square_Well.pdf-Today (1) Review of time dependent
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Lecture 10 - Finite Square Well.pdf...
Lecture_10_-_Finite_Square_Well.pdf-Today (1) Review of time dependent
##### Page 12
12
PE for electrons with most PE. “On top”
as more electrons fill in, potential energy for later ones gets
flatter and flatter.
For top ones, is VERY flat.
+
+
+
+
+
+
+
+ +
+
+
+
+
+
+ +
How could you find out how deep the pit is for the top
electrons in copper wire?
This is just the energy needed to remove them from the metal.
That is the work function!!
work function of
copper = 4.7 eV

##### Page 13
13
L
0
0 eV
0
L
4.7 eV
Energy
x
x<0,
V(x) = 4.7 eV
x> L,
V(x) = 4.7 eV
0<x<L, V(x) =0
How to solve?
a. mindless mathematician approach- find
in each region,
make solutions match at boundaries, normalize.
Works, but bunch of math.
x
)
(
)
(
)
(
)
(
2
2
2
2
x
E
x
x
V
x
x
m

##### Page 14
14
b. Clever physicist approach. Reasoning to simplify how to solve.
Electron energy not much more than ~kT=0.025 eV.
Where is electron likely to be?
What is chance it will be outside
of well?
a. zero chance, b. very small chance, c. small,
d. likely
mathematically
V(x) = 4.7 eV
for x<0 and x>L
V(x) = 0 eV
for
0>x<L
0 eV
0
L
4.7 eV
b.
0.025 eV<< 4.7eV.
So very small chance an electron could
have enough energy get out.
What does that say about boundary condition on
(x) ?
a.
(x) must be same x<0, 0<x<L, x>L,
b.
(x<0)~0,
(x>0)
0
c.
(x) ~0 except for 0<x<L
ans c.
So pretty good approximation to say never gets out

##### Page 15
15
0
Energy
x
)
(
)
(
2
2
2
2
x
E
x
x
m
x<0, V(x) ~ infinite
x> L, V(x) ~ infinite
0<x<L, V(x) =0
0
L
so clever physicist just has to
solve
with boundary conditions,
(0)=
(L) =0
solution a lot like microwave & guitar string

##### Page 16
16
A solution to this diff. eq. is
A cos(kx) and
B sin (kx), so sum is also solution.
Check-- plug in.
kx
EA
kx
A
k
m
cos
cos
2
2
2
)
(
)
(
2
2
2
2
x
E
x
x
m
solution if
E
k
m
2
2
2

##### Page 17
17
kx
A
x
cos
)
(
)
(
)
(
2
2
2
2
x
E
x
x
m
E
k
m
2
2
2
makes sense,
because
k
p
so condition on k is just saying that (p
2
)/2m = E.
V=0, so E= KE = ½ mv
2
= p
2
/2m
This result implies that the total energy of electron is
a. quantized according to E
n
= const. x n
2
,
n= 1,2, 3, …
b. quantized according to E
n
= const x n
c. quantized according to E
n
= const. x 1/n
2
d. quantized according to some other condition but
don’t know what it is.
e. not quantized, energy can take on any value.
e. no boundary, not quantized, energy can take on any value.

##### Page 18
18
functional form of solution:
)
sin(
)
cos(
)
(
kx
B
kx
A
x
0 eV
0
L
Apply boundary conditions
x=0
?
A
)
0
(
A=0
0
)
sin(
)
(
kL
B
L
x=L
kL=n
(n=1,2,3,4 …)
2
k
n
L
2
k=n
/L
)
(
)
(
2
2
2
2
x
E
x
x
m
?
1
2

##### Page 19
19
Does this L dependence make sense?
2
2
2
2
2
2
2
mL
n
m
p
E
)
/
(
L
n
k
p
n
L
2
Energy quantized by boundary
conditions.
Why bound electrons
have only certain discrete energies,
unbound can have any energy.
leave it to you to calculate L when
E = kT room temp
How many atomic diameters?
What if cooled to 1 Kelvin?

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