Table Of Content2 Solutions
a constant. In this manner, given a Boltzmann distribution of energies at
equilibrium, any mode having hf much greater than k T is not excited.
B
2. Describe the photoelectric e⁄ect experiments and Einstein(cid:146)s explana-
tion for both the number and energy of the emitted electrons as a function
of the frequency of the incident light.
Solution: Shine light on a metal. Visible light, no electrons given o⁄.
Increase intensity, still no electrons given o⁄. With uv (ultraviolet light)
electrons given o⁄. Increase intensity of uv light, more electrons given o⁄,
but the maximum kinetic energy of the electrons is unchanged. Einstein
explanation was that light consists of photons. Radiation at frequency f is
made up of photons having energy hf. They can give up their energy to
the electrons in an all or nothing fashion. Therefore electrons are given o⁄
only if light having frequency f > W=h(W =work function of the metal)
is incident. Maximum kinetic energy =hf W:
(cid:0)
3. What were Bohr(cid:146)s postulates in his theory of the hydrogen atom?
Howdothesepostulatesexplainthespectrumandstabilityofthehydrogen
atom?
Solution: The postulates are (1) The electron orbits the nucleus in
circular orbits having discrete values of angular momentum given by
L=mvr =n~;
wheren=1;2;3;4,....and(2)Theelectronsradiateenergyonlywhenthey
(cid:147)jump(cid:148)from a larger to a smaller orbit. The frequency of the radiation
emitted is given by
E E
f = n2 (cid:0) n1;
n2;n1 h
Stability is explained since the electron only radiates in changing orbits. If
itisinitslowestenergystate,thereisnolowerenergystateforittogoto.
The (cid:133)rst postulate of quantized angular momentum plus Newton(cid:146)s Second
Lawledtoquantizedenergylevels.Thesecondpostulatethenexplainsthe
discrete spectrum.
4. Draw an energy level diagram for the hydrogen atom and indicate
on it the energy of the four lowest energy states. Also indicate the radius,
electron velocity and angular momentum of these states as given by the
Bohr theory.
Solution: r =n2a =5:29 10 11n2 (m) v =(cid:11)c=n=2:19 10 6=n
n 0 (cid:0) n (cid:0)
(cid:2) (cid:2)
(m/s) En = (1=2)(cid:11)2(mc2)=n2 = 13:6 eV=n2. For n = 1;L = ~;r =
(cid:0) (cid:0)
a0;v = (cid:11)c;E = 13:6 eV; for n = 2;L = 2~;r = 4a0;v = (cid:11)c=2;E = 3:4
(cid:0) (cid:0)
eV; for n = 3;L = 3~;r = 9a0;v = (cid:11)c=3;E = 1:51 eV; for n = 4;L =
(cid:0)
4~;r =16a0;v =(cid:11)c=4;E = 0:85 eV;
(cid:0)
5. Calculate the wavelength of the n = 7 to n = 2 and of the n = 3 to
n = 2 transitions in hydrogen. What frequency of radiation is needed to
excite the hydrogen atom from its n=1 to n=3 state? How much energy
is required to ionize a hydrogen atom from its n=2 state?
0.1 Chapter 1. Introduction 3
Solution: E = 13:6=49 = :278 eV E = 13:6=4 = 3:4eV E =
7 2 3
(cid:0) (cid:0) (cid:0) (cid:0)
13:6=9= 1:51eV.
(cid:0) (cid:0)
f =[ :278 ( 3:4)]eV=4:14 10 15eV s=7:54 1014Hz;(cid:21)=c=f =
72 (cid:0)
(cid:0) (cid:0) (cid:0) (cid:2) (cid:1) (cid:2)
3:97 10 7m=397nm:
(cid:0)
(cid:2)
f =[ 1:51 ( 3:4)]eV=4:14 10 15eV s=4:56 1014Hz;(cid:21)=c=f =
32 (cid:0)
(cid:0) (cid:0) (cid:0) (cid:2) (cid:1) (cid:2)
6:57 10 7m=657nm:
(cid:0)
(cid:2)
E E =12:1eV. This is the energy needed to excite the transition.
3 1
(cid:0)
f = (E E )=h = 2:92 1015Hz. To ionize the hydrogen atom, we
3 1
(cid:0) (cid:2)
must provide enough energy to give the electron a total energy of zero, in
which case it is no longer bound to the proton. If it starts in n = 2 with
E = 3:4eV, 3.4 eV of energy is needed.
3
(cid:0)
6. Radiation from the n=2 to n=1 state of hydrogen is incident on a
metal having a work function of 2.4 eV. What is the maximum energy of
the emitted electrons from the metal?
Solution: E E = 10:2eV =photon energy= hf. Maximum K.E.
2 1
(cid:0)
=hf W(work function)=10:2 2:4=7:8eV:
(cid:0) (cid:0)
7. What is the signi(cid:133)cance of the equation (cid:21) = h=p? What is meant
dB
bythewaveparticledualityofmatter?Whatdetermineswhenmatteracts
as a particle and when it acts as a wave?
Solution: (cid:21) = h=p implies that a wavelength can be associated with
dB
matter. The wave-particle duality refers to the fact that sometimes matter
acts as a wave and sometimes as a particle. It will act as a wave if its de
Broglie wavelength is larger than or of the order of a characteristic length
in the problem (such as beam size, obstacle size, slit width, orbit radius,
etc.)
8. Calculate the de Broglie wavelength for a particle of mass 1.0 gm
moving with a speed of 1:0 cm/year. Calculate the de Broglie wavelength
for the electron in the n=1 state of the Bohr atom.
Solution: 1cm/yr=3:2 10 10m/s.
(cid:0)
(cid:2)
(cid:21) = h=p = h=mv = 6:63 10 34Js=[10 3kg3:2 10 10m/s] = 2:1
dB (cid:0) (cid:0) (cid:0)
(cid:2) (cid:1) (cid:1) (cid:2) (cid:2)
10 11m.
(cid:0)
In the n=1 state v =(cid:11)c=c=137=2:2 106m/s:
(cid:2)
(cid:21) = h=p = h=mv = 6:63 10 34Js=[9:1 10 31kg2:2 106m/s] =
dB (cid:0) (cid:0)
(cid:2) (cid:1) (cid:2) (cid:1) (cid:2)
3:3 10 10m.
(cid:0)
(cid:2)
Note that the de Broglie wavelength is larger than the size of the orbit
which is 5:3 10 11m. The electron acts as a wave.
(cid:0)
(cid:2)
9. In general terms, discuss the measurement process in quantum me-
chanics. Why is it necessary to make measurements on a large number of
identically prepared systems to obtain (r;t)2? How does this di⁄er from
j j
Newtonian mechanics? Why is a single particle in a superposition state an
intrinsically quantum object?
Solution: Since the Schr(cid:246)dinger equation describes probability waves,
measuringoneparticlecannotgiveyoutheprobabilitydistribution (r;t)2.
j j
By measuring the position of the particle on a large number of identi-
callypreparedsystems,youcandeducewhat (r;t)2 is.Themeasurement
j j
4 Solutions
FIGURE 1. Problem 1.10
process in each case irreversibly changes the wave function of the particle.
In Newtonian mechanics, it is assumed that measurements can be made
on the position and velocity of a particle without signi(cid:133)cantly a⁄ecting its
motion. There is no classical analogue of a single particle in a superposi-
tion state. It simultaneously has the possibility of yielding a measurement
having di⁄erent values of a dynamic variable.
10. In a two-slit experiment, particles are sent into the apparatus one
particle at a time (see (cid:133)gure). How many detectors does a single particle
trigger?Iftheexperimentisrepeatedmanytimes,whatwillagraphofthe
number of counts in a detector vs detector position look like? Explain.
Solution: One particle, one detector. (r;t)2 just before the detectors
j j
willlooklikethetwo-slitinterferencepattern.Aftermanytrialsthispattern
will be recorded by the detectors.
11. When light is incident on a glass slab, some of the light is re(cid:135)ected.
This is a wave-like phenomenon (if a classical particle encounters a change
in potential, it simply slows down or speeds up with no re(cid:135)ection), even
though we are in a geometrical optics limit (neglect of di⁄raction). Why
does a wave-like e⁄ect occur in this case? Is there any connection of this
with the rainbow?
Solution:Re(cid:135)ectionisawave-likephenomenonsinceaclassicalparticle
that sees asteppotential will either speedup or slow down as it enters the
potential region. A wave-like e⁄ect occurs because the index of refraction
changesinadistancethatissmallcomparedwithawavelength.Ifwewere
tochangetheindexcontinuouslyoveradistancelargecomparedtoawave-
length, the re(cid:135)ection would be negligible. Similarly, although described by
geometricaloptics,therainbowisawavee⁄ect.Itoccursatananglewhere
the (cid:133)rst internal re(cid:135)ection gives rise to a maximum scattered signal. The
angle di⁄ers slightly for di⁄erent wavelengths, giving rise to the rainbow.
0.1 Chapter 1. Introduction 5
FIGURE 2. Problem 1.12.
12. The blackbody spectrum as a function of frequency u (f) can be
f
obtained using u (f)df = u(!)d! = 2(cid:25)u(2(cid:25)f)df. Plot [cu (f)=4(cid:25)] 1018
f f
(cid:2)
asafunctionoffrequencyforT =2:73 Kand(cid:133)ndthemaximumfrequency
(cid:14)
[cu (f)=4(cid:25)isthepowerperunitareaperunitfrequencyperunitsolidangle
f
- this corresponds to the (cid:135)ux incident on a detector per unit frequency
per steradian (sr)]. This Planck distribution corresponds to the cosmic
microwave background. What is the energy per unit volume of the cosmic
microwave background?
Solution: A plot of
cuf(f) = 1~(2(cid:25)f)3 1
4(cid:25) 2 (cid:25)2c2 e2(cid:25)~f=kBT 1
(cid:0)
(cid:0) (cid:1)
is shown. The maximum occurs at f 160 GHz, in the radio-frequency
(cid:25)
part of the spectrum. The energy per unit volume is equal to 4(cid:27)T4=c =
4:20 10 14 J/m3.
(cid:0)
(cid:2)
6 Solutions
0.2 Chapter 2. Mathematical Preliminaries
1. Evaluate ei(cid:25), ei(cid:25)=2, and e2:3i. If
eiaxe gx2=2
(x)= (cid:0) =u+iv =rei(cid:18);
x+ib
(cid:133)nd u;v;r;(cid:18), assuming that x;a;b;g;r;(cid:18) are real. Evaluate (x)2.
j j
Solution: ei(cid:25) =cos(cid:25)+isin(cid:25) = 1; ei(cid:25)=2 =cos((cid:25)=2)+isin((cid:25)=2)=i;
(cid:0)
e2:3i =cos(2:3)+isin(2:3)= 0:67+0:75i
(cid:0)
eiaxe gx2=2x ib eiaxe gx2=2(x ib)
(x) = (cid:0) (cid:0) = (cid:0) (cid:0)
x+ib x ib x2+b2
(cid:0)
[cos(ax)+isin(ax)]e gx2=2(x ib)
= (cid:0) (cid:0)
x2+b2
[xcos(ax)+bsin(ax)]+i[ bcos(ax)+xsin(ax)] e gx2=2
= f (cid:0) g (cid:0) :
x2+b2
Therefore
[xcos(ax)+bsin(ax)]e gx2=2
u = (cid:0) ;
x2+b2
[ bcos(ax)+xsin(ax)]e gx2=2
v = (cid:0) (cid:0) ;
x2+b2
e gx2
(x)2 = (cid:0) ;
j j x2+b2
e gx2=2
r = u2+v2 = (x)2 = (cid:0) ;
j j px2+b2
q
p
bcos(ax)+xsin(ax)
(cid:18) = tan(cid:0)1(v=u)=tan(cid:0)1 (cid:0) :
xcos(ax)+bsin(ax)
(cid:20) (cid:21)
In general, for any complex function f =u+iv, where u and v are real,
f = u iv;
(cid:3)
(cid:0)
f +f
Re[f] = (cid:3);
2
f f
Im[f] = (cid:0) (cid:3):
2i
2. Given the function
N
f(x)= ;
b2+x2
(cid:133)nd N such that f2(x) is normalized; that is, (cid:133)nd N such that
1 f2(x)dx=1;
Z(cid:0)1
0.2 Chapter 2. Mathematical Preliminaries 7
assuming that b is real. Find the Fourier transform a(k) of f(x): Evaluate
(cid:1)x2 = 1 (x x(cid:22))2f2(x)dx
(cid:0)
Z(cid:0)1
(cid:1)k2 = 1 k k(cid:22) 2 a(k)2dk
(cid:0) j j
Z(cid:0)1(cid:0) (cid:1)
andtheproduct(cid:1)x(cid:1)k.Does(cid:1)x(cid:1)k =1=2forthesefunctions?Toevaluate
theintegrals,youcanuseintegraltablesorMathematica,Maple,orMatlab.
Solution: Assume that b>0: If b<0, replace b by b in the equations
j j
below.
dx (cid:25)
1 f2(x)dx = N2 1 =N2 =1;
(b2+x2)2 2b3
Z(cid:0)1 Z(cid:0)1
2
N = b3=2;
(cid:25)
r
1 1 2 e ikx
a(k) = 1 f(x)e ikxdx= b3=2 1 (cid:0) dx
p2(cid:25) (cid:0) p2(cid:25) (cid:25) b2+x2
Z(cid:0)1 r Z(cid:0)1
= b1=2e kb:
(cid:0)j j
Both x(cid:22) and k(cid:22) equal zero. Thus
2b3 x2dx 2b3 (cid:25)
(cid:1)x2 = 1 x2f2(x)dx= 1 = =b2;
(cid:25) (b2+x2)2 (cid:25) 2b
Z(cid:0)1 Z(cid:0)1
b 1
(cid:1)k2 = 1 k2 a(k)2dk =b 1 k2e 2kbdk = = ;
j j (cid:0) j j 2b3 2b2
Z(cid:0)1 Z(cid:0)1
1
(cid:1)x(cid:1)k = >1=2:
p2
This is not a minimum uncertainty packet.
3-4. The k space amplitude for a free particle is given by
a =1=p2k k k k
a(k)= 0 0 (cid:0) 0 (cid:20) (cid:20) 0 :
0 other wise
(cid:26)
The wave function (x;0) is the Fourier transform of a(k). Plot both
k a(k)2 as a function of k=k and (x;0)2=k as a function of k x.
0 0 0 0
j j j j
By (cid:147)eyeballing(cid:148)the graphs, estimate (cid:1)x, (cid:1)k, and their product. Now cal-
culate (cid:1)k and show that (cid:1)x is in(cid:133)nite.
Solution:
You are given
a =1=p2k k k k
a(k)= 0 0 (cid:0) 0 (cid:20) (cid:20) 0 :
0 other wise
(cid:26)
