Download (8.04) Spring 2005 Solutions to Problem Set 1

Massachusetts Institute of Technology
Quantum Mechanics I (8.04) Spring 2005
Solutions to Problem Set 1
By Kit Matan
1. Photon energy scale I (20 points)
According to quantum mechanics, electromagnetic radiation of frequency ν can be regarded as
consisting of photons of energy hν, where h=6.626 × 10−34 J·s is Planck’s constant.
(a) What is the frequency range of visible photons (400 nm to 700 nm)? What is the energy
range of visible photons (both in J and in eV)?
§ Frequency range of visible photons:
c
λ
2.998 × 108
λ = 400 nm =⇒ ν =
= 7.50 × 1014 Hz
−9
400 × 10
2.998 × 108
λ = 700 nm =⇒ ν =
= 4.28 × 1014 Hz
700 × 10−9
Energy range of visible photons.
In Joule:
ν=
(1)
(2)
(3)
λ = 400 nm =⇒ E = hν = 6.626 × 10−34 · 7.495 × 1014 = 4.97 × 10−19 J
(4)
λ = 700 nm =⇒ E = hν = 6.626 × 10−34 · 4.283 × 1014 = 2.84 × 10−19 J
(5)
In eV:
You can either use h = 4.136 × 10−15 eV · s or 1 eV = 1.602 × 10−19 J .
λ = 400 nm =⇒ E = hν = 4.136 × 10−15 · 7.495 × 1014 = 3.10 eV
(6)
λ = 700 nm =⇒ E = hν = 4.136 × 10−15 · 4.283 × 1014 = 1.77 eV
(7)
(b) How many photons per second does a low power (1 mW) He-Ne laser (λ=633 nm) emit?
A cell phone that emits 0.4 W of 850 MHz radiation? A microwave oven operateing at 2.45
GHz generating a microwave power of 750 W? How many photos of the latter frequency
have to be absorbed to heat up a glass of water (0.2 L, heat capacity of water 4.18 kJ
kg−1 K−1 ) by 10◦ C?
§ He-Ne laser: Energy of a single photon of λ=633 nm is:
E=
hc
6.626 × 10−34 · 2.998 × 108
=
= 2.96 × 10−19 J .
λ
633 × 10−9
1
(8)
A number of photons per second the He-Ne laser emits, Nlaser , is:
(Note: 1 mW = 10−3 J·s−1 .)
Nlaser =
10−3
= 3.38 × 1015 photons/ s .
2.96 × 10−19
(9)
Cell phone: Energy of a single photon of ν=850 MHz is:
E = hν = 6.626 × 10−34 · 850 × 106 = 5.63 × 10−25 J .
(10)
A number of photons per second the cellphone emits, Ncellphone , is:
Ncellphone =
0.4
= 7.10 × 1023 photons/s .
5.63 × 10−25
(11)
Microwave: Energy of a single photon of ν= 2.45 GHz is:
E = hν = 6.626 × 10−34 · 2.45 × 109 = 1.62 × 10−24 J .
(12)
A number of photons per second the microwave emits, Nmicrowave , is:
Nmicrowave =
750
= 4.62 × 1026 photons/s .
1.62 × 10−24
(13)
Energy required to heat 0.2 L of water by 10◦ C is:
Note: 1 L of water weighs 1 kg.
Eheat = CmT = 4.18 × 103 · 0.2 · 10 = 8.36 × 103 J .
Nheat =
Eheat
8.36 × 103
=
= 5.16 × 1027 photons .
Ephoton
1.62 × 10−24
(14)
(15)
(c) At a given power of an electromagnetic wave, do you expect a classical wave description
to work better for radiofrequencies, or for X-rays? Why? At what He-Ne laser power do
you expect quantum effects to become important?
§ We expect a classical wave description to work better for radiofrequencies. The classical
electromagnetic description of photons works fine when a number of photons is large.
However, this description breaks down when we try to describe a single photon. At a
given power of an electromagnetic wave, a number of photons for radiofrequencies in the
detection window is much larger than that for X-rays. Therefore, the wave description of
photons is adequate for radiofrequencies but is not adequate for X-rays. We can estimate
a ratio of a number of photons for radiofrequencies and for X-rays at a given power (i.e.
1 W). Energy of radiofrequencies is about 10−7 eV and that of X-rays is about 104 eV.
R=
Nradio
104
≈ −7 = 1011 .
NX−ray
10
2
(16)
Therefore, at a given power, for every X-ray photon, there are about 1011 radiofrequency
photons.
Assume that a relaxation time of a photon detector is about 1 ps (10−12 s). Our detector
can detect a single photon if it arrives at the detector at a rate of 1 photon per 1 ps. This
time scale determines our detection-window time scale. Therefore, the power of the He-Ne
laser in which we expect quantum effect to become important is:
Ephoton
2.96 × 10−19
P =
=
= 2.96 × 10−7 W .
−12
1 ps
10
(17)
(d) An adapted human eye (person has spent 30 min in the dark) can see 1 ms flashes of
power 4×10−14 W at 510 nm with 60% reliability. Assuming that 10% of the incident
power reaches the retina, how many photons at the receptors generate the signal that the
test person recognize as a flash of light?
§ Assuming that 10% of the incident power reaches the retina, a number of photons that
reach the retina, N, in 1 ms is:
1
P
4 × 10−14
· t = 0.1 ·
· 10−6 = 10.26 photons .
10 Ephoton
6.626 × 10−24 · 3 × 108 /(510 × 10−9 )
(18)
1
However, with only 60% reliability, the human eye would need 0.6 of the total number of
photons in order to tinkle a neuron with 100% reliability. Therefore, a number of photons
the human eye needs to generate a signal , N0 , is:
N=
N0 =
1
N = 17.10 photons .
0.6
(19)
2. Photon energy scale II (20 points)
(a) The thermal energy scale is kB T, where kB =1.38×10−23 J/K is Boltzmann’s constant, and
T is the absolute temperature. What energy does room temperature (20◦ C) correspond
to? What would be the frequency and wavelength of the corresponding photons? Is it
reasonable that a hot body starts to grow around 1000◦ C?
§ Room temperature:
ET = (274 + 20) · 1.38 × 10−23 = 4.04 × 10−21 J .
(20)
λ and ν of the corresponding photons are:
hc
=
ET
ET
=
ν=
h
λ=
6.626×10−34 ·2.998×108
4.04×10−21
4.04×10−21
6.626×10−34
= 49.2 µm
(21)
= 6.10 × 1012 Hz .
(22)
◦
The corresponding energy of a hot body around 1000 C is about 1.76×10−20 J, which
corresponds to photons with λ=11.3 µm. This wavelength is much higher than that of
3
Figure 1: Blackbody radiation spectrum
visible light, so one would not expect this hot body to glow. However, if we look at
◦
◦
◦
blackbody radiation at 20 C and 1000 C, we can see that the hot body of 1000 C starts
to radiate red light while the radiation intensity of the room-temperature body stays at
very low value as shown in Fig 1. Therefore, it is reasonable for a hot body to glow red.
(b) What is the photon flux (rate of arriving photons per unit area) at 1 m distance from a 60
W light bulb, if you assume that the bulb conversion efficiency (electrical power to light)
is 10% and take the photon wavelength as 500 nm?
§ Energy of a single photon with λ=500 nm is
Ephoton =
hc
6.626 × 10−34 · 2.998 × 108
=
= 3.97 × 10−19 J .
λ
500 × 10−9
Suppose that 10% of electric power is converted into light. Plight =
Therefore, the rate in which this light bulb emits photons is,
n=
1
P
10 electric
6
Pphoton
=
= 1.51 × 1019 photons/s .
Ephoton
3.97 × 10−19
(23)
= 6 W.
(24)
The photon flux, S, is:
S=
n
1.51 × 1019
n
=
=
= 1.20 × 1018 photons/(m2 · s) .
A
4πr2
4π · 12
4
(25)
(c) A photodiode measure light power by converting incident photons into electron-hole pairs,
such that the electron current is proportional to the incident light power. The quantum
efficiency is defined as the probability than as incident photon generates an electron. If a
typical photodiode has a responsivity of 0.5 A/W for infrared light at 850 nm, what is the
quantum efficiency of the device? If the quantum efficiency is independent of frequency,
what responsivity do you expect for blue light at 400 nm?
§ First, we need to calculate a number of photons emitted per second.
P
1
=
= 4.28 × 1018 photons/s .
−34
8
−9
hc/λ
6.626 × 10
· 2.998 × 10 /(850 × 10 )
(26)
With a responsivity, ζ, 0.5 A/W, 0.5 A is generated from 1 W of light. Therefore, the
quantum efficiency, ξ, is:
Nphoton =
ξ=
0.5
Nelectron
−19
= 1.602×10 18 = 0.73 electrons/photon .
Nphoton
4.28 × 10
(27)
Now suppose that the quantum efficiency is independent of frequency. For blue light at
400 nm, the responsivity is:
ζ =ξ·
I
1.602 × 10−19
= 0.73 ×
= 0.32 A/W .
P
6.626 × 10−34 · 2.998 × 108 /(400 × 10−9 )
(28)
3. The spacing of atoms in a sodium chloride (15 points)
The crystal lattice of sodium chloride has ions of sodium (Na+ ) and ions of chloride (Cl− )
arranged alternately in a simple cubic array. Thus the crystal volume per ion is d3 , where d is
the distance between the centers of adjacent ions. The molecular weight of NaCl is 58.45 and
its density is 2.164 g/cm3 .
(a) Calculate the volume of 1 mole of crystalline NaCl.
§ We know from the molecular weight that 1 mole of NaCl weighs 58.45 g. Therefore, from
the density, s, we have:
m
58.45
V =
=
= 27.01 cm3 .
(29)
s
2.164
(b) Calculate the spacing d of the ions in the lattice.
§ First, calculate a volume of ion, v
v = d3 =
1 V
27.01
=
= 4.49 × 10−23 cm3 .
23
2 NA
6.02 × 10
Note that there are two ions per unit cell.
Therefore, we obtain
d = v 1/3 = 2.82 × 10−8 cm = 2.82 Å .
5
(30)
(31)
4. The electric field of a single photon (20 points)
Using or refreshing your E&M knowledge about the Poynting vector, calculate the electric field
amplitude for a single photon of wavelength 850 nm contained in a 100 ns long pulse of area
(100 µm)2
§ Poynting vector, S, is the energy transported by the electromagnetic field per unit time per
unit area, or S is the energy flux density,
S=
1
(E × B) .
µ0
(32)
For electromagnetic wave, B=E/c, where E=E0 e−iωt . Therefore, hSi= 2µ10 c E0 2 , where a factor
of 21 is from taking time average.
For a single photon for wavelength 850 nm in a 100 ns pulse of area (100µ m)2 , the energy flux
density is:
hc 1
λ t·A
6.626 × 10−34 · 2.998 × 108
1
=
−9
−9
850 × 10
100 × 10 · (100 × 10−6 )2
hSi = 2.34 × 10−4 J/(m2 · s) .
hSi =
(33)
(34)
(35)
Therefore,
E0 = (2µ0 chSi)1/2 = (2 · 4π × 10−7 · 2.998 × 108 · 2.34 × 10−4 )1/2 = 0.42 V/m .
(36)
5. Radiative collapse of the classical atom (25 points)
The classical atom has a stability problem. We model the hydrogen atom as an electron in a
classical circular orbit about a proton. We know that a non-relativistic accelerating electric
q 2 a2
= −P = − 32 4πε
charge radiates energy at a rate given by the Larmor formula: dE
3 , where q
dt
0 c
is the electric charge and a is the magnitude of the acceleration.
(a) Show that the energy lost per revolution is small compared to the electron’s kinetic energy.
§ For an electron circling the nucleus, the acceleration is given by
F = ma =
1 e2
4πε0 r2
=⇒
a=
1 e2
4πε0 mr2
(37)
The electron’s velocity v can be related to the radius from the relation
1 e2
1 e2
mv 2
2
=
=⇒
v
=
r
4πε0 r2
4πε0 mr
so the expression for the energy of electron in the orbit of radius r is
F =
1 e2
1 e2
1
=−
E = mv 2 −
2
4πε0 r
4πε0 2r
6
(38)
(39)
Combining the equations above we get
dE
1 e2 dr
2 1 e2
=
=
−
dt
4πε0 2r2 dt
3 4πε0 c3
Ã
1 e2
4πε0 mr2
!2
(40)
For a circular orbit, one period is given by
s
2rπ
2rπ
mr3
T =
=q
=
2π
4πε
0
1 e2
v
e2
(41)
4πε0 mr
Since the power is approximately constant during one period, total radiated power is given
by
ZT
E1 =
0
dE
dE
2 1 e2
dt ≈
×T =
dt
dt
3 4πε0 c3
Ã
1 e2
4πε0 mr2
!2
s
√
Ã
!5/2
mr3
16 2π
1 e2
×2π 4πε0 2 = 3 3/2
e
3c m
4πε0 2r
(42)
to be compared with the kinetic energy
mv 2
1 e2
EK =
=
2
4πε0 2r
The ratio of the two is given by
√ Ã
!3/2
µ
¶
E1
Eorbit 3/2
16 2π
1 e2 /2r
=
= 23.7
EK
3
4πε0 mc2
Erest
(43)
(44)
For a typical atom, Eorbit is on the order of 10 eV while the rest energy of the order of
105 eV so the energy ratio
√ factor gives a suppression of around 6 orders of magnitude,
while the first factor 16 2π/3 is at most of order of 10−1 to 101 making the energy ratio
on the order of 10−5 to 10−7 . Hence, it is a good approximation to regard the orbit as
circular at any instant.
For hydrogen, Eorbit = 13.6 eV in the ground state and the rest energy of electron, Erest ,
is 0.511 MeV, so
µ
¶3/2
E1
13.6
= 23.7 ·
= 3.253 × 10−6 .
(45)
EK
0.511 × 106
(b) Using the typical size for the atom (1 Å) and the nucleus (10−5 Å), calculate how long it
would take for the electron to spiral into the proton.
§ From equation (40) we get the differential equation for the time change of the radius
µ
dr
1
=−
dt
4πε0
¶2
4e4
.
3m2 r2 c3
(46)
Integrating this we get the expression for the time it takes for the atom to collapse
τ=
Z τ
0
dt = −(4πε0 )2
2 3 ³
´
m2 c3 Z rf 2
3
3
2m c
−
r
r
3r
dr
=
(4πε
)
0
f
i
4e4 ri
4e4
7
(47)
For a typical atom ri ≈ 10−10 m is five orders of magnitude larger than rf ≈ 10−15 m and
rf can be neglected so the final formula is
τ = (4πε0 )2
Using the formula α =
e2
/(h̄c)
4πε0
m2 c3 3
r
4e4 i
(48)
and mc2 = 0.511 MeV again we get
2
(0.511 MeV)
³
´3
2.998 × 108 m s−1
m2 c 3
τ =
r =
4 h̄2 α2 i
4 ∗ (6.58265 × 10−16 eV s)2
τ = 1.05 × 10−10 s .
³
1
137.0362
∗ 10−10 m
´3
(49)
(50)
8