Physics 43 Homework Set 9 Chapter 40 Key
1.
The wave function for an electron that is confined to x ≥ 0 nm is
a. Find the normalization constant.
b. What is the probability of finding the electron in a 0.010 nm-wide region at x = 1.0 nm?
+∞
gives b 0.559nm −1
a) Normalize: =
∫ ψ dx 1=
2
−∞
1.005nm
b) P =
∫
0.995nm
2
1.005nm
ψ dx =
∫
(.559)2 exp( −2 x / 3.2)dx =0.002286 =.23% 17%
0.995nm
2. An electron in an infinite square well has a wave function that is given by ψ ( x ) =
2
 2πx 
sin 

L  L 
for 0 ≤ x ≤ L and is zero otherwise.
a. Sketch the probability density function for this state.
b. Find the most probable position(s) of the electron in the well.
c. Find the expectation value of x.
d. Why or why not should the results of b and c be the same or different? That is, should the most
probable position(s) be the expectation value? (Hint: Use your sketch)
Solution in glass case.
3. A particle in an infinite square well has a wave function that is given by
2
 πx 
sin  
L L
ψ 1 (x ) =
for 0 ≤ x ≤ L and is zero otherwise. (a) Determine the probability of finding the particle between x = 0 and
x = L/3. (b) Use the result of this calculation and symmetry arguments to find the probability of finding
the particle between x = L/3 and x = 2L/3. Do not re-evaluate the integral. (c) What If? Compare the result
of part (a) with the classical probability.
L3
(a)
The probability is
P
=
∫
0
L3
2
ψ dx
=
∫
0
2
2
 π x
sin 2  =
dx
 L 
L
L
2π x 
x 1
P=
 L − 2π sin L 
L3
0
L3
∫
0
2π x 
1 1
 2 − 2 cos L  dx
2π   1
3
1 1
sin  =
=
−
=0.196 .
 −

3 2π
3   3 4π 
L
.
2
Thus, the probability of finding the particle between
2L
and x = L is the same 0.196. Therefore, the
x=
3
L
2L
probability of finding it in the range ≤ x ≤
is
3
3
P=
1.00 − 2( 0.196) =
0.609 .
(b)
The probability density is symmetric about x =
(c)
FIG. P41.21(b)
Classically, the electron moves back and forth with constant speed between the walls,
and the probability of finding the electron is the same for all points between the walls.
Thus, the classical probability of finding the electron in any range equal to one-third of
the available space is Pclassical =
1
.
3
4. A particle in an infinitely deep square well has a wave function given by
ψ 2 (x ) =
2
 2πx 
sin 

L  L 
for 0 ≤ x ≤ L and zero otherwise. (a) Determine the expectation value of x. (b) Determine the probability of
finding the particle near L/2, by calculating the probability that the particle lies in the range 0.490L ≤ x ≤
0.510L. (c) What If? Determine the probability of finding the particle near L/4, by calculating the
probability that the particle lies in the range 0.240L ≤ x ≤ 0.260L. (d) Argue that the result of part (a) does
not contradict the results of parts (b) and (c).
L
(a)
L
4π x 
2
2
 2π x 
1 1
x
x sin 2  =
x  − cos
=
 dx
 dx
L
L
L 
 L 
2 2
0
0
∫
∫
1 x2
x =
L 2
L
−
0
1 L2
L 16π 2
4π x
4π x 
L
 4π x
 L sin L + cos L  =
2

0
L
0.510L
(b)
Probability
=
4π x 
2
1 L
 2π x 
1
dx  x −
sin 2 
sin
=
L
L 4π
L  0.490L
 L 
L
0.490L
0.510L
∫
0.020 −
Probability =
1
5.26 × 10−5
( sin 2.04π − sin 1.96π ) =
4π
4π x 
x 1
Probability  −
=
sin
3.99 × 10−2

L  0.240L
 L 4π
0.260L
(c)
(d)
In the n = 2 , it is more probable to find the particle either
L
3L
than at the center, where the
near x = or x =
4
4
probability density is zero.
Nevertheless, the symmetry of the distribution means that
L
the average position is .
2
5. An electron is trapped somewhere in a molecule which is 7.13 nm long. Calculate the
minimum kinetic energy of the electron.
h
h
6.33x10−34 Js
∆x∆p ≥
→=
∆p
≥
= 7.06 x10−27 kgm / s
−9
4π
4π∆x 4π (7.13x10 m )
p 2 (7.06 x10−27 kgm / s ) 2
=
K =
= 2.74 x10−23=
J .0002eV
−31
2m
2(9.11x10 kg )
6.
The wave function for a particle is
ψ (x ) =
a
π x + a2
(
2
)
for a > 0 and –∞ < x < +∞. Determine the probability that the particle is located somewhere between x = –a
and x = +a.
a
P41.2
a
a
 a   1
 x
Probability=
P ∫=
dx     tan −1  
ψ ( x)
∫ π x 2=
2





a
π a
+a
−a
−a
P
=
2
(
)
1 π  π  
 tan −1 1 − tan −1 ( −1=
−− =
) 


π
π  4  4  
1
a
−a
1
2
7. The wave function of a particle is given by
ψ (x ) = A cos(kx ) + B sin (kx )
where A, B, and k are constants. Show that ψ is a solution of the Schrödinger equation, assuming the
particle is free (U = 0), and find the corresponding energy E of the particle.
8. Consider a particle moving in a one-dimensional box for which the walls are at x = –L/2 and x = L/2. (a)
Write the wave functions and probability densities for n = 1, n = 2, and n = 3. (b) Sketch the wave
functions and probability densities.
A particle is confined in a rigid one-dimensional box of length L.
Visualize:
Solve: (a) ψ (x) is zero because it is physically impossible for the particle to be there because the box is rigid.
(b) The potential energy within the region −L/2 ≤ x ≤ L/2 is U(x) = 0 J. The Schrödinger equation in this region is
d 2ψ ( x )
2m
=
− 2 Eψ ( x ) =
− β 2ψ ( x )
dx 2

where β = 2mE  2 .
(c) Two functions ψ (x) that satisfy the above equation are sinβ x and cosβ x . A general solution to the Schrödinger
equation in this region is
ψ (x) = Asinβ x + Bcosβ x
where A and B are constants to be determined by the boundary conditions and normalization.
(d) The wave function must be continuous at all points. ψ = 0 just outside the edges of the box. Continuity requires that ψ
also be zero at the edges. The boundary conditions are ψ (x = −L/2) = 0 and ψ (x = L/2) = 0.
(e) The two boundary conditions are
 βL 
 βL 
 βL 
 βL 
− A sin 
0
 + B cos  −
=
 + B cos 
=
2
2
2






 2 
ψ ( −L 2) =
A sin  −
 βL 
 βL 
 + B cos 
 =0
 2 
 2 
ψ ( L 2 ) = A sin 
These are two simultaneous equations. Unlike the boundary conditions in the particle in a box problem of Section 41.4,
there are two distinct ways to satisfy these equations. The first way is to add the equations. This gives
 βL 
2 B cos 
 = 0 ⇒ B = 0 ⇒ ψ (x) = Asinβ x
 2 
To finish satisfying the boundary conditions,
 βL 
sin 
 = 0 ⇒ β L = 2π , 4π , 6π , … = 2nπ with n = 1, 2, 3, …
 2 
With this restriction on the values of β , the wave function becomes ψ ( x ) = A sin ( 2nπ x L ) . Using the definition of β
from part (b), the energy is
2nπ )  2
(=
2
En
=
2
2mL
( 2n )
2
h2
8mL2
n = 1, 2, 3, …
The second way is to subtract the second equation from the first. This gives
 βL 
−2 A cos 
0 ⇒ A = 0 ⇒ ψ (x) = Bcosβ x
=
 2 
To finish satisfying the boundary conditions,
 βL 
cos 
 = 0 ⇒ β L = π , 3π , 5π , … = (2n – 1)π
 2 
n = 1, 2, 3, …
With this restriction on the values of β , the wave function becomes
=
ψ ( x ) B cos ( 2 ( n − 1) π x L ) . Using the definition of
β , the energy is
=
En
2
2
( 2n − 1) π 2=
( 2n − 1)
2
2mL
2
h2
8mL2
n = 1, 2, 3, …
Summarizing this information, the allowed energies and the corresponding wave functions are
2

 ( 2n − 1) π x 
2 h
E1 , 9 E1 , 25E1 , 
=
( 2n 1)
 B cos 
 En =−
L
8mL2



ψ ( x) = 
2
x
2 h
 sin  ( 2n ) π=
2
4 E1 , 16 E1 , 36 E1 , 
=
A
E
n
(
)


n

8mL2
 L 

where E1 = h2/8mL2.
(f) The results are actually the same as the results for a particle located at 0 ≤ x ≤ L. That is, the energy levels are the same
and the shapes of the wave functions are the same. This has to be, because neither the particle nor the potential well have
changed. All that is different is our choice of coordinate system, and physically meaningful results can’t depend on the
choice of a coordinate system. The new coordinate system forces us to use both sines and cosines, whereas before we
could use just sines, but the shapes of the wave functions in the box haven’t changed.
9. An electron is contained in a one-dimensional box of length 0.100 nm. (a) Draw an energy-level
diagram for the electron for levels up to n = 4. (b) Find the wavelengths of all photons that can be emitted
by the electron in making downward transitions that could eventually carry it from the n = 4 state to the n
= 1 state.
(a)
We can draw a diagram that parallels our treatment of standing
mechanical waves. In each state, we measure the distance d from one node to
another (N to N), and base our solution upon that:
h
λ
Since
and λ =
dN to N =
p
2
h
h
.
p =
=
λ 2d
2

−34
p2
1  6.626 × 10 J⋅ s 
h2
Next,
K =
.
=
=
2me 8med2 d2  8 9.11 × 10−31 kg 


(
Evaluating,
K=
(
)
)
6.02 × 10−38 J⋅ m 2
d2
In state 1,
3.77 × 10−19 eV ⋅ m 2
.
d2
K1 = 37.7 eV .
=
d 1.00 × 10−10 m
In state 2,
=
d 5.00 × 10−11 m
K=
−11
K 2 = 151 eV .
In state 3,
=
d 3.33 × 10
m
K 3 = 339 eV .
In state 4,
=
d 2.50 × 10−11 m
K 4 = 603 eV .
FIG. P41.5
10. A particle is described by the wave function
(a) Determine the normalization constant A. (b) What is the probability that the particle will be found
between x = 0 and x = L/8 if its position is measured?
∞
(a)
∫ψ
2
dx = 1 becomes
−∞
L4
A
2
 2π x 
 L
cos2 
A2 
 dx =
L 
 2π

−L 4
∫
or A 2 =
(b)
 4π x  
 π x 1
 L  π 
+ sin 
=
=
1
A2 



 L 4
 2π   2 
 L  −L 4
L4
2
4
and A =
.
L
L
The probability of finding the particle between 0 and
L8
∫
0
L8
ψ dx =A 2
2
1 1
 2π x 
= 0.409
 dx = +
4 2π
L 
∫ cos 
2
0
L
is
8