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Physics 202H - Introductory Quantum Physics I
Final Exam - Solutions
Fall 2004 Name:
Monday 2004/12/20 Student Number:
This examination paper includes 4 pages and 11 questions. You are responsible for ensuring that your copy of the paper is complete. Bring any discrepancy to the attention of your invigilator.
Special Instructions:
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Of possible use: ∫ sin ax dx = − cos^ ax a
cos ax dx = sin^ ax ∫ a sin^2 ax dx =
x 2 −^
sin 2ax 4 a
cos^2 ax dx =
x 2 +
sin 2ax ∫^4 a x sin ax dx =
sin ax a^2 −^
x cos ax a
x cos ax dx =
cos ax a^2 +^
x sin ax a ∫ x^2 sin ax dx =
2 x a^2 sin^ ax^ +
a^3 −^
x^2 a
cos ax
x^2 cos ax dx =
2 x a^2 sin^ ax^ +
a^3 −^
x^2 a
cos ax
∫ x^2 cos^2 bx dx =
4 b^3 x^3 + 3(2b^2 x^2 − 1) sin 2bx + 6bx cos 2bx 24 b^3 ∫ (^) ∞
0
e−ax^2 dx =^12
π a x^3 − y^3 = (x − y)(x^2 + xy + y^2 )
c = 2. 99792458 × 108 m/s e = 1. 602176462 × 10 −^19 Coul h = 6. 626068 × 10 −^34 J · s = 4. 1356668 × 10 −^15 eV · s ℏ = 1. 05457148 × 10 −^34 J · s = 6. 58211814 × 10 −^16 eV · s me = 9. 10938188 × 10 −^31 kg = 0.510998903 MeV/c^2 mp = 1. 67262158 × 10 −^27 kg = 938.271996 MeV/c^2 mn = 1. 6749286 × 10 −^27 kg = 939.565630 MeV/c^2 σ = 5. 670400 × 10 −^8 W/m^2 ◦K^4 λc = h/(mec) = 2. 4263106 × 10 −^12 m (4π 0 )−^1 = 8. 988 × 109 N · m^2 /Coul^2
There is only this sentence on this page.
- A region of space has a potential step such that particles have a wave function given by
Ψ(x, t) =
√^5 a 2
ei(k^1 x−Et/ℏ)^ + √^3 a 2
ei(−k^1 x−Et/ℏ), x < 0 ,
√^8 a 2
ei(k^2 x−Et/ℏ), x > 0.
The incident particles, initially at x 0, are initially travelling in the positive x direction. (a) What fraction of the incident particles will be reflected? [5] Solution: The given solution matches the format we have used in the course with wave amplitudes of A, B, and C. The reflection amplitude R will give us the fraction of incident particles reflected:
R = v^1 B
∗B
v 1 A∗A
= B
∗B
A∗A
√^3 a 2
√^3 a 2
( (^5) a √ 2
) ( (^5) a √ 2
Of the incident particles, 36% or 9/25 of them will be reflected. (b) What is k 2 /k 1? [5] Solution: Since 9/25 of the incident particles will be reflected, 16/25 or 64% will be transmitted. Using the relationship for the transmission coefficient T we have
T = 1 − R = v^2 C
∗C
v 1 A∗A
= k^2 C
∗C
k 1 A∗A
1 − 9 25
k 2
√^8 a 2
√^8 a 2
k 1
( (^5) a √ 2
) ( (^5) a √ 2
= k^2 (8)(8) k 1 (5)(5)
= k^264 k 125 16 = 64
k 2 k 1 k 2 k 1 =
Alternatively, one could use the continuity of the first derivative of the eigenfunction at x = 0 to get [ dψ 1 (x) dx
]
x=
[
dψ 2 (x) dx
]
x= k 1 √^5 a 2
− k 1 √^3 a 2
= k 2 √^8 a 2 2 k 1 = 8k 2 k 2 k 1 =
- In a photoelectric effect experiment on a certain metal, it is observed that incident light of wavelength 413 nm causes electrons to be ejected from the metal’s surface with a maximum kinetic energy of 3. 2 × 10 −^19 J. What is the longest wavelength of light that will eject electrons from this metal? [5] Solution: The photon has energy of hν = hc/λ, so the kinetic energy is related to the work function of the material by K = hν − w 0
= hc λ
− w 0
w 0 = hcλ − K
=
(6. 626068 × 10 −^34 J · s)(2. 99792458 × 108 m/s) (413 nm) −^ (3.^2 ×^10
− 19 J)
= (4. 8098 × 10 −^19 J) − (3. 2 × 10 −^19 J)
w 0 = 1. 6098 × 10 −^19 J The longest wavelength will occur when the kenetic energy of the emitted electron is a mini- mum, zero. 0 = hνmin − w 0
=
hc λmax^ −^ w^0 w 0 =
hc λmax λmax =
hc w 0 = (6.^626068 ×^10
− (^34) J · s)(2. 99792458 × 108 m/s) (1. 6098 × 10 −^19 J) = 1. 23397 × 10 −^6 m ≈ 1. 23 μm = 1230 nm. The longest wavelength of light that will eject electrons from this metal is about 1230 nm
- A particle is in an region of space where it has a wave function given by
Ψ(x, t) =
0 , x < 0 , Ae−κxeiEt/ℏ, x > 0. What is the value of A? [5] Solution: We need to normalize the wave function by setting the integral over all space of P (x, t)dx = Ψ∗(x, t)Ψ(x, t)dx to 1.
1 =
−∞
P (x, t)dx =
−∞
Ψ∗(x, t)Ψ(x, t)dx
0
Ae−κxe−iEt/ℏ
Ae−κxeiEt/ℏ
dx = A^2
0
e−^2 κxdx
= A^2
[
e−^2 κx − 2 κ
]+∞
0
= A^2
[
e−^2 κ(∞) − 2 κ
− e
− 2 κ(x) − 2 κ
]
= A^2
[
2 κ
]
1 = A
2 2 κ
=⇒ A =
2 κ
For Ψ(x, t) to be properly normalized, A must have a value of
2 κ.
- A particle moves in an infinite potential well described by
V (x) =
0 , |x| > a/ 2 , ∞, |x| ≤ a/ 2. The eigenfunctions are of the form ψn(x) = An cos (knx), or ψn(x) = Bn sin (knx), depending on the value of n. For n = 3, ψ 3 (x) = (
2 /a) cos (3πx/a) for |x| ≤ a/2 and ψ 3 (x) = 0 for |x| > a/2. (a) What are the expectation values of x and x^2 in the n = 3 state. [5] Solution: The expectation value of x is calculated by integrating ψ 3 ∗ xψ 3 dx over all space, but since the wave function is zero outside of ±a/2 we only need to integrate in this region.
x =
∫ (^) +a/ 2
−a/ 2
ψ∗(x)xψ(x)dx
∫ (^) +a/ 2
−a/ 2
a cos
3 πx a
x
a cos
3 πx a
dx
=^2
a
∫ (^) +a/ 2
−a/ 2
x cos^2
3 πx a
dx u =^3 πx a
, du =^3 π a
dx
= (^92) πa 2
∫ (^) +3π/ 2
− 3 π/ 2
︷︸︸︷odd u
︷ ︸︸ ︷^ even ︸ ︷︷^ cos 2 u︸ odd
du = 0
x = 0.
The expectation value of x^2 is calculated by integrating ψ∗ 3 x^2 ψ 3 dx over all space, but since the wave function is zero outside of ±a/2 we only need to integrate in this region.
x^2 =
∫ (^) +a/ 2
−a/ 2
ψ∗(x)x^2 ψ(x)dx
∫ (^) +a/ 2
−a/ 2
a
cos
3 πx a
x^2
a
cos
3 πx a
dx
=^2 a
∫ (^) +a/ 2
−a/ 2
x^2 cos^2
3 πx a
dx u =^3 πxa , du =^3 aπ dx
= 2 a
2 27 π^3
∫ (^) +3π/ 2
− 3 π/ 2
u︸ 2 cos︷︷ 2 u︸ even
du = 4 a
2 27 π^3
∫ (^) +3π/ 2
0
u^2 cos^2 u du
4 a^2 27 π^3
[
4 b^3 x^3 + 3(2b^2 x^2 − 1) sin 2bx + 6bx cos 2bx 24 b^3
]+3π/ 2
0
b = 1
= a
2 (27)(6)π^3
[
4 x^3 + 3(2x^2 − 1) sin 2x + 6x cos 2x
]+3π/ 2 0
= a
2 (27)(6)π^3
[
3 π 2
3 π 2
− 1) sin 2
3 π 2
3 π 2
cos 2
3 π 2
)]
= a
2 (27)(6)π^3
[
3 π 2
3 π 2
)]
= a
2 (27)(6)π^3
[
27 π^3 2 −^9 π
]
= a
2 6
3 π^2
= a
2 12
− a
2 18 π^2
≈ (0.0777)a^2
(b) What are the expectation values of p and p^2 in the n = 3 state. [5] Solution: To calculate the expectation value for momentum, we need to use the mo- mentum operator −iℏ(∂/∂x).
p =
∫ (^) +a/ 2
−a/ 2
ψ∗(x)
−iℏ ∂ ∂x
ψ(x)dx
∫ (^) +a/ 2
−a/ 2
a
cos
3 πx a
−iℏ ∂ ∂x
) √^
a
cos
3 πx a
dx
a
∫ (^) +a/ 2
−a/ 2
cos
3 πx a
iℏ 3 π a sin
3 πx a
dx u =
3 πx a ,^ du^ =
3 π a dx
= −iℏ^2 a
∫ (^) +3π/ 2
− 3 π/ 2
︷ ︸︸ ︷even cos u
︷︸︸︷^ odd ︸ ︷︷^ sin u︸ odd
du = 0
p = 0.
To calculate the expectation value for p^2 we need to use −ℏ^2 (∂^2 /∂x^2 ).
p^2 =
∫ (^) +a/ 2
−a/ 2
ψ∗(x)
−ℏ^2 ∂
2 ∂x^2
ψ(x)dx
∫ (^) +a/ 2
−a/ 2
a
cos
3 πx a
−ℏ^2 ∂
2 ∂x^2
) √^
a
cos
3 πx a
dx
a
∫ (^) +a/ 2
−a/ 2
−ℏ^2
− 9 π^2 a^2
cos^2
3 πx a
dx
=^18 π
a^3
∫ (^) +a/ 2
−a/ 2
cos^2
3 πx a
dx u =^3 πx a
, du =^3 π a
dx
6 πℏ^2 a^2
∫ (^) +3π/ 2
− 3 π/ 2
cos︸ ︷︷ ︸^2 u even
du =
12 πℏ^2 a^2
∫ (^) +3π/ 2
0
cos^2 u du
=^12 πℏ
2 a^2
[
u 2
]+3π/ 2
0
=^12 πℏ
2 a^2
[
3 π 4
]
p^2 =
9 π^2 ℏ^2 a^2 =
3 πℏ a
3 h 2 a
iii. What angle φ does the path of the scattered electron make with the direction of the incident photon? [5] Solution: The kinetic energy of the recoil electron is less than 10% of its rest mass energy, so non-relativistic relations will be fairly accurate. The momentum of the electron in the y direction, perpendicular to the initial photon’s direction of travel, must be equal and opposite to the momentum of the scattered photon in the y direction, since there is zero total momentum in the y direction. Thus
pipy = 0 = prey + pspy prey = −pspy pre sin φ = −psp sin θ √ 2 meEre sin φ = −
Esp c sin^ θ sin φ = − √ Esp 2 mec^2 Ere
sin θ
= −
(120 keV) √ 2(510.998903 keV)(40 keV)
sin (1.635422407)
= − 0. 5922686084 φ = sin−^1 (− 0 .5922686084) = − 0 .6338714937 rad ≈ − 36. 3 ◦.
The electron is scattered about 36.3◦from the path of the incident photon, with the negative sign indicating that the electron is scattered on the one side of the x axis as defined by the direction of the incident photon, while the scattered photon is on the opposite side.
(b) Electrons are accelerated through an electric potential V and then fall on a pair of slits that have a separation of 100 nm. The resultant interference pattern indicates that the electrons have a wavelength of 1.0 nm. i. What is the momentum of one of the electrons? [5] Solution: The momentum is related to the de Broglie wavelength by
p =
h λ =
(6. 626068 × 10 −^34 J · s) (1. 0 × 10 −^9 m) = (6.^626068 ×^10
− (^25) kg · m/s) ≈ 6. 63 × 10 − (^25) kg · m/s.
The momentum of one of the electrons is about 6. 63 × 10 −^25 kg · m/s. ii. What is the accelerating electric potential V? [5] Solution: With a momentum of about 6. 63 × 10 −^25 kg · m/s and a mass of me =
- 10938188 × 10 −^31 kg, the electron’s velocity of v = p/m = 7. 274 × 105 m/s c, thus we can use non-relativistic relations. The kinetic energy gives us
K = eV =
p^2 2 me^. V = p
2 2 eme
= (6.^626068 ×^10
− (^25) kg · m/s) 2 2(1. 602176462 × 10 −^19 Coul)(9. 10938188 × 10 −^31 kg) = 1.5041201366 V ≈ 1 .50 V.
The accelerating electric potential is about 1.50 V.
iii. Using the uncertainty principle for the electrons after passing through the slits, what is the minimum spread in the electron’s momentum in the direction parallel to the plane of the slits and perpendicular to the average path of the electron? [5] Solution: Since the slits are separated by 100 nm, the uncertainty in the position in the direction parallel to the plane of the slits is ∆x = 100 nm. Thus
∆x∆px ≥ ℏ 2 ∆px ≥ ℏ 2∆x
= (1.^05457148 ×^10
− (^34) J · s) 2(1. 0 × 10 −^7 m) ≥ 5. 2728574 × 10 −^28 kg · m/s ≈ 5. 27 × 10 −^28 kg · m/s.
The minimum spread in the electron’s momentum in the direction parallel to the plane of the slits is about 5. 27 × 10 −^28 kg · m/s.
