PHY 251 Fall 2009: homework problem set 7, due in the PHY 251 drop box in room A129 by
noon on Friday, Oct. 30.
1. Serway 6.15
Answer:
(a) The potential energy from the Coulomb force is
U
=
q
1
q
2
4
πǫ
0
r
. The
q
1
q
2
terms are all going
to have an electron charge times either 1 (electron) or +1 (proton), and the distances
r
are
going to be
1
d
,
2
d
, or
3
d
. We can then work out the potential seen by the particles from
left to right, being careful not to doublecount (that is, don’t count again the potential of a
particle to the left of whatever one we’re now considering):
Δ
U
1
=
e
2
4
πǫ
0
d
bracketleftBigg
(
−
1)(+1)
1
+
(
−
1)(
−
1)
2
+
(
−
1)(+1)
3
bracketrightBigg
=
e
2
4
πǫ
0
d
bracketleftbigg
−
1 +
1
2
−
1
3
bracketrightbigg
Δ
U
2
=
e
2
4
πǫ
0
d
bracketleftBigg
(+1)(
−
1)
1
+
(+1)(+1)
2
bracketrightBigg
=
e
2
4
πǫ
0
d
bracketleftbigg
−
1 +
1
2
bracketrightbigg
Δ
U
3
=
−
e
2
4
πǫ
0
d
bracketleftBigg
(
−
1)(+1)
1
bracketrightBigg
=
e
2
4
πǫ
0
d
[
−
1]
so that the net potential energy is
U
=
Δ
U
1
+ Δ
U
2
+ Δ
U
3
=
e
2
4
πǫ
0
d
bracketleftbigg
−
1 +
1
2
−
1
3
−
1 +
1
2
−
1
bracketrightbigg
=
e
2
4
πǫ
0
d
bracketleftbigg
−
6
6
+
3
6
−
2
6
−
6
6
+
3
6
−
6
6
bracketrightbigg
=
e
2
4
πǫ
0
d
bracketleftbigg
−
14
6
bracketrightbigg
=
−
e
2
4
πǫ
0
d
7
3
=
−
7
e
2
12
πǫ
0
d
.
(b) If we say that the electrons are restricted to a 1D box of width
3
d
, that’s like saying
L
= 3
d
in the infinite quantum well solution, and
n
= 1
for the lowest energy quantum state.
Therefore the kinetic energy of allowed electron wavefunctions in the box is
E
k
=
E
1
= 1
2
π
2
¯
h
2
2
m
(3
d
)
2
=
h
2
72
md
2
which should be doubled for two electrons.
(c) The total energy is
U
+
E
k
; to find the value of
d
which minimizes the total energy we
set the derivative with respect to
d
to be zero:
∂
∂d
(
U
+
E
k
) = 0
=
∂
∂d
parenleftBigg
−
7
e
2
12
πǫ
0
d
+
h
2
36
md
2
parenrightBigg
=
parenleftBigg
−
(
−
1)
7
e
2
12
πǫ
0
d
2
+ (
−
2)
h
2
36
md
3
parenrightBigg
7
e
2
12
πǫ
0
d
2
=
h
2
18
md
3
d
=
4
πǫ
0
h
2
42
me
2
=
4
π
(8
.
85
×
10

12
)
·
(6
.
63
×
10

34
)
2
42
·
(9
.
11
×
10

31
)
·
(1
.
602
×
10

19
)
2
=
4
.
98
×
10

11
meters
= 0
.
0498
nm
.
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 Fall '01
 Rijssenbeek
 Physics, Energy, Force, Potential Energy, Work, ev, hω, Serway 6.28

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