Download Common Exam - 2011 Department of Physics University of Utah

Common Exam - 2011
Department of Physics
University of Utah
August 20, 2011
Examination booklets have been provided for recording your work and your solutions.
Please note that there is a separate booklet for each numbered question (i.e., use
booklet #1 for problem #1, etc.).
To receive full credit, not only should the correct solutions be given, but a sufficient
number of steps should be given so that a faculty grader can follow your reasoning.
Define all algebraic symbols that you introduce. If you are short of time it may be helpful
to give a clear outline of the steps you intended to complete to reach a solution. In some
of the questions with multiple parts you will need the answer to an earlier part in order to
work a later part. If you fail to solve the earlier part you may represent its answer with an
algebraic symbol and proceed to give an algebraic answer to the later part. This is a
closed book exam: No notes, books, or other records should be consulted. YOU MAY
ONLY USE THE CALCULATORS PROVIDED. The total of 250 points is divided
equally among the ten questions of the examination.
All work done on scratch paper should be NEATLY transferred to answer booklets.
SESSION 1 COMMON EXAM DATA SHEET
Physical constants and units
e = −1.602 × 10−19 C = −4.803 × 10−10 esu
1eV = 1.602 × 10−19 J
c = 299792458m · s−1 ≈ 3 × 108 m · s−1
h = 6.626 × 10−34 J · s = 6.626 × 10−27 erg · s = 4.135 × 10−15 eV · s
hc = 1.240 × 104 eV · Å
h̄ = 1.054 × 10−34 J · s = 1.054 × 10−27 erg · s = 6.582 × 10−16 eV · s
k = 1.380 × 10−23 J · K −1 = 1.380 × 10−16 erg · K −1 = 8.617 × 10−5 eV · K −1
G = 6.674 × 10−11 N · m2 · kg −2
g = 9.80 m · s−2
NA = 6.022 × 1023
!0 = 8.854 × 10−12 F · m−1
µ0 = 4π × 10−7 H · m−1
MElectron = 9.109 × 10−31 kg = 5.4858 × 10−4 amu = 511keV /c2
MP roton = 1.673 × 10−27 kg = 1.007276amu = 938.3M eV /c2
MN eutron = 1.675 × 10−27 kg = 1.008665amu = 939.5M eV /c2
1mile = 1609 m
1m = 3.28 f t
Some useful integrals
a, b, n > 0
!
!
√ x
dx
a+bx
√1
dx
x2 a+bx
! (a+bx)n/2
x
=−
√
√
−2(2a−bx) a+bx
2
3b
a+bx
ax
x
!
=
x
!
!
√
a − bx2 −
√ 1
dx
a2 +x2
! √a2 +x2
dx
!
!
=
√
b
2a
√
"
= ln ""x +
!
√1
dx
x a+bx
! (a+bx)(n−2)/2
x
√
"
a2 + x2 ""
"
"
= arcsin xa
1
dx
(x2 ±a2 )3/2
=
ñx
a2 x2 ±a2
1
dx
(a2 −x2 )3/2
=
√x
a2 a2 −x2
√
sin2 xdx = 12 (x − sin x cos x)
2
e−x dx =
! ∞ n−1 −x
x e dx
0
! ∞ n+1 −a2 x2
x e
dx
=
√
π
= Γ(n)
1
2an+2
#
Γ
"
a2 +x2 "
"
x
cos2 xdx = 12 (x + sin x cos x)
!∞
dx
"√ √
"
" a+ a−bx2 "
"
x
a ln "
a2 + x2 − a ln " a+
√ 1
dx
a2 −x2
−∞
0
−
dx = n2 (a + bx)n/2 + a
! √a−bx2
dx
!
=
n+2
2
$
Maxwell’s equations (SI unit system)
∇·D
=
ρ
∇·B
=
0
∇ × E = − ∂B
∂t
∇×H
=
J+
∂D
∂t
Maxwell’s equations (Gaussian unit system)
∇·D
=
4πρ
∇·B
=
0
∇ × E = − 1c ∂B
∂t
∇×H
=
4π
1 ∂D
J+
c
c ∂t
Simplest material equations
D = $·E
B = µ·H
J = σ·E
Common Exam 2011, Part 1
Problem 1
Classical Mechanics 1
A box of total mass M and length L is free to slide on a friction-free horizontal surface, as shown
in the diagram. At the left end of the box, there is a cannon that is pre-loaded with a projectile of
mass m. All system is in gravitational field with free-fall acceleration g. The box is initially at
rest. At a particular time the cannon launches the projectile to the right with non-relativistic
speed u (relative to the laboratory reference frame) from an initial height H. At a later time the
projectile embeds itself into the right-hand wall at a final height h. (The total mass M of the box
does not include the mass of the projectile.)
(a)
(3 points) What is the speed of the box, v, while the projectile is in flight?
(b)
(3 points) How far does the box move between the time the projectile is launched and the
time it embeds into the wall?
(c)
(4 points) How much energy is dissipated when the projectile embeds into the wall?
(d)
(7 points) What is the minimum initial speed of the projectile such that it embeds into the
right-hand wall before it hits the bottom surface of the box?
(e)
(8 points) Now imagine that the projectile is actually a photon of wavelength λ ‘launched’
by a single-photon light source from the left side of the box to the right side where it is
absorbed 1 . How much mass is transported by this massless projectile from the left to the
right side of the box?
1
The momentum of a photon is h / λ , where h is Planck’s constant and λ is the wavelength.
4
Problem 2
Classical Mechanics
A box of total mass M and length L is free to slide on a friction-free horizontal surface,
as shown in the diagram. At the left end of the box, there is a cannon that is pre-loaded
with a projectile of mass m. The box is initially at rest. At a particular time the cannon
launches the projectile to the right with non-relativistic speed u from an initial height H.
At a later time the projectile embeds itself into the right-hand wall at a final height h.
L
H
Initial
h
Final
(a) (3 points) What is the speed of the box, v, while the projectile is in flight?
As the box and projectile are initially at rest, the total momentum is zero. After the
projectile is launched, the total momentum is conserved:
pbox + pproj = 0 =⇒ (M − m)v = −mu
m
=⇒ v = −
u,
M −m
(8)
(9)
and the box recoils to the left.
(b) (3 points) How far does the box move between the time the projectile is launched and
the time it embeds into the wall?
Let ∆ = |v|dt be the distance the box moves to the left during the time dt the projectile
is in the air, and let δ = u dt be the distance the projectile moves to the right in the
same time. Since the box moves to the left, the distance traveled by the bullet is less
than L:
δ = L − ∆ = u dt
∆
M −m
m
= u
=u ∆
=⇒ ∆ = L .
|v|
mu
M
(10)
(11)
(c) (4 points) How much energy is dissipated when the projectile embeds into the wall?
When the projectile is in flight, both it and the box are in motion, so both have kinetic
energy. After the projectile embeds into the wall, neither it nor the box are in motion,
so this energy must have been dissipated during the collision. Thus, we simply must
compute the total kinetic energy:
Ktot = Kbox + Kproj =
1
(M − m)v 2 + mu2 + mg(H − h),
2
(12)
5
where the term mg(H − h) is the kinetic energy gained by the projectile due to its
change in gravitational potential energy. Inserting the relevant expressions gives:
m2
1
2
2
(M − m)
u + mu + mg(H − h)
(13)
Ktot =
2
(M − m)2
1
m
=
mu2
+ 1 + mg(H − h)
(14)
2
M −m
1
M
2
=
+ mg(H − h).
(15)
mu
2
M −m
We see that if m M , then the energy dissipated is due almost entirely to the kinetic
energy of the projectile.
(d) (7 points) What is the minimum initial speed of the projectile such that it embeds into
the right-hand wall before it hits the bottom surface of the box?
The time it takes for the projectile to drop from a height H to the bottom of the box is
dt. If the projectile hits in the bottom right corner of the box, then the box will travel
a distance ∆ = vdt in this time interval. Thus we have dt = ∆/v and we can solve the
problem:
!2
2
Lm
1
g
∆
g
M
H = g(dt)2 =
=
(16)
2
2 v
2 Mmu
−m
r
gL2 M − m 2
g
M −m
=
=⇒ u = L
.
(17)
2u2
M
2H
M
An initial speed larger than this will result in the projectile hitting the right-hand wall
before it hits the bottom of the box.
(e) (8 points) Now imagine that the projectile is actually a photon of wavelength λ ‘launched’
by a single-photon light source from the left side of the box to the right side where it is
absorbed2 . How much mass is transported by this massless projectile from the left to
the right side of the box?
Since the box slides on a friction-free surface and since the forces associated with the
‘launching’ and absorption of the photon are internal to the system, the center-of-mass
of the entire system must not move. Thus, since the box itself will clearly move to the
left when any projectile is launched from left to right, the photon must transport a
mass to the right side of the box:
∆
,
(18)
δ
where ∆ and δ are as above and µ is the mass transported by the photon. Now, ∆ = v dt
and δ = c dt, so
M v dt
h/λ
µ=
=
,
(19)
c dt
c
where the substitution, M v = h/λ has been made, which follows from conservation of
momentum after the photon is emitted.
M ∆ = µδ =⇒ µ = M
2
The momentum of a photon is h/λ, where h is Planck’s constant and λ is the wavelength.
Problem 2 Electricity and Magnetism – 1
The picture shows an infinite, non-conducting plane of uniform surface charge density σ. This
plane of charge lies in the yz plane. An infinite, non-conducting line of uniform linear charge
density λ lies along the x-axis, and is moving with constant speed u in the +x direction. The
picture also shows a rectangle of width w=0.400 m (along x) and height b=0.300 m (along y)
situated in the xy plane, with corners A, B, C, and D as indicated. The base of the rectangle is at
a distance a=0.100m above the x-axis. At the moment depicted, a charge Q is located at point B
and is moving in the –x direction with speed v. The electric field at B lies in the xy plane and
points at angle α from the +x axis.
The permittivity of free space is ε0 = 8.85×10−12 C2/(Nm2).
The permeability of free space is μ0 = 4π×10−7 T/(mA).
In parts (a)-(c), ignore the charge Q, you are given: λ = 5.00×10−5 C/m, and α = 30.0°.
a) (5 points) Starting with the value of the angle α given above for the direction of the electric
field, find the magnitude E of this electric field at point B.
b) (4 points) Find σ, the surface charge density of the infinite plane.
c) (8 points) Find the change in electrostatic potential from point A to point C: ΔVAC=VC-VA.
For part (d) ONLY, you are given: u = 8.00 ×105 m/s, Q = −1.00×10−6 C, v = 5.00×106 m/s
d) (8 points) Find the magnitude and direction of the magnetic force on the charge Q at the
moment shown. Indicate the direction in terms of an angle q counter-clock-wise from the +x
direction.
Problem 3 Modern Physics
Consider the unstable nuclear isotope 60Co, which undergoes beta-decay to 60Ni and has a halflife of 1925.1 days. Co has an atomic number of 27 and 60Co a mass of 59.933822u. The
dominant (>99%) decay chain consists of an electron with an endpoint energy of 0.31 MeV, and
two photons with energies of 1.17 MeV and 1.33 MeV.
a) (8 points) Estimate the rate of activity in a pure 1-g sample of 60Co. Give the rate in Bq
(decays per second).
b) (2 points) How many protons and how many neutrons are in the final state, 60Ni?
c) (7 points) The electron in the decay is not mono-energetic, having a distribution of
kinetic energies as shown in the figure below. Estimate the average energy of the
electron. (Hint: Approximate the spectrum as a straight line)
d) (3 points) The end-point of the electron spectrum is at 0.31 MeV. When the electron has
an energy below the end-point energy, the energy of the electron and two photons does
not add up to the difference in energy between the two nuclei. Where does this energy
go?
e) (5 points) In the 1-g sample considered in part (a), assume the two photons escape the
sample without interacting, but that the electron is absorbed. Estimate the rate heat is
absorbed by the sample due to the nuclear decays (give answer in Watts).
Problem 4 Quantum Mechanics 1
Consider a one-dimensional motion of two particles with the same mass, m. The particles are
bound by a spring with elastic constant, κ. The system resides in a confining potential,
U ( x) =
Qx 2
, shown in the figure below. Coordinates of the particles are x1 and x2 .
2
The goal of the problem is to find the energy levels of the system.
(a)
(5 points) Set up the Hamiltonian of the system.
(b)
(8 points) Set Q = 0 (No confining potential is present). By introducing the relative
coordinates or otherwise, find the ground-state energy.
(c)
(7 points) For finite Q reduce the Hamiltonian to two harmonic oscillators.
(d)
(5 points) Find the energy levels.
Problem 5 Thermodynamics
A non-ideal fluid has the following thermodynamic properties:
(i) The heat capacity of one mole of the fluid at constant volume depends on temperature as
CV = α T β in the range of temperatures between T1 and T2 ( α , β > 0 are constants, so within the
parameters of the problem CV does not depend on volume of the system ).
(ii) Work done by one mole of the substance during isothermal expansion with temperature T in
2
⎡ ⎛ V ⎞⎤
the range T1 < T < T2 from volume V1 to volume V2 is W = γ T ⎢ln ⎜ 2 ⎟ ⎥ , where γ is a constant.
⎣ ⎝ V1 ⎠ ⎦
(1) (7 points). Find the change in internal energy E of one mole of the substance when
temperature of the substance changes from T1 to T2 . (Remember this is not an ideal gas)
(2) (7 points) Find the equation of state of the substance.
(3) (2 points) Give thermodynamic definition of entropy.
(4) (9 points) Find the change in the entropy of one mole of the substance when it is heated from
temperature T1 to T2 at constant volume.
Problem #5 Thermodynamics
A non-ideal fluid has the following thermodynamic properties:
(i) The heat capacity of one mole of the fluid at constant volume depends on temperature as
CV   T  in the range of temperatures between T1 and T2 (  ,   0 are constants, so within the
parameters of the problem CV does not depend on volume of the system ).
(ii) Work done by one mole of the substance during isothermal expansion with temperature T in
 V
the range T1  T  T2 from volume V1 to volume V2 is W   T ln  2
  V1
2

  , where  is a constant.

(1) (7 points). Find the change in internal energy E of one mole of the substance when
temperature of the substance changes from T1 to T2 . (Remember this is not an ideal gas)
(2) (7 points) Find the equation of state of the substance.
(3) (2 points) Give thermodynamic definition of entropy.
(4) (9 points) Find the change in the entropy of one mole of the substance when it is heated from
temperature T1 to T2 at constant volume.
Solution: (1) From the first law of thermodynamics ( dE   Q   W ) we have, for a process at constant volume,
 Q 

dE  dQ . Then, using the definition of CV , CV  
   T , we have
 T V
T2
T2
T1
T1
E   CV dT    T  dT 

T2  1  T1 1  1


(2) In this question either V1 or V2 must be assumed to be a constant; otherwise the equation of state is not uniquely defined. Assuming V1  const , pressure of the gas is a derivative over the upper limit of the work written as an integral W 
V2
dW
 V2   T
 P(V )dV . We have then P  dV2  2ln  V1  V2
. Equation of state V1
 V  T
is therefore P  2ln  
 V1  V
dQ
. The entropy of a system approaches a constant value as temperature approaches zero. (3) dS 
T
Specific heat of a system approaches zero as temperature approaches zero.  Q 

(4) By definition CV  
   T . Using thermodynamic definition of entropy we obtain 
T

V
S 
T2

T1
T
T
2

dQ 2 CV dT

   T  1dT  T2   T1 . 
T
T
T
T
1
1


Common Exam - 2011
Department of Physics
University of Utah
August 20, 2011
Examination booklets have been provided for recording your work and your solutions.
Please note that there is a separate booklet for each numbered question (i.e., use
booklet #1 for problem #1, etc.).
To receive full credit, not only should the correct solutions be given, but a sufficient
number of steps should be given so that a faculty grader can follow your reasoning.
Define all algebraic symbols that you introduce. If you are short of time it may be helpful
to give a clear outline of the steps you intended to complete to reach a solution. In some
of the questions with multiple parts you will need the answer to an earlier part in order to
work a later part. If you fail to solve the earlier part you may represent its answer with an
algebraic symbol and proceed to give an algebraic answer to the later part. This is a
closed book exam: No notes, books, or other records should be consulted. YOU MAY
ONLY USE THE CALCULATORS PROVIDED. The total of 250 points is divided
equally among the ten questions of the examination.
All work done on scratch paper should be NEATLY transferred to answer booklets.
SESSION 1 COMMON EXAM DATA SHEET
Physical constants and units
e = −1.602 × 10−19 C = −4.803 × 10−10 esu
1eV = 1.602 × 10−19 J
c = 299792458m · s−1 ≈ 3 × 108 m · s−1
h = 6.626 × 10−34 J · s = 6.626 × 10−27 erg · s = 4.135 × 10−15 eV · s
hc = 1.240 × 104 eV · Å
h̄ = 1.054 × 10−34 J · s = 1.054 × 10−27 erg · s = 6.582 × 10−16 eV · s
k = 1.380 × 10−23 J · K −1 = 1.380 × 10−16 erg · K −1 = 8.617 × 10−5 eV · K −1
G = 6.674 × 10−11 N · m2 · kg −2
g = 9.80 m · s−2
NA = 6.022 × 1023
!0 = 8.854 × 10−12 F · m−1
µ0 = 4π × 10−7 H · m−1
MElectron = 9.109 × 10−31 kg = 5.4858 × 10−4 amu = 511keV /c2
MP roton = 1.673 × 10−27 kg = 1.007276amu = 938.3M eV /c2
MN eutron = 1.675 × 10−27 kg = 1.008665amu = 939.5M eV /c2
1mile = 1609 m
1m = 3.28 f t
Some useful integrals
a, b, n > 0
!
!
√ x
dx
a+bx
√1
dx
x2 a+bx
! (a+bx)n/2
x
=−
√
√
−2(2a−bx) a+bx
2
3b
a+bx
ax
x
!
=
x
!
!
√
a − bx2 −
√ 1
dx
a2 +x2
! √a2 +x2
dx
!
!
=
√
b
2a
√
"
= ln ""x +
!
√1
dx
x a+bx
! (a+bx)(n−2)/2
x
√
"
a2 + x2 ""
"
"
= arcsin xa
1
dx
(x2 ±a2 )3/2
=
ñx
a2 x2 ±a2
1
dx
(a2 −x2 )3/2
=
√x
a2 a2 −x2
√
sin2 xdx = 12 (x − sin x cos x)
2
e−x dx =
! ∞ n−1 −x
x e dx
0
! ∞ n+1 −a2 x2
x e
dx
=
√
π
= Γ(n)
1
2an+2
#
Γ
"
a2 +x2 "
"
x
cos2 xdx = 12 (x + sin x cos x)
!∞
dx
"√ √
"
" a+ a−bx2 "
"
x
a ln "
a2 + x2 − a ln " a+
√ 1
dx
a2 −x2
−∞
0
−
dx = n2 (a + bx)n/2 + a
! √a−bx2
dx
!
=
n+2
2
$
Maxwell’s equations (SI unit system)
∇·D
=
ρ
∇·B
=
0
∇ × E = − ∂B
∂t
∇×H
=
J+
∂D
∂t
Maxwell’s equations (Gaussian unit system)
∇·D
=
4πρ
∇·B
=
0
∇ × E = − 1c ∂B
∂t
∇×H
=
4π
1 ∂D
J+
c
c ∂t
Simplest material equations
D = $·E
B = µ·H
J = σ·E
Common Exam 2011, Part 2
Problem 6
Electricity and Magnetism 2
Consider a conducting medium with zero charge density (ρ=0) everywhere and (non-zero)
r
r
conductivityσ, such that J = σ E gives the current density everywhere. This medium is neither
polarizable (relative dielectric constant = 1) nor magnetic (relative permeability = 1).
r
r
a) (4 points) Write down the four Maxwell equations in E and B appropriate for this medium.
r
r
r
r
r r
r
b) (4 points) Using the identity ∇ × ∇ × V ≡ −∇ 2 V + ∇ (∇ • V ) , write down a wave equation for
r
r
the electric field E that includes a non-homogeneous term with ∂E / ∂t .
c) (6 points) Assume a linearly polarized (in the y direction) plane wave solution propagating
r
in the +x direction, of the form, E( x, t ) = yˆ E 0 exp i( Kx − ω t ) with a complex wave number
K=k+iα. Here k is the “real” wave number. Show that this leads to a propagating wave with
exponentially decaying amplitude E 0 exp( − x / b) . How is b related to α ?
d) (7 points) Find the decay length b in terms of other quantities, such as σ, ω, c, and/or ε0 and
μ0.
e) (2 points) Simplify your expression for b above in the limit of a very good conductor
(σ → ∞).
f) (2points) Alternatively simplify your answer to (d) in the high frequency limit (ω → ∞).
Problem 7 Statistical Mechanics
Part (a) A gas of N particles with spin 3/2 is confined in a volume V. Particles are placed in a
uniform magnetic field H directed along axis z, no gravitational field is present. As a result of
the Zeeman splitting the energy of a particle with spin projection S z is E = −ξ HS z , where ξ is a
constant. The system of particles is in thermal equilibrium at temperature T.
3
2
(a1) (8 points) Find average number of particles that have projection of the spin S Z = + .
Part (b) The same gas of N 3/2-spin particles is confined in a rectangular container with sizes
indicated in the figure below. The gas is in thermal equilibrium at low temperature T. Strong
non-uniform field with direction along z-axis and dependence H Z = H 0 + γ x is applied to the
system. At these conditions, we assume that ξ H 0 >> kT and therefore only particles with
3
S Z = + are present in the system.
2
(b1) (5 points) Find ratio
n( x )
, where n( x) is the volume concentration of particles at the distance
n(0)
x from zero.
(b2) (7 points) Compute n(0) .
(c) The same gas of N 3/2-spin particles at low temperature T is placed in the same strong nonuniform magnetic field H Z = H 0 + γ x and is confined in a container that has a triangular shape
with sizes shown in the figure below.
(c1) (5 points) Find the ratio of volume concentrations
n( x )
.
n(0)
Problem #7 Statistical Mechanics (Common 2011) (a) A gas of N particles with spin 3/2 is confined in a volume V. Particles are placed in a uniform magnetic field H directed along axis Z, no gravitational field is present. As a result of the Zeeman splitting the energy of a particle with spin projection S z is E   HS z , where  is a constant. The system of particles is in thermal equilibrium at temperature T. 3
(a1) (8 points) Find average number of particles that have projection of the spin S Z   . 2
(b) The same gas of N 3/2‐spin particles is confined in a rectangular container with sizes indicated in the figure below. The gas is in thermal equilibrium at low temperature T. Strong non‐uniform field with direction along z‐axis and dependence H Z  H 0   x is applied to the system. At these conditions, we 3
assume that  H 0  kT and therefore only particles with S Z   are present in the system. 2
n( x )
, where n( x ) is the volume concentration of particles at the distance x
(b1) (5 points) Find ratio
n(0)
from zero. (b2) (7 points) Compute n(0) . (c) The same gas of N 3/2‐spin particles at low temperature T is placed in the same strong non‐uniform magnetic field H Z  H 0   x and is confined in a container that has a triangular shape with sizes shown in the figure below . n( x )
(c1) (5 points) Find the ratio of volume concentrations . n(0)
Solutions A1) Partition function for a 3/2 spin in magnetic field at temperature T is 1
 3

 1

 1

 3

Z  exp    H   exp    H   exp    H   exp    H  , where  
. kT
 2

 2

 2

 2

3
The average number of particles that have projection of the spin SZ   is then 2
3


exp    H 
2

 N1  N
Z
B1) The ratio of concentrations has a form of the Boltzmann distribution  3

exp    ( H 0   x) 
n( x )
 2
  exp   3  x  


n(0)
 3

 2

exp    H 0 
 2

B2) The constant n(0) is found from the normalization condition. b
 3

N   dVn( x)  n(0)a 2  dx exp    x  2


0
1
3 
 3
 
n(0)  N
exp    b   1 2 
2a 
 2
 
C1) The ratio is the same as in the question (B1)  3

exp    ( H 0   x) 
n( x )
2

  exp   3  x  


n(0)
 3

 2

exp    H 0 
 2

Problem 8 General Physics Consider the following electrical circuit
At t < 0, S1 and S 2 are both disconnected; at t = 0, S1 is connected while S 2 remains disconnected.
a) (3 points) Find the voltage and current at A immediately after S1 is connected.
b) (3 points) Find the current at A after a very long time.
c) (4 points) Write a differential equation describing the current at A as a function of time.
d) (4 points) Solve the differential equation you found in part c).
After the current at A has stabilized, S 2 is connected and S1 is disconnected. For the following
parts, the components have the following values: V = 9 V, R1 = 100 Ω, R 2 = 1 Ω, L = 100 μH, C = 1 μF .
e) (4 points) Write a differential equation for the current at A .
f) (4 points) Neglect the damping due to the resistor, what is the period of oscillation of the current?
g) (3 points) Find the quality factor of the oscillator.
Problem 9 – Quantum Mechanics
Consider a motion of a particle of mass, m, in a one-dimensional potential well shown in the
figure.
a
⎧
⎪⎪ 0, x > 2
U ( x) = ⎨
⎪−U , x < a
⎪⎩ 0
2
In the first part of the problem you will be looking for the relation between the depth of the well,
U0, and the width of the well, a, for which the bound ground state of a particle has energy
E0 = −
(a)
(b)
(c)
U0
.
2
(5 points) Find the wave function, ψ ( x) , outside the well.
(5 points) Find ψ ( x) inside the well.
(7 points) From continuity of ψ ( x) and dψ ( x) / dx find the relation between U0 and a.
Let P be the probability to find the particle in the ground state inside the well.
(d)
(e)
(2 points) Express the probability, P, in terms ofψ ( x) .
(6 points) Evaluate P.
Problem 10 Special Relativity
A particle with mass m is moving in the lab frame S such that its kinetic energy is twice the rest energy.
(a)
(4 points) What is the total energy E of the particle in frame S?
(b)
(4 points) What is the speed u of the particle in frame S ?
(c)
(4 points) What is the relativistic momentum p of the particle in frame S ?
Now consider the motion of the particle from the point of view of an observer in frame S′, which
is moving with v = 0.8c relative to S in a direction opposite to the motion of the particle in S, as
shown in the figure below.
(d)
(9 points) What are the total energy E′ and relativistic momentum p′ of the particle in frame
S′ ? (Hint: you can either find the velocity of the particle in frame S′ first (part (e)), or you
can use a Lorentz transformation 1 to find E′ and p′ in frame S′ first and then the velocity.)
(e)
(4 points) What is the speed of the particle u′ in frame S′ ?
1
For frame S′ moving in the +x direction with velocity v relative to frame S, the Lorentz
transformation matrix is given by:
⎛ γv
⎜
⎜ − βvγ v
⎜ 0
⎜
⎝ 0
−βvγ v
γv
0
0
0
0
1
0
0⎞
⎟
0⎟
0⎟
⎟
1⎠
2
Where β v = v / c , γ v = 1 / 1 − β v , and c is the speed of light in vacuum.
Common Exam 2011 Solutions
Problem 1
Special Relativity
A particle with mass m is moving in the lab frame S such that its kinetic energy is twice
the rest energy.
(a) (4 points) What is the total energy E of the particle in frame S?
The total energy is the sum of rest and kinetic energies:
E = mc2 + K = mc2 + 2mc2 = 3mc2 .
(1)
Note that the total energy is also given by E = γu mc2 [or K = (γu − 1)mc2 ], so the
Lorentz factor associated with the velocity u of the particle in the frame S is γu = 3.
(b) (4 points) What is the speed u of the particle in frame S?
Using γu = 3, solve for u:
√
8
u
γu = p
=
= 0.943.
= 3 =⇒
2
c
3
1 − (u/c)
1
(2)
Alternate solution: Calculate the relativistic momentum pc as below and then divide
by the total energy (E = γu mc2 ):
√
pc
γu mu c
u
8mc2
=
=
=
= 0.943.
E
γu mc2
c
3mc2
(c) (4 points) What is the relativistic momentum p of the particle in frame S?
√
p = γu mu = 3m
√
8
c = 8mc.
3
(3)
Alternate solution: Start with the energy-momentum invariant, E 2 − (pc)2 = (mc2 )2 ,
and solve for pc:
p
p
√
pc = E 2 − (mc2 )2 = (3mc2 )2 − (mc2 )2 = 8mc2 .
(4)
Now consider the motion of the particle from the point of view of an observer in frame S 0 ,
which is moving with v = 0.8c relative to S in a direction opposite to the motion of the
proton in S, as shown in the figure above.
1
2
y�
y
v
p
u
S
S�
x
x�
(d) (9 points) What are the total energy E 0 and momentum p0 of the particle in frame S 0 ?
(Hint: you can either find the velocity of the particle in frame S 0 first (part (e)), or
you can use a Lorentz transformation1 to find E 0 and p0 in frame S 0 first and then the
velocity.)
Calculate the factors needed to perform a Lorentz transformation (as given in the
footnote) from the S frame to the S 0 frame:
βv = v/c = 4/5,
5
1
= ,
γv = p
2
3
1 − βv
4 5
4
β v γv =
· = .
5 3
3
Note that the subscripts on β and γ are a reminder that they pertain to the relative
velocity between the two reference frames, and not to the velocity of the particle in a
particular frame. Using the direction convention shown in the diagram, px = −p, and
py = pz = 0, so we have:
0 E (5/3) − pc (4/3)
γv
−βv γv
E
E
.
(5)
=
=
−E (4/3) + pc (5/3)
pc
p0 c
−βv γv
γv
√
From parts (a) and (b) above, we have E = 3mc2 and pc = − 8mc2 in the S frame (it
is negative because the particle moves to the left as shown in the diagram), so in the
S 0 frame we have:
√
0 E
3(5/3) + √8(4/3)
8.77
2
mc =
mc2 .
(6)
p0 c
−8.71
−3(4/3) − 8(5/3)
Note that this part can also be solved by finding the velocity of the particle u0 in the S 0
frame using the Lorentz velocity addition rules and then computing E 0 = γu0 mc2 and
p0 = γu0 mu0 .
For frame S 0 moving in the +x direction with velocity v relative to frame S, the Lorentz transformation
matrix is given by:
0
1
γv
−βv γv 0 0
B −βv γv
γv
0 0 C
C
B
@
0
0
1 0 A
0
0
0 1
p
where βv = v/c, γv = 1/ 1 − βv2 , and c is the speed of light in vacuum.
1
3
(e) (4 points) What is the speed of the particle u0 in frame S 0 ?
As above, we use the expressions for the relativistic (p0 = γu0 mu0 ) and total energy
(E 0 = γu0 mc2 ):
γu0 mu0 c
u0
p0 c
8.71
=
=
=−
= −0.993.
(7)
0
2
E
γu0 mc
c
8.77
So we see that in S 0 , the particle’s total energy, momentum, and velocity are all greater
than in the laboratory frame. This makes sense since the particle and S 0 frame are
moving in opposite directions.