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Transcript
```1
Problem 8.1
A particle with mass m moves in a 3D box with edges L1 = L, L2 = 2L, and
L3 = 2L. Find the energies of the six lowest states. Which ones are degenerate?
1.1
Solution
We get the wavenumbers the usual way, using the boundary conditions.
r
2mE1
n1 π
k1 =
=
L
~2
r
n2 π
2mE2
k2 =
=
2L
~2
r
n3 π
2mE3
=
k3 =
2L
~2
Solving for the energies gives
E1 =
~2 π 2 2
n
2mL2 1
E2 =
~2 π 2 n22
2mL2 4
E3 =
~2 π 2 n23
2mL2 4
Or
~2 π 2
E = E1 + E2 + E3 =
2mL2
n21
n2
n2
+ 2+ 3
4
4
The ground state is when n1 = n2 = n3 = 1 leading to
~2 π 2
12
12
3~2 π 2
2
E=
1
+
+
=
2mL2
4
4
4mL2
There is a twofold degeneracy in the first excited state: n1 = n2 = 1; n3 = 2
or n1 = n3 = 1; n2 = 2
~2 π 2
E=
2mL2
22
12
+
1 +
4
4
2
~2 π 2
=
2mL2
12
22
+
1 +
4
4
2
=
9~2 π 2
8mL2
The second excited state is unique and will be n1 = 1; n2 = n3 = 2.
22
22
3~2 π 2
~2 π 2
2
E=
1
+
+
=
2mL2
4
4
mL2
1
Again, there is a twofold degeneracy in the third excited state n1 = 1; n2 =
2; n3 = 3 or n1 = 1; n2 = 3; n3 = 2
~2 π 2
E=
2mL2
22
32
1 +
+
4
4
2
~2 π 2
=
2mL2
32
22
17~2 π 2
2
1 +
+
=
4
4
8mL2
These are the lowest six states. 1 ground + 2 first excited + 1 second excited
+ 2 third excited = 6.
2
Problem 8.3
A particle of mass m is in a 3D cube with sides L. It is in the third excited
state, corresponding to n2 = 11.
(a) Calculate the energy of the particle.
(b) The possible combinations of n1 , n2 , and n3
(c) The wavefunctions for these states.
2.1
2.1.1
Solution
Part (a)
Just plug in n2 = 11 to the 3D box’s energy.
E=
2.1.2
11~2 π 2
2mL2
Part (b) and Part (c)
We’ll need
n2 = n21 + n22 + n23
There are three ways to do this
11 = (32 + 12 + 12 ) = (12 + 32 + 12 ) = (12 + 12 + 32 )
Corresponding to the three states and their wavefunctions
n1 = 3; n2 = 1; n3 = 1
→
ψ(x, y, z) =
3/2
πy πz 2
3πx
sin
sin
sin
L
L
L
L
n1 = 1; n2 = 3; n3 = 1
→
ψ(x, y, z) =
3/2
πx πz 2
3πy
sin
sin
sin
L
L
L
L
n1 = 1; n2 = 1; n3 = 3
→
ψ(x, y, z) =
3/2
πy πx 2
3πz
sin
sin
sin
L
L
L
L
2
3
Problem 8.4
A particle of mass m is stuck in a 2D box of length L.
(a) What are the wavefunctions?
(b) What are the energies of the ground state and the first excited state?
3.1
Solution
3.2
Part (a)
We start with a general wavefunction see if separation of variables works.
ψ(x, y) = ψ1 (x)ψ2 (y)
Plugging this into the Schrodinger Equation and mixing around a few terms
around we get
~2
1 ∂ 2 ψ1 (x)
1 ∂ 2 ψ2 (y)
−
+
=E
2m ψ1 (x) ∂x2
ψ2 (y) ∂y 2
This separates into two equations
−
~2 ∂ 2 ψ1 (x)
= E1 ψ1 (x)
2m ∂x2
−
~2 ∂ 2 ψ2 (y)
= Eψ2 (y)
2m ∂y 2
This will have solutions and energies of
r
n πx 2
1
sin
ψ1 (x) =
L
L
r
n πy 2
2
ψ2 (y) =
sin
L
L
3.3
E1 =
~2 π 2 2
n
2mL2 1
E2 =
~2 π 2 2
n
2mL2 2
Part (b)
We can see that the total energy will be
~2 π 2 2
(n + n22 )
2mL2 1
The ground state will be at n1 = n2 = 1 and will have energy
E = E1 + E2 =
~2 π 2
mL2
There will be a degeneracy in the first excited state between n1 = 2; n2 = 1
and n1 = 1; n2 = 2 which will have energy
E=
3
~2 π 2 2
5~2 π 2
~2 π 2 2
(2 + 12 ) =
(1 + 22 ) =
2
2
2mL
2mL
2mL2
E=
4
Problem 8.7
(a) Normalize the wavefunction for a particle in a cube of length L.
(b) What if it is a rectangular prism with sides L1 6= L2 6= L3
4.1
Solution
4.1.1
Part (a)
We know that the wavefunction will be
n πy n πz n πx 2
3
1
sin
sin
ψ(x, y, z) = A sin
L
L
L
We apply the normalization rule
Z +∞
1=
ψ ∗ (~r)ψ(~r)dV
−∞
Z
1 = A2
L
Z
x=0
1=A
2
L
Z
y=0
L
sin2
z=0
n πy n πz n πx 2
3
1
sin2
sin2
dxdydz
L
L
L
Z L
n πz n πx Z L
3
1
2 n2 πy
dx
dy
dz
sin
sin2
sin
L
L
L
y=0
z=0
x=0
Z
L
2
3
L
2
3/2
2
A=
L
1 = A2
4.2
Part (b)
The wavefunction (and its square) will be the same as before but with a minor
tweak
1=A
2
Z
L1
L2
Z
L3
2
sin
x=0
1 = A2
Z
Z
y=0
L1
x=0
sin2
z=0
n1 πx
L1
n1 πx
L1
Z
sin
L2
dx
2
sin2
y=0
4
n2 πy
L2
n2 πy
L2
2
sin
Z
n3 πz
L3
L3
dy
z=0
sin2
dxdydz
n3 πz
L3
dz
L1
L2
L3
2
2
2
r
8
A=
L1 L2 L3
1 = A2
5
Problem 8.11
The orbital angular momentum of the Earth is 4.83 × 1031 kg· m2 /s. Calculate
~ when l → l + 1
l and the fractional change in kLk
5.1
Solution
~ is
The quantization rule for kLk
~ =
kLk
p
l(l + 1)~
It will be proven in the second part of this problem that l ≈ l + 1. Using
this we get
~
kLk
~
Since ~ = 1.05 × 10−34 kg· m2 /s we see
l=
4.83 × 1031 kg · m2 /s
≈ 4.6 × 1065
1.05 × 10−34 kg · m2 /s
That is a HUGE quantum number! To get a sense of how big this is, look
at the second part of the problem
l=
~l k
kL~l+1 k − kL
=
~l k
kL
p
p
√
√
(l + 1)(l + 2)~ − (l)(l + 1)~
l+2− l
2
p
√
=
≈ 65 ≈ 10−65 ≈ 0
10
l
(l)(l + 1)~
This is essentially zero! This is why quantum mechanics isn’t necessary
for examining macroscopic objects and the natural world ”feels” continuous.
When you work with numbers that are so big, the quantization of nature isn’t
noticeable.
6
Problem 8.18
A hydrogen atom is in the 6g state.
(a) What is the principle quantum number?
(b) What is the energy of the atom?
~
(c) What are the possible values for l and kLk?
(d) What are the possible values of ml ? (e) What are the possible values of Lz
~ make with the z-axis?
and for each of these what angle will L
5
6.1
6.1.1
Solution
Part (a)
n=6
6.1.2
Part (b)
In general, the energy for a hydrogen-like atom in the n-th energy state is
En = −
ke2 Z 2
2a0 n2
In this case Z = 1 and n = 6 so
E6 = −
6.1.3
ke2
72a0
Part (c)
The rule for l is that it must be an integer between
0≤l<n
This makes the allowed values 0, 1, 2, 3, 4, and 5.
6.1.4
Part (d)
The rule for ml is that it must be an integer between
−l ≤ ml ≤ l
This makes the allowed values 0, ±1, ±2, ±3, ±4, and ±5. The z component
of the angular momentum is given by
Lz = ml ~
So the allowed values will be 0, ±~, ±2~, ±3~, ±4~, and ±5~. The values of
~ are given by
kLk
p
~ = l(l + 1)~
kLk
In general we will have
θlm = ± sin−1
kLz k
~
kLk
!
= ± sin−1
All possible values of θlm are listed below
θ00 = 90◦
6
km k
p l
l(l + 1)
!
θ1±1 = ± sin−1
θ10 = 90◦
◦
θ20 = 90
−1
θ2±1 = ± sin
θ3±2 = ± sin−1
θ40 = 90
√
θ50 = 90
3
√
4·5
−1
√
≈ ±24
3
5·6
√
1
5·6
= ±45◦
−1
◦
≈ ±13
√
1
3·4
√
−1
θ5±5 = ± sin
5
√
5·6
4
√
4·5
−1
◦
θ5±2 = ± sin
θ5±4 = ± sin−1
= ±33◦
≈ ±55◦
= ±60◦
θ4±2 = ± sin
θ4±4 = ± sin
≈ ±11
3
3·4
−1
◦
2
√
2·3
≈ ±17◦
−1
θ2±2 = ± sin
◦
1
1·2
θ3±3 = ± sin−1
= ±42
θ5±1 = ± sin
θ5±3 = ± sin−1
1
√
4·5
√
≈ ±35◦
θ4±1 = ± sin
θ4±3 = ± sin
◦
2
3·4
−1
−1
1
√
2·3
θ3±1 = ± sin−1
θ30 = 90◦
◦
√
4
5·6
2
√
4·5
≈ ±27◦
= ±63◦
√
2
5·6
≈ ±21◦
= ±47◦
= ±66◦
One cool thing to notice is that θll increases as l increases. This agrees with
~ which
our prediction that in the limit l >> 1 we will be able to have Lz = kLk,
is usually what we deal with classically.
7
Problem 8.20
Using a semiclassical approximation, the energy of a circular orbit becomes
L̂2
kZe2
−
2mr2
r
Use this to show that we must have l ≤ n − 1
E=
7
(1)
7.1
Solution
The formula for energy is
En = −
ke2 Z 2
2a0 n2
The formula for angular momentum is
p
~ = l(l + 1)~
kLk
According to Bohr, the allowed values of r for an atom with Z protons is
a0 n2
Z
Plugging these into Equation 1 we get:
r=
−
l(l + 1)~2 Z 2
Z
ke2 Z 2
=
− kZe2
2
2
2
2a0 n
2m
(a0 n )
a0 n2
Shuffling a few terms around will yield
l(l + 1)~2
−ke2 n2
=
− ke2 n2
2
2ma0
Remember that
~2
mke2
Using this will simplify our equation to
a0 =
−n2
l(l + 1)
=
− n2
2
2
n2 = l(l + 1)
Remember that we were dealing with the largest l possible. With this in
mind we can see that we must have
l ≤n−1
8
Problem 8.22
R2p (r) = Are−r/2a0
Calculate hri for an electron in this state.
8
8.1
Solution
First we need to normalize the wavefunction.
Z +∞
∗
1=
ψ2p
(~r)ψ2p (~r)dV
−∞
reality, there will be an angular dependence as well, but we can still consider
solely the radial portion since r̂ = r and the wavefunction is separable.
Z +∞
∗
1=
R2p
(r)R2p (r)[r2 dr]
r=0
Z
+∞
∗
R2p
(r)R2p (r)r2 dr
1=
−∞
∞
Z
A2 r4 e−r/a0 dr
1=
0
1 = A2 (24a50 )
1
24a50
A2 =
Now we can calculate the expectation value for r.
Z +∞
∗
hri =
R2p
(r)r̂R2p (r)[r2 dr]
r=0
hri =
1
24a50
hri =
∞
Z
r5 e−r/a0 dr
0
1
(120a60 )
24a50
hri = 5a0
9
```
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