[User]
Question A2
An ideal (but not necessarily perfect monatomic) gas undergoes the following cycle.

   • The gas starts at pressure P0 , volume V0 and temperature T0 .

   • The gas is heated at constant volume to a pressure αP0 , where α > 1.

   • The gas is then allowed to expand adiabatically (no heat is transferred to or from the gas) to
     pressure P0

   • The gas is cooled at constant pressure back to the original state.

The adiabatic constant γ is defined in terms of the specific heat at constant pressure Cp and the
specific heat at constant volume Cv by the ratio γ = Cp /Cv .

  a. Determine the efficiency of this cycle in terms of α and the adiabatic constant γ. As a
     reminder, efficiency is defined as the ratio of work out divided by heat in.

  b. A lab worker makes measurements of the temperature and pressure of the gas during the
     adiabatic process. The results, in terms of T0 and P0 are

                             Pressure    units of P0   1.21   1.41   1.59   1.73   2.14
                           Temperature   units of T0   2.11   2.21   2.28   2.34   2.49

     Plot an appropriate graph from this data that can be used to determine the adiabatic constant.

  c. What is γ for this gas?

                    Copyright c 2012 American Association of Physics Teachers

2012 Semifinal Exam                              Part A                                                  4

Solution:
Question A2
An ideal (but not necessarily perfect monatomic) gas undergoes the following cycle.
   • The gas starts at pressure P0 , volume V0 and temperature T0 .
   • The gas is heated at constant volume to a pressure αP0 , where α > 1.
   • The gas is then allowed to expand adiabatically (no heat is transferred to or from the gas) to
     pressure P0
   • The gas is cooled at constant pressure back to the original state.
The adiabatic constant γ is defined in terms of the specific heat at constant pressure Cp and the
specific heat at constant volume Cv by the ratio γ = Cp /Cv .
  a. Determine the efficiency of this cycle in terms of α and the adiabatic constant γ. As a
     reminder, efficiency is defined as the ratio of work out divided by heat in.
  b. A lab worker makes measurements of the temperature and pressure of the gas during the
     adiabatic process. The results, in terms of T0 and P0 are
                        Pressure      units of P0      1.21   1.41   1.59   1.73   2.14
                      Temperature     units of T0      2.11   2.21   2.28   2.34   2.49
     Plot an appropriate graph from this data that can be used to determine the adiabatic constant.
  c. What is γ for this gas?

                                             Solution
  a. Label the end points as 0, 1, and 2. The ideal gas law yields T1 = αT0 . The quantity P V γ is
     conserved along the adiabatic process 1 → 2, so
                                            1
                        P1 V1γ = P2 V2γ =     P1 V2γ   ⇒      V2 = V1 α1/γ = V0 α1/γ .
                                            α
     Again using the ideal gas law
                                                T2 = α1/γ T0 .
     Now, heat enters the gas during the isochoric process 0 → 1, so
                                     Qin = nCv ∆T = nCv (α − 1)T0 .
     Heat exits the system during the process 2 → 0, so
                                   Qout = nCp ∆T = nCp (α1/γ − 1)T0 .

     The work done is the difference,
                          W = Qin − Qout = nCv (α − 1)T0 − nCp (α1/γ − 1)T0
     and the efficiency is then
                               W     Cv (α − 1) − Cp (α1/γ − 1)      α1/γ − 1
                        η=         =                            =1−γ
                               Qin           Cv (α − 1)               α−1
     where we used the definition γ = Cp /Cv .

                  Copyright c 2012 American Association of Physics Teachers

2012 Semifinal Exam                            Part A                                              6

  b. For an adiabatic process, the quantity
                                                       γ
                                                    T
                                        PV γ                 = P 1−γ T γ
                                                   PV

     is constant, which implies that
                                               P ∝ T γ/(γ−1) .
     Thus, if we make a graph with log T on the horizontal axis and log P on the vertical, we will
     find a line with slope γ/(γ − 1). To get full credit, all five data points should be plotted and
     used, though one can get an approximate result with just two.

  c. Using the data given, we find a slope of about 3.5, giving γ ≈ 1.4, as expected for a diatomic
     gas. Note that credit will not be given for simply writing this number down; some degree of
     data analysis is necessary.

                  Copyright c 2012 American Association of Physics Teachers

2012 Semifinal Exam                              Part A                                                  7

USAPhO 2012 Question A2


[User]
2. Two blocks of masses m1 =1.0 kg and m2 = 2.0 kg are connected by a massless elastic
   spring and are at rest on a smooth horizontal surface with the spring at its natural length.
   A horizontal force of constant magnitude F = 6.0 N is applied to the block m1 for a certain
   time t in which m1 suffers a displacement ∆x1 = 0.1 m and ∆x2 = 0.05 m. Kinetic energy
   of the system with respect to center of mass is 0.1 J. The force F is then withdrawn. [Marks: 6]

                                  m1                           m2
                       F
                                  x1                           x2

Que. 3                      INPhO - 2014 Solutions
    (a) Calculate t.                                                                             [11/2]

    (b) Calculate the speed and the kinetic energy of the center of mass after the force is        [2]
        withdrawn.

    (c) Calculate the energy stored in the system.                                               [21/2]

 3. (a) Five vibrations of equal amplitudes are superposed first with all in phase agreement       [4]
        yielding intensity I1 and then with successive phase difference 30◦ yielding intensity
        I2 . Calculate I1 /I2 .

    (b) The trajectory of a ray in a non homogeneous medium is represented by x = A sin(y/B)       [6]
        where A and B are positive constants. Compute the index of refraction n in the space
        between the planes x = A and x = −A, assuming that n depends only on x and has
        the value n0 at x = 0. Plot n(x) for x ∈ [−A, A].

Page 2 of 8                          HBCSE-TIFR

                            INPhO - 2014 Solutions                                       Que. 4

Solution:
2. Two blocks of masses m1 =1.0 kg and m2 = 2.0 kg are connected by a massless elastic
   spring and are at rest on a smooth horizontal surface with the spring at its natural length.
   A horizontal force of constant magnitude F = 6.0 N is applied to the block m1 for a certain
   time t in which m1 suffers a displacement ∆x1 = 0.1 m and ∆x2 = 0.05 m. Kinetic energy
   of the system with respect to center of mass is 0.1 J. The force F is then withdrawn. [Marks: 6]

                                  m1                           m2
                       F
                                  x1                           x2

Que. 3                      INPhO - 2014 Solutions
    (a) Calculate t.                                                                             [11/2]

          Solution:                               r
                                                2(m1 ∆x1 + m2 ∆x2 )
                                        t =
                                                        F
                                           = 0.26 s

    (b) Calculate the speed and the kinetic energy of the center of mass after the force is        [2]
        withdrawn.

          Solution:                                          p
                                                                2F (m1 ∆x1 + m2 ∆x2 )
                                         Speed =
                                                                     m1 + m2
                                                         =   0.52 m.s−1
                                                             F (m1 ∆x1 + m2 ∆x2 )
                             Kinetic energy =
                                                                    m1 + m2

                                       E-                =   0.40 J

                                         20
    (c) Calculate the energy stored in the system.

          Solution:
                                                                                [21/2]

                                           14
                  Kinetic energy w.r.t. CM + Energy stored =
                                                                         m2 (∆x1 − ∆x2 )
                                                                             m1 + m2
                                                                       = 0.20 J
                                                         Energy stored = 0.10 J

          Alternatively : 0.60 J is also accepted as correct answer.
                                                                                F

                        CS
 3. (a) Five vibrations of equal amplitudes are superposed first with all in phase agreement
        yielding intensity I1 and then with successive phase difference 30◦ yielding intensity
        I2 . Calculate I1 /I2 .
                                                                                [4]

              HB
          Solution: Let a be the amplitude of incident wave.

                                       I1  A2
                                          = 12 =
                                                     25a2
                                                      √      = 1.8
                                       I2  A2    (2 + 3)2 a2

          Here A1 , A2 are amplitudes of resultant waves in two cases respectively.

    (b) The trajectory of a ray in a non homogeneous medium is represented by x = A sin(y/B)       [6]
        where A and B are positive constants. Compute the index of refraction n in the space
        between the planes x = A and x = −A, assuming that n depends only on x and has
        the value n0 at x = 0. Plot n(x) for x ∈ [−A, A].

                                h           2
                                                  i 12
          Solution: n(x) = n0       1 − A2x+B 2

Page 2 of 8                             HBCSE-TIFR

                            INPhO - 2014 Solutions                                       Que. 4

         Plot of n(x):

                                                  n(x)
                                                 n0

                                    h        2
                                                 i 21
                                  n0 1 − A2A+B 2

                                                                            x
                             −A                   (0,0)                 A

                                  E-

INPhO 2014 Problem 2


[User]
Let’s illustrate some consequences of the previous example.

     (a) For a monatomic ideal gas, we have U = (3/2)nRT and the equation of state P V = nRT .
         Evaluate CV and CP using the results above.
     (b) In a real gas, there are short-ranged attractive intermolecular forces. If they are relatively
         weak, they yield a small negative contribution to the energy, of the form
                                                   3      an2
                                                U = nRT −     .
                                                   2       V
         Explain qualitatively why the change in energy is proportional to n2 /V .
     (c) We will show in problem 4 that this modifies the equation of state to
                                                 an2

                                             P + 2 V = nRT.
                                                 V
         Using these results, show that to first order in a, we have
                                         3                             2an2
                                     CV = nR,       CP − CV ≈ nR +          .
                                         2                              TV

           Qualitatively explain why the new term has a positive sign.

         Idea 4: Equality of Mixed Partial Derivatives

         If we take two distinct partial derivatives of a state function f , then the order doesn’t matter,

                                             ∂ ∂f       ∂ ∂f
                                                      =          .
                                             ∂x y ∂y x ∂y x ∂x y

         If this identity doesn’t hold, it means that f is not actually a state function. Imposing this
         requirement for various choices of independent variables x and y yields a variety of nontrivial
         identities, collectively called Maxwell relations.

Solution:
(a) Using the results of the previous example, we have

                     ∂U        3                        ∂U         ∂V      nR
               CV =         = nR,         C P − CV =         +P         =P    = nR.
                     ∂T V      2                        ∂V T       ∂T P    P

     Combining these gives CP = (5/2)nR, just as expected.

 (b) Since the interaction is short-ranged, the number of molecules near a single given molecule is
     proportional to the average density n/V . We then multiply by the total number of molecules,
     which is proportional to n, to get a contribution proportional to n2 /V . The sign of this
     contribution is negative since the force is attractive.
     Note that we implicitly assumed that the density of molecules near a given molecules was the
     same as the density in the whole container. But if the interaction is sufficiently strong, that
     won’t be true, because the molecules will clump up. This dramatic behavior is associated
     with a phase transition to a liquid or solid state, and will be discussed further in T3. For now,
     we’re just concerned with the case of weak interactions.

 (c) The first result follows directly from the definition of the heat capacity. Next,

                                                                     an2 ∂V

                                         ∂U          ∂V
                          CP − CV =           +P            = P+ 2               .
                                         ∂V T        ∂T P            V     ∂T P

     To get ∂V /∂T |P , differentiate the equation of state with respect to T at constant P ,

                                                  an2 2an2 ∂V

                                     nR = P + 2 −                     .
                                                  V       V2    ∂T P
     Plugging this result in and expanding to lowest order in a gives the result. The reason the
     new term has a positive sign is that the interaction is attractive, so when the gas expands
     under heating, we need to put in extra heat to get the molecules further away from each other.

  Idea 4: Equality of Mixed Partial Derivatives

  If we take two distinct partial derivatives of a state function f , then the order doesn’t matter,
                                      ∂ ∂f       ∂ ∂f
                                               =          .
                                      ∂x y ∂y x ∂y x ∂x y

  If this identity doesn’t hold, it means that f is not actually a state function. Imposing this
  requirement for various choices of independent variables x and y yields a variety of nontrivial

       identities, collectively called Maxwell relations.

Kevin Zhou Physics Olympiad Handouts — educational use, attribute to Kevin Zhou (knzhou.github.io). LICENSE FLAG: must reconfirm permissions before public release.


[User]
5. Rotating disk (7 points)

A lamp is attached to the edge of a disk, which moves (slides) rotating
on ice. The lamp emits light pulses: the duration of each pulse is negli-
gible, the interval between two pulses is τ = 100ms. The first pulse is
of orange light, the next one is blue, followed by red, green, yellow, and
again orange (the process starts repeating periodically). The motion of
the disk is photographed using so long exposure time that exactly four
pulses are recorded on the photo (see figure). Due to the shortness of
the pulses and small size of the lamp, each pulse corresponds to a colo-
red doton the photo. The colors of the dots are provided with lettering:
O—orange,S —blue, P —red,R —green, andK —yellow). Thefriction
forces acting on the disk can be neglected.
1) Mark on the figure by numbers (1–4) the order of the pulses (dots).
Motivateyouranswer. Whatcanbesaidaboutthevalueoftheexposure
time?
2) Using the provided figure, find the radius of the disk R, the velocity
of the center of the disk v and the angular velocity ω (it is known that
ω < 60rad/s). The scale of the figure is provided by the image of a line
of length l = 10cm;

Solution:
5. Rotating disk

1) We notice that there is no image of the orange pulse, hence it must
have taken place immediately before the shutter release. So the blue
pulse is first, red — the second etc. The exposure time must have been
triple and quadruple flash interval, 300 ms < t < 500 ms.
2)Thedisplacementofthelampbetweentwosubsequent pulses canbe
representedasthesumoftwocomponents: ~ ri = ~ vτ +2Rsin(ωτ/2)~ ei,
where each next unitvector~ ei+1 is rotatedwithrespecttothe previous
one (~ ei) by angle ωτ. So, if the starting points of the displacement vec-
tors~ ri coincide, then the end-points mustbe on a circle, at equal angu-
lar distances ωτ from each other,see figure.
Inourcaseweredrawthedisplacementvectors1,2and3asvectors
withcommonorigin, ~ PA, ~ PB,and ~ PC. Since thestartingpointsofthe
vectors 2Rsin(ωτ/2)~ ei are brought together to the point O, their end-
points lay on the circle, the center of which can be found as the center
of the circle drawnaround the triangle ABC.
1
2
3
1
2
3
A
B
C
O
P
L
The velocity of center of the disk is found as the ratio of the
length PO and the interval τ: v ≈ 65cm/s. The angular velocity
is found as the ratio of the angle 6 AOB = 6 BOC and the inter-
val τ: ω ≈ 23rad/s. Radius of the disk is found from the length
|OA| = 2Rsin(ωτ/2) = 2Rsin6 BOC ≈ 1.5R; using the scale of
the figure, 1.5R ≈ 8cm and R ≈ 5cm.

Rotating disk


[User]
Some exercises involving square roots.
                                            √
     (a) Manually find the Taylor series for 1 + x up to second order, and verify they agree with the
         binomial theorem.
                      √
     (b) Approximate 1 + 2x + x2 for small x using the binomial theorem. Does the result match
         what you expect? If not, how can you correct it?

       Example 11: Birthday Paradox

       If you have n people in a room, around how large does n have to be for there to be at least
       a 50% chance of two people sharing the same birthday?

       Solution
       Imagine adding people one at a time. The second person has a 1/365 chance of sharing a
       birthday with the first. If they don’t share a birthday, the third person has a 2/365 chance
       of sharing a birthday with either, and so on. So a decent estimate for n is the n where

                                        1           2            n  1
                                   1−         1−         ... 1 −       ≈ .
                                       365        365            365      2
       The surprising point of the birthday paradox is that n ≪ 365. So we can use the binomial
       theorem in reverse, approximating the left-hand side as
                                                                           2
                                          1 2             1 n          1 n /2

                         1−          1−         ... 1 −        ≈ 1−
                             365         365             365          365
       which is valid since n/365 is small. It’s tempting to use the binomial theorem again to write
                                                 2
                                             1 n /2           n2

                                       1−            ≈1−            =
                                            365             2 · 365   2
       which gives n = 19. However, this is a bad approximation, because the binomial theorem only
       works if (n2 /2)(1/365) is very small, but here we’ve set it to 1/2, which isn’t particularly small.
       Since the series expansion variable is 1/2, each term in the series expansion is roughly 1/2 as
       big as the last (ignoring numerical coefficients), so we expect to be off by about (1/2)2 = 25%.

       The binomial theorem is an expansion for (1 + x)y which works when xy is small. Here xy
       isn’t small, and we instead want an approximation that works when only x is small. One
       trick to dealing with an annoying exponent is to take the logarithm, since that just turns it
       into a multiplicative factor. Note that
                                      log((1 + x)y ) = y log(1 + x) ≈ yx
       by Taylor series, which implies that
                                                 (1 + x)y ≈ eyx
       when x is small, an important fact which you should remember. So we have
                                               2
                                           1 n /2

                                                        2          1
                                     1−            ≈ e−n /2(365) =
                                          365                      2
       and solving gives n = 22.5. We should round up since n is actually an integer, giving n = 23,
       which is indeed the exact answer.

       Remark
       Precisely how accurate is the approximation (1 + x)y ≈ eyx ? Note that the only approximate
       step used to derive it was taking log(1 + x) ≈ x, which means we can get the corrections by
       expanding to higher order. If we take the next term, log(1 + x) ≈ x − x2 /2, then we find

                                            (1 + x)y ≈ eyx e−x y/2 .

       Note that because we are approximating the logarithm of the quantity we want, the next
       correction is multiplicative rather than additive. Our approximation has good fractional
       precision as long as x2 y ≪ 1. In the previous example, x2 y/2 = (22.5/365)2 /4 = 0.1%, so
       our answer was quite accurate.

Solution:
(a) The binomial theorem gives 1 + x/2. By differentiating, we get 1/(2 1 + x) and
         −1/(4(1 + x)3/2 ). Then
                                        √             1    1
                                          1 + x = 1 + x − x2 + O(x3 ).
                                                      2    8
         The first two terms agree with the usual form of the binomial theorem. For the third term,
         note that the coefficient should be

                                            1/2     (1/2)(−1/2)    1
                                                 =              =−
                                             2           2         8

           which is indeed what we find.

     (b) Of course, the result is 1 + x, so we want the O(x2 ) term to vanish. On the other hand,
         applying the binomial theorem gives
                                   p                 1                     x2
                                    1 + 2x + x2 ≈ 1 + (2x + x2 ) = 1 + x +
                                                     2                     2

    which is wrong! The reason is that the first order binomial theorem isn’t good enough, because
    the second order term in the binomial theorem will also contribute a second order term to the
    answer. Using the result of part (a),
                  p                     1               1
                     1 + 2x + x2 = 1 + (2x + x2 ) − (2x + x2 )2 + O((2x + x2 )3 )
                                        2               8
                                             x2 1
                                 =1+x+           − (2x + x2 )2 + O(x3 )
                                              2     8
                                             x 2    1
                                 =1+x+           − (2x)2 + O(x3 )
                                              2     8

                                 = 1 + x + O(x )
    as desired.

  Example 11: Birthday Paradox

  If you have n people in a room, around how large does n have to be for there to be at least
  a 50% chance of two people sharing the same birthday?

  Solution
  Imagine adding people one at a time. The second person has a 1/365 chance of sharing a
  birthday with the first. If they don’t share a birthday, the third person has a 2/365 chance
  of sharing a birthday with either, and so on. So a decent estimate for n is the n where

                                   1           2            n  1
                              1−         1−         ... 1 −       ≈ .
                                  365        365            365      2

  The surprising point of the birthday paradox is that n ≪ 365. So we can use the binomial
  theorem in reverse, approximating the left-hand side as
                                                               2
                                  1 2           1 n        1 n /2

                     1−       1−       ... 1 −       ≈ 1−
                        365      365           365        365

  which is valid since n/365 is small. It’s tempting to use the binomial theorem again to write
                                          2
                                      1 n /2       n2

                                  1−         ≈1−         =
                                     365         2 · 365   2

  which gives n = 19. However, this is a bad approximation, because the binomial theorem only
  works if (n2 /2)(1/365) is very small, but here we’ve set it to 1/2, which isn’t particularly small.
  Since the series expansion variable is 1/2, each term in the series expansion is roughly 1/2 as
  big as the last (ignoring numerical coefficients), so we expect to be off by about (1/2)2 = 25%.

  The binomial theorem is an expansion for (1 + x)y which works when xy is small. Here xy
  isn’t small, and we instead want an approximation that works when only x is small. One
  trick to dealing with an annoying exponent is to take the logarithm, since that just turns it
  into a multiplicative factor. Note that

                                 log((1 + x)y ) = y log(1 + x) ≈ yx

       by Taylor series, which implies that

                                                (1 + x)y ≈ eyx

       when x is small, an important fact which you should remember. So we have
                                               2
                                           1 n /2

                                                       2          1
                                       1−         ≈ e−n /2(365) =
                                          365                     2

       and solving gives n = 22.5. We should round up since n is actually an integer, giving n = 23,
       which is indeed the exact answer.

       Remark
       Precisely how accurate is the approximation (1 + x)y ≈ eyx ? Note that the only approximate
       step used to derive it was taking log(1 + x) ≈ x, which means we can get the corrections by
       expanding to higher order. If we take the next term, log(1 + x) ≈ x − x2 /2, then we find

                                           (1 + x)y ≈ eyx e−x y/2 .

       Note that because we are approximating the logarithm of the quantity we want, the next
       correction is multiplicative rather than additive. Our approximation has good fractional
       precision as long as x2 y ≪ 1. In the previous example, x2 y/2 = (22.5/365)2 /4 = 0.1%, so
       our answer was quite accurate.


[User]
Question A4
A tape recorder playing a single tone of frequency f0 is dropped from rest at a height h. You stand directly
underneath the tape recorder and measure the frequency observed as a function of time. Here t = 0 s is the
time at which the tape recorder was dropped.
                                                 t (s)   f (Hz)
                                                  2.0     581
                                                  4.0     619
                                                  6.0     665
                                                  8.0     723
                                                 10.0     801

    The acceleration due to gravity is g = 9.80 m/s2 and the speed of sound in air is vsnd = 340 m/s. Ignore
air resistance. You might need to use the Doppler shift formula for co-linear motion of sources and observers
in still air,
                                                     vsnd ± vobs
                                              f = f0
                                                     vsnd ± vsrc
where f0 is the emitted frequency as determined by the source, f is the frequency as detected by the observer,
and vsnd , vsrc , and vobs are the speed of sound in air, the speed of the source, and the speed of the observer.
The positive and negative signs are dependent upon the relative directions of the motions of the source and
the observer.

  a. Determine the frequency measured on the ground at time t, in terms of f0 , g, h, and vsnd . Consider
     only the case where the falling tape recorder doesn’t exceed the speed of sound vsnd .
  b. Verify graphically that your result is consistent with the provided data.
   c. What (numerically) is the frequency played by the tape recorder?

  d. From what height h was the tape recorder dropped?

                        Copyright c 2008 American Association of Physics Teachers

2008 Semifinal Exam                            Part A                             7

   STOP: Do Not Continue to Part B
If there is still time remaining for Part A, you should review your work for
    Part A, but do not continue to Part B until instructed by your exam
                                  supervisor.

                      Copyright c 2008 American Association of Physics Teachers

2008 Semifinal Exam                                 Part B                                                    8

Part B

Solution:
Question A4
A tape recorder playing a single tone of frequency f0 is dropped from rest at a height h. You stand directly
underneath the tape recorder and measure the frequency observed as a function of time. Here t = 0 s is the
time at which the tape recorder was dropped.
                                                 t (s)     f (Hz)
                                                  2.0       581
                                                  4.0       619
                                                  6.0       665
                                                  8.0       723
                                                 10.0       801

    The acceleration due to gravity is g = 9.80 m/s2 and the speed of sound in air is vsnd = 340 m/s. Ignore
air resistance. You might need to use the Doppler shift formula for co-linear motion of sources and observers
in still air,
                                                     vsnd ± vobs
                                              f = f0
                                                     vsnd ± vsrc
where f0 is the emitted frequency as determined by the source, f is the frequency as detected by the observer,
and vsnd , vsrc , and vobs are the speed of sound in air, the speed of the source, and the speed of the observer.
The positive and negative signs are dependent upon the relative directions of the motions of the source and
the observer.

  a. Determine the frequency measured on the ground at time t, in terms of f0 , g, h, and vsnd . Consider
     only the case where the falling tape recorder doesn’t exceed the speed of sound vsnd .
  b. Verify graphically that your result is consistent with the provided data.

   c. What (numerically) is the frequency played by the tape recorder?
  d. From what height h was the tape recorder dropped?

                                                 Solution
  a. The position of the tape recorder above the ground at a time t is given by
                                                           1
                                                    y = h − gt2
                                                           2
      and the speed of the tape recorder is given by

                                                         vsrc = −gt

                        Copyright ©2008 American Association of Physics Teachers

2008 Semifinal Exam                                       Part A                                            8

     The observer “hears” the sound emitted from the tape recorder a time δt earlier, since it takes time
     for the sound to travel to the listener. In this case,

                                                           y = vsnd δt

     So at time t the listener is hearing the tape recorder when it had emitted at time t0 = t − δt, or
                                                              h       g
                                                  t0 = t −        +       (t0 )2
                                                             vsnd   2vsnd
     Solve this for t0 , first by rearranging,
                                              g 0 2
                                                (t ) − vsnd t0 + (vsnd t − h) = 0
                                              2
     and the by applying the quadratic formula
                                                          p
                                          0      vsnd ±       vsnd 2 + 2gh − 2gvsnd t
                                         t =                                          .
                                                                     g

     In the limit of small h and large vsnd , this reduces to the expected t0 = t if one keeps the negative
     result. Consequently,                   p
                                     vsrc = vsnd 2 + 2gh − 2gvsnd t − vsnd
     gives the velocity of that source had when it emitted the sound heard at time t. This result is negative,
     indicating motion down, and toward the observer, so one must use the positive sign in the denominator
     of the Doppler shift formula. Applying the Doppler shift formula,
                                                                   vsnd
                                              f = f0 p                              .
                                                          vsnd 2 + 2gh − 2gvsnd t

  b. First, note that in the limit of large vsnd and small h, f reduces to

                                                              g
                                                f = f0 1 +        t .
                                                             vsnd

     However, this is too inaccurate an approximation to get a reasonable answer. Keeping to the exact
     expression, we can rearrange it as

                                        1     1          2gh      2g
                                           =        1 +        −      t
                                        f2   f0 2       vsnd 2   vsnd

     which would graph as a straight line by plotting t horizontally and 1/f 2 vertically. The slope and
     intercept are
                                            2g        1        2gh
                                       −           ,       1+
                                         vsnd f0 2   f0 2     vsnd 2
     respectively. Now we may plot the following data:

                                              t (s)   f (Hz)      1/f 2 (×10−6 s2 )
                                               2.0     581               2.96
                                               4.0     619               2.61
                                               6.0     665               2.26
                                               8.0     723               1.91
                                              10.0     801               1.56

     When graphed, this indeed produces a line.

                        Copyright ©2008 American Association of Physics Teachers

2008 Semifinal Exam                                  Part A                                              9

  c. The slope is −1.75 × 10−7 s. Then
                                            s
                                                       2(9.8)
                                     f0 =                           Hz = 574 Hz.
                                                (1.75 × 10−7 )(340)

  d. The intercept is 3.31 × 10−6 s2 . This yields a height

                                                    (3.31)   (340)2
                                      h = (340)            −        = 533 m.
                                                   (0.175)   2(9.8)

     Clearly, an impressive building, and a more impressive tape recorder, that it could be heard from such
     a distance!

                       Copyright ©2008 American Association of Physics Teachers

2008 Semifinal Exam                           Part A                             10

    STOP: Do Not Continue to Part B
If there is still time remaining for Part A, you should review your work for
    Part A, but do not continue to Part B until instructed by your exam
                                  supervisor.

                      Copyright ©2008 American Association of Physics Teachers

2008 Semifinal Exam                                Part B                                                   11

Part B

USAPhO 2008 Question A4


[User]
Question A3
Electroneering
An electron is a particle with charge −q, mass m, and magnetic moment µ. In this problem we will
explore whether a classical model consistent with these properties can also explain the rest energy
E0 = mc2 of the electron.
    Let us describe the electron as a thin spherical shell with uniformly distributed charge and
radius R. Recall that the magnetic moment of a closed, planar loop of current is always equal to
the product of the current and the area of the loop. For the electron, a magnetic moment can be
created by making the sphere rotate around an axis passing through its center.

a. If no point on the sphere’s surface can travel faster than the speed of light (in the frame of the
   sphere’s center of mass), what is the maximum magnetic moment that the sphere can have?
   You may use the integral:               Z π
                                                           4
                                               sin3 θ dθ = .
                                            0              3

b. The electron’s magnetic moment is known to be µ = q~/2m, where ~ is the reduced Planck
   constant. In this model, what is the minimum possible radius of the electron? Express your
   answer in terms of m and fundamental constants.

c. Assuming the radius is the value you found in part (b), how much energy is stored in the electric
   field of the electron? Express your answer in terms of E0 = mc2 and the fine structure constant,

                                                 q2       1
                                          α=           ≈     .
                                               4π0 ~c   137

d. Roughly estimate the total energy stored in the magnetic field of the electron, in terms of E0 and
   α. (Hint: one way to do this is to suppose the magnetic field has roughly constant magnitude
   inside the sphere and is negligible outside of it, then estimate the field inside the sphere.)

e. How does your estimate for the total energy in the electric and magnetic fields compare to E0 ?

In parts (a) and (b), you can also give your answers up to a dimensionless multiplicative constant
for partial credit.

                   Copyright ©2021 American Association of Physics Teachers

2021 USAPhO                                   Part B                                               1

Part B

Solution:
Question A3
Electroneering
An electron is a particle with charge −q, mass m, and magnetic moment µ. In this problem we will
explore whether a classical model consistent with these properties can also explain the rest energy
E0 = mc2 of the electron.
    Let us describe the electron as a thin spherical shell with uniformly distributed charge and
radius R. Recall that the magnetic moment of a closed, planar loop of current is always equal to
the product of the current and the area of the loop. For the electron, a magnetic moment can be
created by making the sphere rotate around an axis passing through its center.
a. If no point on the sphere’s surface can travel faster than the speed of light (in the frame of the
   sphere’s center of mass), what is the maximum magnetic moment that the sphere can have?
   You may use the integral:               Z π
                                                           4
                                               sin3 θ dθ = .
                                            0              3

                                               Solution
     A point on the sphere’s equator moves at a speed ωR, where ω is the angular velocity of
     rotation. Setting ωR = c gives ω = c/R.
     The spinning sphere can be thought of as a stack of infinitesimal current loops, all of which
     have a magnetic moment pointing in the same direction. Consider making a thin, circular
     slice of the sphere’s surface, corresponding to polar angles in the range (θ, θ + dθ). This
     slice has a radius R sin θ, so that the surface area of the slice is

                                           ds = 2πR sin θ R dθ.

     The charge of the slice is
                                           q ds       q sin θ
                                        dQ = − 2
                                                 =−           .
                                          4πR            2
     Since the charge dQ moves around the rotation axis one time per period T = 2π/ω, the
     corresponding current is
                                         dQ       ωq sin θ
                                    dI =      =−           .
                                          T          4π
     The magnitude of the magnetic moment of this slice is
                                                         1
                                  dµ = π(R sin θ)2 |dI| = qωR2 sin3 θ dθ.
                                                         4
     Using the provided integral, the total magnetic moment is
                                       Z π
                                           1                 1
                                  µ=         qωR2 sin3 θ dθ = qcR.
                                        0 4                  3

     If you weren’t able to do this, you could also have given the answer µ ∼ qcR, which can be
     derived by dimensional analysis, for partial credit.
     Alternative solution: Note that for a uniformly charged ring of mass dm, charge dq,
     and radius r, rotating with angular velocity ω, the ratio of the magnetic moment and the

                   Copyright ©2021 American Association of Physics Teachers

2021 USAPhO                                    Part A                                                 9

     angular momentum is
                                      µ   πr2 (ω dq/2π)   1 dq
                                        =      2
                                                        =      .
                                      L     (r dm)ω       2 dm
     The ratio is independent of r and ω. Since the sphere can be decomposed into such rings,
     the total magnetic moment and total angular momentum must have the same ratio,
                                                µ   1 q
                                                  =     .
                                                L   2m
     Finally, we know that L = (2/3)mR2 ω for a spherical shell. Plugging this in and using
     ω = c/R gives µ = qcR/3 as before, but with no integration required.

b. The electron’s magnetic moment is known to be µ = q~/2m, where ~ is the reduced Planck
   constant. In this model, what is the minimum possible radius of the electron? Express your
   answer in terms of m and fundamental constants.

                                              Solution
     Since the magnetic moment is fixed, and we want the radius to be small, we want the
     electron to be spinning as fast as possible. Thus, the magnetic moment has the value found
     in part (a), and equating this to the known value gives

                                                    3 ~
                                               R=        .
                                                    2 mc
     Again, you can get R ∼ ~/mc by dimensional analysis.

c. Assuming the radius is the value you found in part (b), how much energy is stored in the electric
   field of the electron? Express your answer in terms of E0 = mc2 and the fine structure constant,
                                                 q2       1
                                          α=           ≈     .
                                               4π0 ~c   137

                                              Solution
     For a collection of charges, the total energy stored in the electrostatic field is
                                                   1X
                                            UE =      qi V i
                                                   2
                                                      i

     where Vi is the electric potential at qi . In this case, the total charge is q, and all of the
     charge is at potential q/4π0 R, so
                                                      q2
                                             UE =          .
                                                    8π0 R
     Using the result of part (b),
                                                 1
                                            UE = αE0 .
                                                 3
     Note that you can’t get this answer by dimensional analysis alone, since α is dimensionless.
     (However, if you found R by dimensional analysis, and additionally reasoned that UE could

                   Copyright ©2021 American Association of Physics Teachers

2021 USAPhO                                         Part A                                          10

     depend only on q, 0 , and R, then you could derive UE ∼ αE0 , for partial credit.)

d. Roughly estimate the total energy stored in the magnetic field of the electron, in terms of E0 and
   α. (Hint: one way to do this is to suppose the magnetic field has roughly constant magnitude
   inside the sphere and is negligible outside of it, then estimate the field inside the sphere.)

                                                   Solution
     Following the hint, we can estimate

                                                B2

                                                                4 3
                                            UB ∼ 0                πR
                                                2µ0             3

     where B0 is the typical magnetic field inside the sphere. Actually finding the value of B0
     would require doing some complicated integrals. To get a rough estimate, note that if we
     replaced the sphere with a ring of charge, then at the center of the ring,
                                                        µ0 I   µ0 qc
                                             B0 ∼            ∼       .
                                                         R      R2
     Thus, we have
                                                   2
                                                                µ0 q 2 c2   µ0 mc3 q 2

                                  1        µ0 qc
                             UB ∼                       R3 ∼              ∼            .
                                  µ0        R2                     R            ~
     To get this in terms of the fine structure constant, we use c2 = 1/µ0 0 , giving

                                                           q2
                                        UB ∼ mc2                ∼ αE0 .
                                                          0 ~c
     An even faster way to get this result is to note that in general, the energy stored in magnetic
     fields tends to be a factor of order (v/c)2 smaller than the energy stored in electric fields,
     where v is the speed of the charge. In this problem the charge is all moving relativistically,
     so we must have UB ∼ UE .

e. How does your estimate for the total energy in the electric and magnetic fields compare to E0 ?

                                                   Solution
     Both UE and UB are much smaller than E0 , by a factor of α  1. Thus, this classical
     model cannot explain the origin of the electron’s rest energy.
     There were many attempts to make classical models of the electron in the early 20th century,
     but they all ran into difficulties like this one. For more on this subject, see chapter II-28 of
     the Feynman lectures.

In parts (a) and (b), you can also give your answers up to a dimensionless multiplicative constant
for partial credit.

                   Copyright ©2021 American Association of Physics Teachers

2021 USAPhO                                     Part B                                              1

Part B

USAPhO 2021 Question A3


[User]
Question 1

The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and H.
Stormer at Bell Labs in 1981. In the experiment electrons were confined in two
dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs
heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the
two-dimensional electron layer). A strong uniform magnetic field 𝐵𝐵 was applied
perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when
a current 𝐼𝐼 was passing through the sample, the voltage 𝑉𝑉H across the current path
exhibited an unexpected quantized plateau (corresponding to a Hall resistance 𝑅𝑅H =
3ℎ/𝑒𝑒 2 ) at sufficiently low temperatures. The appearance of the plateau would imply
the presence of fractionally charged quasiparticles in the system, which we analyze
below. For simplicity, we neglect the scattering of the electrons by random potential,
as well as the electron spin.

(a) In a classical model, two-dimensional electrons behave like charged billiard balls
     on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is
     reduced to an effective mass 𝑚𝑚∗ due to their interaction with ions.
     (i) (2 point) Write down the equation of motion of an electron in perpendicular
        electric field 𝐸𝐸�⃗ = −𝐸𝐸𝑦𝑦 𝑦𝑦� and magnetic field 𝐵𝐵
                                                           �⃗ = 𝐵𝐵𝑧𝑧̂ .

    (ii) (1 point) Determine the velocity 𝑣𝑣s of the electrons in the stationary case.
    (iii) (1 point) Which direction is the velocity pointing at?

(b) (2 points) The Hall resistance is defined as 𝑅𝑅H = 𝑉𝑉H /𝐼𝐼. In the classical model,
     find 𝑅𝑅H as a function of the number of the electrons 𝑁𝑁 and the magnetic flux
     𝜙𝜙 = 𝐵𝐵𝐵𝐵 = 𝐵𝐵𝐵𝐵𝐵𝐵 , where 𝐵𝐵 is the area of the sample, and 𝐵𝐵 and 𝐵𝐵 the
     effective width and length of the sample, respectively.

(c) (2 points) We know that electrons move in circular orbits in the magnetic field. In
     the quantum mechanical picture, the impinging magnetic field 𝐵𝐵 could be
     viewed as creating tiny whirlpools, so-called vortices, in the sea of
     electrons—one whirlpool for each flux quantum ℎ/𝑒𝑒 of the magnetic field,
     where ℎ is the Planck's constant and 𝑒𝑒 the elementary charge of an electron.
     For the case of 𝑅𝑅H = 3ℎ/𝑒𝑒 2 , which was discovered by Tsui and Stormer, derive
    the ratio of the number of the electrons 𝑁𝑁 to the number of the flux quanta 𝑁𝑁𝜙𝜙 ,

                                                                                1/4

    known as the filling factor ν.

  Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a
  current 𝐼𝐼 is passing through a two-dimensional electron system in the longitudinal direction with
  an effective length 𝐵𝐵. The Hall voltage 𝑉𝑉H is measured in the transverse direction with an effective
  width 𝐵𝐵. In addition, a uniform magnetic field 𝐵𝐵 is applied perpendicular to the plane. The
  direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall
  resistance 𝑅𝑅H versus 𝐵𝐵 at four different temperatures (curves shifted for clarity) in the original
  publication on the FQHE. The features at 𝑅𝑅H = 3ℎ/𝑒𝑒 2 are due to the FQHE.

(d) (2 points) It turns out that binding an integer number of vortices (𝑛𝑛 > 1) with
     each electron generates a bigger surrounding whirlpool, hence pushes away all
     other electrons. Therefore, the system can considerably reduce its electrostatic

                                                                                2/4

    Coulomb energy at the corresponding filling factor. Determine the scaling
    exponent 𝛼𝛼 of the amount of energy gain for each electron Δ𝑈𝑈(𝐵𝐵) ∝ 𝐵𝐵 α .

(e) (2 points) As the magnetic field deviates from the exact filling ν = 1/𝑛𝑛 to a
     higher field, more vortices (whirlpools in the electron sea) are being created.
     They are not bound to electrons and behave like particles carrying effectively
     positive charges, hence known as quasiholes, compared to the negatively charged
     electrons. The amount of charge deficit in any of these quasiholes amounts to
     exactly 1/𝑛𝑛 of an electronic charge. An analogous argument can be made for
     magnetic fields slightly below ν and the creation of quasielectrons of negative
     charge 𝑒𝑒 ∗ = −𝑒𝑒/𝑛𝑛. At the quantized Hall plateau of 𝑅𝑅H = 3ℎ/𝑒𝑒 2 , calculate the
     amount of change in 𝐵𝐵 that corresponds to the introduction of exactly one
     fractionally charged quasihole. (When their density is low, the quasiparticles are
     confined by the random potential generated by impurities and imperfections,
     hence the Hall resistance remains quantized for a finite range of 𝐵𝐵.)

(f) In Tsui et al. experiment,
     the magnetic field corresponding to the center of the quantized Hall plateau
         𝑅𝑅H = 3ℎ/𝑒𝑒 2 , 𝐵𝐵1/3 = 15 Tesla,

    the effective mass of an electron in GaAs, 𝑚𝑚∗ = 0.067 𝑚𝑚𝑒𝑒 ,
    the electron mass, 𝑚𝑚𝑒𝑒 = 9.1 × 10−31 kg,
    Coulomb's constant, 𝑘𝑘 = 9.0 × 109 N ∙ m2 /C2 ,
    the vacuum permittivity, ε0 = 1/4π𝑘𝑘 = 8.854 × 10−12 F/m,
    the relative permittivity (the ratio of the permittivity of a substance to the vacuum
         permittivity) of GaAs, ε𝑟𝑟 = 13,
    the elementary charge, 𝑒𝑒 = 1.6 × 10−19 C,
    Planck's constant, ℎ = 6.626 × 10−34 J ∙ s, and
    Boltzmann's constant, 𝑘𝑘B = 1.38 × 10−23 J/K.
    In our analysis, we have neglected several factors, whose corresponding energy
    scales, compared to Δ𝑈𝑈(𝐵𝐵) discussed in (d), are either too large to excite or too
    small to be relevant.
    (i) (1 point) Calculate the thermal energy 𝐸𝐸th at temperature 𝑇𝑇 = 1.0 K.
    (ii) (2 point) The electrons spatially confined in the whirlpools (or vortices) have
         a large kinetic energy. Using the uncertainty relation, estimate the order of
         magnitude of the kinetic energy. (This amount would also be the additional
         energy penalty if we put two electrons in the same whirlpool, instead of in
         two separate whirlpools, due to Pauli exclusion principle.)

(g) There are also a series of plateaus at 𝑅𝑅H = ℎ/𝑖𝑖𝑒𝑒 2 , where 𝑖𝑖 = 1, 2, 3, … in Tsui et
     al. experiment, as shown in Figure 1(b). These plateaus, known as the integer
     quantum Hall effect (IQHE), were reported previously by K. von Klitzing in 1980.
     Repeating (c)-(f) for the integer plateaus, one realizes that the novelty of the
     FQHE lies critically in the existence of fractionally charged quasiparticles. R.

                                                                                3/4

de-Picciotto et al. and L. Saminadayar et al. independently reported the
observation of fractional charges at the ν = 1/3 filling in 1997. In the
experiments, they measured the noise in the charge current across a narrow
constriction, the so-called quantum point contact (QPC). In a simple statistical
model, carriers with discrete charge 𝑒𝑒 ∗ tunnel across the QPC and generate
charge current IB (on top of a trivial background). The number of the carriers 𝑛𝑛τ
arriving at the electrode during a sufficiently small time interval τ obeys Poisson
probability distribution with parameter λ

                                                    λ𝑘𝑘 e−λ
                                    P(𝑛𝑛τ = 𝑘𝑘) =
                                                      𝑘𝑘!

where 𝑘𝑘! is the factorial of 𝑘𝑘. You may need the following summation
                                               ∞
                                         λ
                                                    λ𝑘𝑘
                                        e =�
                                                    𝑘𝑘!
                                              k=0

(i) (2 point) Determine the charge current 𝐼𝐼B , which measures total charge per
      unit of time, in terms of λ and τ.
(ii) (2 points) Current noise is defined as the charge fluctuations per unit of time.
      One can analyze the noise by measuring the mean square deviation of the
      number of current-carrying charges. Determine the current noise 𝑆𝑆𝐼𝐼 due to
      the discreteness of the current-carrying charges in terms of λ and τ.
(iii) (1 point) Calculate the noise-to-current ratio 𝑆𝑆𝐼𝐼 /𝐼𝐼B , which was verified by R.
      de-Picciotto et al. and L. Saminadayar et al. in 1997. (One year later, Tsui
      and Stormer shared the Nobel Prize in Physics with R. B. Laughlin, who
      proposed an elegant ansatz for the ground state wave function at ν = 1/3.)

                                                                                4/4

Solution:
Question 1

The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and
H. Stormer at Bell Labs in 1981. In the experiment electrons were confined in
two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs
heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the
two-dimensional electron layer). A strong uniform magnetic field B was applied
perpendicular to the two-dimensional electron system. As illustrated in Figure
1, when a current I was passing through the sample, the voltage VH across the
current path exhibited an unexpected quantized plateau (corresponding to a Hall
resistance RH = 3h/e2 ) at sufficiently low temperatures. The appearance of the
plateau would imply the presence of fractionally charged quasiparticles in the
system, which we analyze below. For simplicity, we neglect the scattering of the
electrons by random potential, as well as the electron spin.

  (a) In a classical model, two-dimensional electrons behave like charged billiard
      balls on a table. In the GaAs/AlGaAs sample, however, the mass of the
      electrons is reduced to an effective m∗ due to their interaction with ions.

       (i) (2 point) Write down the equation of motion of an electron in perpen-
                                  ⃗ = −Ey ŷ and magnetic field B
           dicular electric field E                             ⃗ = B ẑ.

   Solution: An electron with charge −e (e > 0) experiences the Lorentz force
   due to the perpendicular magnetic field and the electric force

                                  d⃗v     !
                                                ⃗ +E
                                                   ⃗
                                                     "
                             m∗       = −e ⃗v × B
                                  dt
   where ⃗v is the velocity of the electron.

   Grading: 1 point for writing down the electric force and the magnetic force
   correct, and 1 point for writing down the effective mass and the acceleration
   correct.

       (ii) (1 point) Determine the velocity vs of the electrons in the stationary

                                         1

         case.

 Solution: In the stationary regime, the acceleration vanishes. Hence
                                      ⃗ +E
                                ⃗vs × B  ⃗ =0

 The velocity can be expressed as

                                          ⃗ ×B
                                          E    ⃗
                                  ⃗vs =
                                            B2
 whose magnitude is simply vs = E/B.

 Grading: Either writing down the correct magnitude of the velocity or its
 vector form is sufficient for the 1 point.

    (iii) (1 point) Which direction is the velocity pointing at?

 Solution: The velocity ⃗vs should be perpendicular to both the magnetic field
                            ⃗ is in the z direction and E
 and the electric field. If B                           ⃗ in the −y direction, as
 given by the problem, ⃗vs is in the −x direction, generating a charge current
 in the x direction.

 Grading: 1 point for the correct direction.

(b) (2 points) The Hall resistance is defined as RH = VH /I. In the classical
    model, find RH as a function of the number of the electrons N and the
    magnetic flux φ = BA = BW L, where A is the area of the sample, and W

                                      2

    and L the effective width and length of the sample, respectively.

 Solution: The Hall voltage VH = Ey W . The current in the −x direction is

                          ∆Q   Ne     Ne Ey    N
                     I=      =      =       = e VH
                          ∆t   L/vs   L B      φ

 Therefore,
                                      VH   1φ
                               RH =      =
                                       I   eN

 Grading: 1 point for the final expression and 1 point for writing down the
 expression for relating I with the number of electrons and their stationary
 velocity (hence the electric field and the magnetic field).

(c) (2 points) We know that electrons move in circular orbitss in the magnetic
    field. In the quantum mechanical picture, the impinging magnetic field B
    could be viewed as creating tiny whirlpools, so-called vortices, in the sea of
    electrons–one whirlpool for each flux quantum h/e of the magnetic field,
    where h is the Planck’s constant and e the elementary charge of an electron.
    For the case of RH = 3h/e2 , which was discovered by Tsui and Stormer,
    derive the ratio of the number of the electrons N to the number of the flux
    quanta Nφ , known as the filling factor ν.

 Solution: The Hall resistance can be rewritten as
                             1φ   h φ/(h/e)   h Nφ
                      RH =      = 2         = 2
                             eN  e    N      e N
 At the plateau, ν = N/Nφ = 1/3.

 Grading: 1 point for the final expression.

(d) (2 points) It turns out that binding an integer number of vortices (n > 1)
    with each electron generates a bigger surrounding whirlpool, hence pushes
    away all other electrons. Therefore, the system can considerably reduce

                                       3

    its electrostatic Coulomb energy at the corresponding filling factor. Deter-
    mine the scaling exponent α of the amount of energy gain for each electron
    ∆U(B) ∝ B α .

 Solution: The average distance between electrons can be written as f l0 ,
 where                       #         #          #
                                LW         φ           h
                        l0 =         =         =
                                 N        NB          νeB
 and f is a dimensionless constant that is determined by the electron distri-
 bution (or, quantum mechanically, wave function). Binding multiple vortices
 with an electron effectively reduces the probability of other electrons getting
 close. Therefore, the electrons optimize their distribution in such a way that
 their average distance increases from f1 l0 to f2 l0 (f1 < f2 ). One expect the
 Coulomb energy gain per electron is proportional to
                                              $         %
                 e2               e2            1   1         e2
                            −                 =   −
            4πε0 εr (f1 l0 ) 4πε0 εr (f2 l0 )   f1 f2       4πε0 εr l0
                              √
 Therefore, ∆(B) ∝ 1/l0 ∝ B, or α = 1/2.

 Grading: The key point here is to realize that the energy scale is determined
 by the Coulomb interaction, which scales inversely with a length scale (e.g.,
 the magnetic length) that characterizes the mean electron distance (and its
 change). 1 point for the final expression and 1 point for writing down the
 correct relation between the length scale and the magnetic field.

(e) (2 points) As the magnetic field deviates from the exact filling ν = 1/m
    to a higher field, more vortices (whirlpools in the electron sea) are being
    created. They are not bound to electrons and behave like particles carry-
    ing effectively positive charges, hence known as quasiholes, compared to
    the negatively charged electrons. The amount of charge deficit in any of
    these quasiholes amounts to exactly 1/m of an electronic charge. An anal-
    ogous argument can be made for magnetic fields slightly below ν and the
    creation of quasielectrons of negative charge e∗ = −e/m. Assume the sam-
    ple has an area A. At the quantized Hall plateau of RH = 3h/e2 , calculate
    the amount of change in B that corresponds to the introduction of exactly
    one fractionally charged quasihole. (When their density is low, the quasi-
    particles are confined by the random potential generated by impurities and

                                      4

   imperfections, hence the Hall resistance remains quantized for a finite range
   of B.)

Solution: The flux change due to the change of the magnetic field is
                                                  h
                             ∆φ = ∆B(W L) =
                                                  e
Therefore, ∆B = h/(eW L).

Grading: 2 points for the final expression.

(f) In Tsui et al. experiment,
      • the magnetic field corresponding to the center of the quantized Hall
        plateau RH = 3h/e2 , B1/3 = 15 Tesla,
      • the effective mass of an electron in GaAs, m∗ = 0.067me ,
      • the electron mass, me = 9.1 × 10−31 kg,
      • Coulomb’s constant, k = 9.0 × 109 N·m2/C2 ,
      • the vacuum permittivity, ε0 = 1/(4πk) = 8.854 × 10−12 F/m,
      • the relative permittivity (the ratio of the permittivity of a substance to
        the vacuum permittivity) of GaAs, εr = 13,
      • the elementary charge, e = 1.6 × 10−19 C,
      • Planck’s constant, h = 6.626 × 10−34 J·s, and
      • Boltzmann’s constant, kB = 1.38 × 10−23 J/K.
   In our analysis, we have neglected several factors, whose corresponding
   energy scales, compared to ∆(B) discussed in (d), are either too large to
   excite or too small to be relevant.
     (i) (1 point) Calculate the thermal energy Eth at temperature T = 1.0 K.

 Solution: The thermal energy

              Eth = kB T = 1.38 × 10−23 × 1.0 = 1.38 × 10−23 J

 Grading: 1 point for the numerical result.

                                      5

     (ii) (2 point) The electrons spatially confined in the whirlpools (or vor-
          tices) have a large kinetic energy. Using the uncertainty relation, es-
          timate the order of magnitude of the kinetic energy. (This amount
          would also be the additional energy penalty if we put two electrons in
          the same whirlpool, instead of in two separate whirlpools, due to Pauli
          exclusion principle.)

 Solution: The size of a vortex is of order
                      #          #
                           h          6.626 × 10−34
               l0 =          =                         = 1.66 × 10−8 m
                          eB         1.6 × 10 −19 × 15
 According to the uncertainty relation, p ∼ ∆p ∼ h/l0 . Therefore, the kinetic
 energy is

                   p2    h2 eB     h eB
                      =         =
                  2m∗   2m∗ h       2 m∗
                        6.626 × 10 −34
                                       × 1.6 × 10−19 × 15
                      =
                             2 × 0.067 × 9.1 × 10−31
                      = 1.3 × 10−20 J

 Grading: 1 point for the final numerical result and 1 point for relating the
 characteristic length to the momentum through the uncertainty relation, and
 hence the kinetic energy. Note this is an estimate problem, hence any final
 numerical result within a factor of 2π can be regarded as correct.

(g) There are also a series of plateau at RH = h/ie2 , where i = 1,2,3,... in
    Tsui et al. experiment, as shown in Figure 1(b). These plateaus, known
    as the integer quantum Hall effect (IQHE), were reported previously by K.
    von Klitzing in 1980. Repeating (c)-(f) for the integer plateaus, one realizes
    that the novelty of the FQHE lies critically in the existence of fractionally
    charged quasiparticles. R. de-Picciotto et al. and L. Saminadayar et al. in-
    dependently reported the observation of fractional charges at the ν = 1/3
    filling in 1997. In the experiments, they measured the noise in the charge
    current across a narrow constriction, the so-called quantum point contact
    (QPC). In a simple statistical model, carriers with discrete charge e∗ tunnel
    across the QPC and generate charge current IB (on top of a trivial back-
    ground). The number of the carriers nτ arriving at the electrode during
    a sufficiently small time interval τ obeys Poisson probability distribution

                                           6

  with parameter λ
                                            λk e−λ
                               P (nτ = k) =        ,                   (1)
                                              k!
  where k! is the factorial of k. You may need the following summation
                                             ∞
                                               λk
                                      eλ =
                                             &
                                                      ,                      (2)
                                             k=0 k!

    (i) (2 point) Determine the charge current IB , which measures total charge
        per unit of time, in terms of λ and τ .

Solution: The current can be calculated by the ratio of the total charge carried
by the averaged nτ quasiparticles to the time interval τ .
                               ∞
                               &                  ∞
                                                  &    λk e−λ
                     ⟨nτ ⟩ =         kP (k) =
                               k=1                k=1 (k − 1)!
                                                          ∞
                                                          &
                                              = λ             P (k)
                                                      k=0
                                              = λ

where we have used       k P (k) = 1. Therefore,
                     '

                                      ⟨nτ ⟩e∗   νeλ
                               IB =           =
                                        τ        τ
Grading: 1 point for correctly calculating the average charge under the Pois-
son distribution and 1 point for the final expression for the charge current.

   (ii) (2 points) Current noise is defined as the charge fluctuations per unit
        of time. One can analyze the noise by measuring the mean square
        deviation of the number of current-carrying charges. Determine the
        current noise SI due to the discreteness of the current-carrying charges

                                        7

        in terms of λ and τ .

Solution: Similarly, the noise can be related to the averaged charge fluctua-
tions during the time interval τ .
             (              )
             (nτ − ⟨nτ )2        = ⟨n2τ ⟩ − ⟨nτ ⟩2
                                     ∞
                                              k 2 P (k) − λ2
                                     &
                                 =
                                     k=1
                                     *        ∞               ∞
                                                                         +
                                              &               &
                                          2
                                 =    λ             P (k) +         kP (k) − λ2
                                              k=0             k=1
                                 = λ

Therefore,                  (                     )
                                (nτ − ⟨nτ )2 (e∗ )2 (νe)2 λ
                     SI =                                 =
                                     τ                 τ
Grading: 1 point for correctly calculating the charge fluctuations under the
Poisson distribution and 1 point for the final expression that relates the current
noise to the charge fluctuations.

  (iii) (1 point) Calculate the noise-to-current ratio SI /IB , which was veri-
        fied by R. de-Picciotto et al. and L. Saminadayar et al. in 1997. (One
        year later, Tsui and Stormer shared the Nobel Prize in Physics with
        R. B. Laughlin, who proposed an elegant ansatz for the ground state
        wave function at ν = 1/3.)

Solution: The noise-to-current ratio SI /IB = e∗ = νe.

Grading: 1 point for the final expression.

                                              8

Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE.
As indicated, a current I is passing through a two-dimensional system in the lon-
gitudinal direction with an effective length L. The Hall voltage VH is measured in
the transverse direction with an effective width W . In addition, a uniform mag-
netic field B is applied perpendicular to the plane. The direction of the current is
given for illustrative purpose only, which may not be correct. (b) Hall resistance
RH versus B at four different temperatures (curves shifted for clarity), adapted
from the original publication on the FQHE. The features at RH = 3h/e2 are due
to the FQHE.

                                         9

APhO 2015 Theory T1


[User]
7. Soapfilm (6 points)

LordRayleighhadin1891alectureabout
taking photos of physical processes. Among others, he showed a
photo of a soap film, which is falling apart (see Fig.). Instead of a
flash,heusedanelectricspark(well,nowadaystheflashesarealso
based on electric sparks). Estimate, how precise must have been
the timing, i.e. estimate the time for a soap film to fall apart. Let
the thickness of the soap film be h = 1 μm, the ring diameter
D = 10cm andthe surfacetension σ = 0.025N/m.
Hint: you may use a model, according to which the already
broken partof the soapfilm gathers into a single front and moves
alltogether towardsthe stillpreservedpartofthe film.

Solution:
7. Soap film

Suppose the area of a broken part of the
soap film is S. The corresponding surface energy is ES = 2σS.
The mass of the soap water that was previously located at the
broken part is m = Shρ, where ρ = 103
kg/m3
. Assuming
that the surface energy is transformed to the kinetic energy of
the moving front, we get an equation for the speed of the front:
2σS = 1
2
mv2
, 4σS = Shρv2
. Therefore v =
q
4σ
hρ
=
q
4 ·0.025
10−6 ·103 m/s = 10m/s and we can estimate the time of
breakupofthe soapfilmto be t = D/v = 1
100
s.

Soapfilm