complete solutions to the physics gre pgre0177

Complete Solutionsto the Physics GRE

Exam #0177

Taylor Faucett

Senior Editor: Taylor Faucett

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BiBTeX:

@BookFaucett0177,

author = Taylor Faucett,

series = Complete Solutions to the Physics GRE,

title = Exam #0177,

pages = 36--38,

year = 2010,

edition = first,

c© 2010 Taylor FaucettThe “Complete Solutions to the Physics GRE” series has been produced strictly for

educational and non-profit purposes. All information contained within this documentmay be copied and reproduced provided that these intentions are not violated.

Contents

1 Physics GRE Solutions 31.1 PGRE0177 #1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 PGRE0177 #2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 PGRE0177 #3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 PGRE0177 #4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.5 PGRE0177 #5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.6 PGRE0177 #6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.7 PGRE0177 #7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.8 PGRE0177 #8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.9 PGRE0177 #9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.10 PGRE0177 #10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.11 PGRE0177 #11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.12 PGRE0177 #12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.13 PGRE0177 #13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.14 PGRE0177 #14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.15 PGRE0177 #15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.16 PGRE0177 #16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.17 PGRE0177 #17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.18 PGRE0177 #18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.19 PGRE0177 #19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.20 PGRE0177 #20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.21 PGRE0177 #21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261.22 PGRE0177 #22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.23 PGRE0177 #23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291.24 PGRE0177 #24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.25 PGRE0177 #25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311.26 PGRE0177 #26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.27 PGRE0177 #27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.28 PGRE0177 #28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361.29 PGRE0177 #29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371.30 PGRE0177 #30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381.31 PGRE0177 #31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391.32 PGRE0177 #32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401.33 PGRE0177 #33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411.34 PGRE0177 #34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421.35 PGRE0177 #35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431.36 PGRE0177 #36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3

CONTENTS CONTENTS

1.37 PGRE0177 #37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451.38 PGRE0177 #38 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461.39 PGRE0177 #39 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471.40 PGRE0177 #40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481.41 PGRE0177 #41 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491.42 PGRE0177 #42 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501.43 PGRE0177 #43 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511.44 PGRE0177 #44 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521.45 PGRE0177 #45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531.46 PGRE0177 #46 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 541.47 PGRE0177 #47 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 551.48 PGRE0177 #48 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561.49 PGRE0177 #49 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571.50 PGRE0177 #50 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581.51 PGRE0177 #51 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 601.52 PGRE0177 #52 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 621.53 PGRE0177 #53 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 631.54 PGRE0177 #54 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 651.55 PGRE0177 #55 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 661.56 PGRE0177 #56 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 671.57 PGRE0177 #57 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 681.58 PGRE0177 #58 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 691.59 PGRE0177 #59 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 701.60 PGRE0177 #60 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 711.61 PGRE0177 #61 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 721.62 PGRE0177 #62 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 731.63 PGRE0177 #63 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 741.64 PGRE0177 #64 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 751.65 PGRE0177 #65 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 761.66 PGRE0177 #66 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771.67 PGRE0177 #67 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 781.68 PGRE0177 #68 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 791.69 PGRE0177 #69 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 801.70 PGRE0177 #70 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 811.71 PGRE0177 #71 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 821.72 PGRE0177 #72 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 831.73 PGRE0177 #73 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 841.74 PGRE0177 #74 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 851.75 PGRE0177 #75 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 861.76 PGRE0177 #76 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 871.77 PGRE0177 #77 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 881.78 PGRE0177 #78 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 891.79 PGRE0177 #79 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 901.80 PGRE0177 #80 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 911.81 PGRE0177 #81 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 921.82 PGRE0177 #82 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 931.83 PGRE0177 #83 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 941.84 PGRE0177 #84 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

4

CONTENTS CONTENTS

1.85 PGRE0177 #85 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 971.86 PGRE0177 #86 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 981.87 PGRE0177 #87 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1001.88 PGRE0177 #88 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1011.89 PGRE0177 #89 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1021.90 PGRE0177 #90 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1041.91 PGRE0177 #91 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1051.92 PGRE0177 #92 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1071.93 PGRE0177 #93 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1091.94 PGRE0177 #94 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1111.95 PGRE0177 #95 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1121.96 PGRE0177 #96 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1141.97 PGRE0177 #97 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1151.98 PGRE0177 #98 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1171.99 PGRE0177 #99 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1181.100PGRE0177 #100 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

5

List of Figures

1.1 Electric potential at point P in relation to ring of radius R . . . . . . . . . . . . . . . 61.2 P-V curve comparison of an Isothermal and Adiabtic process . . . . . . . . . . . . . 91.3 Ray tracing diagram of a concave mirror with a longer focal length than object distance 161.4 Tangential and centripetal acceleration of a particle constrained to a circle . . . . . . 291.5 Relativistic space time diagram for time-like, space-like and light-like reference frames 421.6 3 generic forms of cubic crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

2

Chapter 1

Physics GRE Solutions

1.1 PGRE0177 #1

Recommended Solution

For a moving pendulum in which we can neglect gravity, the only two forces on the the pendulumbob is centripetal acceleration due to its rotation and acceleration due to gravity, g. At point c,the centripetal acceleration should be pointing upwards and acceleration due to gravity must bepointing downwards. Sum these two forces to see that the net force on the pendulum bob can onlypoint up or down, and we can eliminate (B) and (D). Next, check (A) to see that the net forces atpoint a and b should only have a horizontal acceleration in the left direction and, alternatively, ahorizontal acceleration in the right direction for points d and e. Since this is not true in (A), wecan eliminate it. Finally, recall that the centripetal acceleration for a pendulum will be minimizedat its peaks, so the net acceleration should be pointing downward more than left or right, whichmore closely matches (C) than (E).

Correct Answer

(C)

3

1.2. PGRE0177 #2 CHAPTER 1. PHYSICS GRE SOLUTIONS

1.2 PGRE0177 #2


Start with the equation for friction,

f = µFN (1.1)

the normal force will be equal and opposite to the force due to the gravity, making Equation1.1

f = µmg (1.2)

but since the net force in the horizontal will be rotational, f in 1.2 is then

4


f = µmg (1.3)

mv2

r= µmg (1.4)

v2

r= µg (1.5)

r =v2

µg(1.6)

Now, convert velocity into rotational units, by

v = rω (1.7)

= r

(33.3

rev

min

)(2π

rad

rev

)(1 min

60 s

)(1.8)

≈ πrrad

s(1.9)

Finally, substitute Equation 1.9 into Equation 1.6 and solve to get

r =π2r2

µg(1.10)

=µg

π2(1.11)

≈ 3

π2(1.12)

≈ 1

3(1.13)

which is closest to (D).

Correct Answer

(D)

5


1.3 PGRE0177 #3

Figure 1.1: Electric potential at point P in relation to ring of radius R


Recall Kepler’s third law,”The square of the orbital period of a planet is directly proportional to the cube of the semi-

major axis of its orbit”or, equivalently

T 2 ∝ R3 (1.14)

T ∝ R3/2 (1.15)

which is option (D).

Correct Answer

(D)

6


1.4 PGRE0177 #4


Despite the change in energy, we can utilize the conservation of momentum to get the final andinitial momentums

pi = 2mvi (1.16)

pf = 3mvf (1.17)

by conservation, pi = pf and we get

2mvi = 3mvf (1.18)

vf =2

3vi (1.19)

Now, we need to find the initial and final energy of the system, as

Ei =1

2(2m)v2i = mv2i (1.20)

Ef =1

2(3m)v2f =

3

2m

(2

3

)2

v2i =2

3mv2i (1.21)

At which it is clear that the difference between final and initial kinetic energy is 1/3.

Correct Answer

(C)

7


1.5 PGRE0177 #5


From the equipartion theorem, we know that the average energy for an ”ideal gas” with onlytranslational degrees of freedom is

〈H〉 =3

2kBT (1.22)

where the 3 in our 3/2 comes from the 3 degrees of freedom for translational motion. For aharmonic oscillator, we must add to this 2 rotational degrees of freedom and 1 degree of freedomfor its single dimension of oscillations. This brings the grand total to 6 degrees of freedom, makingthe average total energy

〈H〉 =6

2kBT = 3kBT (1.23)

Correct Answer

(D)

8


1.6 PGRE0177 #6


The quickest solution to this problem is simply to be familiar with the P-V curves for adiabaticand isothermal processes. You will likely have seen these in a thermodynamics lab course,

Figure 1.2: P-V curve comparison of an Isothermal and Adiabtic process

which is clearly (E).

Correct Answer

(E)

9


1.7 PGRE0177 #7


From the diagram, we know that the two magnets have similar poles next to one another so weshouldnt have magnetic field lines from one to another, like in (A) (C) and (D). Next, we shouldget some repelling between the field lines which doesnt show up in (E), so we must select (B).

Correct Answer

(B)

10


1.8 PGRE0177 #8


This specific problem is commonly used as one of the more simplistic, but by no means trivial,problems to introduce the method of image charges. Going through the entire proof during thisexam would be a much more lengthy process than we can finish in a reasonable amount of time sowe can really only solve this problem by knowing that a point charge will induce an exactly equaland opposite charge in the grounded plate, in fact choice (D).

Correct Answer

(D)

11


1.9 PGRE0177 #9


With 5 charges set symmetrically and equidistant about the center, we should have 5 equal electricfields canceling out fields pointing in the opposite direction. Because of symmetry, the sum of allelectric fields will equal 0.

Correct Answer

(A)

12


1.10 PGRE0177 #10


Given two capacitors, C1 = 3 microfarad and C2 = 6 microfarad and potential difference V =300 volt, the total energy can be found with

U =1

2CeqV

2 (1.24)

for capacitors in series, the equivalent capacitance is

Ceq =C1C2

C1 + C2(1.25)

=

((3 µF)(6 µF)

(3 µF) + (6 µF)

)(1.26)

=18 µF2

9 µF(1.27)

= 2 µF (1.28)

plug Equation 1.28 into Equation 1.24 and solve

U =1

2(2 µF)(300 volt)2 (1.29)

= 0.09 (1.30)

Correct Answer

(A)

13


1.11 PGRE0177 #11


We start with the thin lens equation

1

do+

1

di=

1

f(1.31)

plug in the given values for do and f1 to solve for di

1

40 cm+

1

di=

1

20 cm(1.32)

1 +40 cm

di= 2 (1.33)

di = 40 cm (1.34)

since the first image is to the right of the second lens by 10 cm, for our next calculation wemust use an object distance of d0 = −10 cm. Again, using the same equation as before, we get

1

−10 cm+

1

di=

1

10 cm(1.35)

1 +−10 cm

di= −1 (1.36)

di = 5 cm (1.37)

which predicts a final image 5.0 cm to the right of the final lens.

Correct Answer

(A)

14


1.12 PGRE0177 #12


Start with the mirror equation

1

d0+

1

di=

1

f(1.38)

1

di=

1

f− 1

d0?? (1.39)

Now, since we know from the image that the focal distance is larger than the object distance, weknow that the RHS of Equation ?? must be negative and, therefore, the image distance is virtualand behind the mirror.

Correct Answer

(E)

Alternate Solution

If we add an object at point O and do some ray tracing, we get Figure 1.3

15


Figure 1.3: Ray tracing diagram of a concave mirror with a longer focal length than object distance

Correct Answer

(E)

16


1.13 PGRE0177 #13


The Rayleigh Criterion can be used to give us the minimum resolution detail of a telescope,

θ = 1.22λ

d(1.40)

so we can solve for d in Equation 1 and plug in our known values to get

d = 1.22λ

θ(1.41)

= 1.22600 nm

3× 10−5 rad(1.42)

= 2.5 cm (1.43)

Correct Answer

(B)

17


1.14 PGRE0177 #14


From the description, we know that the detector is 100% efficient because exactly 50% of thesamples are being detected when pressed up exactly on one of its two sides. Since this is the case,we just need to find out how much fewer gamma rays will hit the detector at a distance of 1 maway. Since the rays will disperse spherically, we want to take the ratio of the samples as it passesthrough a circular surface area (AC = πr2) to that of the surface area of a sphere (AS = 4πr2) ata distance of 1 m = 100 cm. The radius of the circular surface area is 4, so we let AC = 16π cm2.Next, we can find the spherical surface area as

AS = 4πr2 (1.44)

= 4π(100 cm)2 (1.45)

= 40, 000 cm2 (1.46)

Finally, take the ratio of the two areas to get

ACAS

=16π cm2

40, 000 cm2(1.47)

= 4× 10−4 (1.48)

Correct Answer

(C)

18


1.15 PGRE0177 #15


Recall that precision is distinct from accuracy in that accuracy describes how close to the corrector true value a measurement is, while precision is a measurement of how closely grouped or howwell a result can be reproduced. In other words, a group of measurements can be entirely incorrectbut still be ”precise” if they are all extremely similar to one another. Of the plots given, (A)demonstrates the closest grouping of data points.

19


Correct Answer

(A)

20


1.16 PGRE0177 #16


From a laboratory methods course, you probably did some problems on measurement uncertaintyand came up with the uncertainty equation

u =σ√N

(1.49)

where σ is the standard deviation and N is the number of samples. Since our data set is discreteand completed over time, we can find the standard deviation via the Poisson distribution

σ =√x (1.50)

where the average of our current data set appears to be, x = 2. This makes our standarddeviation, σ =

√2. Since we want to be within 1% of the average, we take 1% of 2 to get u = 0.02.

N =

√σ2

u2(1.51)

=2

0.022(1.52)

= 5000 (1.53)

Correct Answer

(D)

21


1.17 PGRE0177 #17


Your first approach to this problem should be to ensure that each of the possible solutions containsall 15 electrons for phosophorous. Checking this, by summing the subscripts, you’ll find 15 foreach. This means that every single solution, except for the correct one, should have an incorrectprogression from our standard energy level diagram.

From the energy level diagram above, it should be clear that the progression is

1s22s22p63s23p3

Correct Answer

(B)

22


1.18 PGRE0177 #18


In the problem, 79.0 eV is given as the ionization energy for both electrons which also tells usthat the ionization for each electron individually must sum to this number. We also know that thefirst electron will be easier to pull from the atom because although it feels the same pull from thepositively charged nucleus, it also has a negatively charged electron trying to push it away. Thus,we know that the first electron will have a lower ionization energy than half of the total. Since24.6 eV is the only choice that meets this criteria, we choose (A).

Correct Answer

(A)

23


1.19 PGRE0177 #19


The sun is powered by nuclear fusion of hydrogen atoms into helium atoms. Because the atomicmass of hydrogen is approximately 1 and the atomic mass of helium is roughly 4, it must be true,by conservation of energy, that 4 hydrogen atoms combine to make 1 helium atom.

Correct Answer

(B)

24


1.20 PGRE0177 #20


Bremsstrahlung radiation refers to E&M radiation generated when a charged particle gets acceler-ated as the result of a collision with another charged particle, i.e choice (E). This is one of thoseproblems that you either know, or you don’t. Unless you speak German, in which case you couldbreak the Bremsstrahlung into its components, bremsen ”to brake” and Strahlung ”radiation”.

Correct Answer

(E)

25


1.21 PGRE0177 #21


The Lyman and Balmer series both refer to different types of transitions of an electron in a hydrogenatom from one radial quantum level (n) to another. The Lyman series is a description of all suchtransitions from n=r to n=1, such that r ≥ 2 and is an integer. The first Lyman transition(commonly called Lyman-α) is n=2 going to n=1, the second (Lyman-β) involves a transition ofn=3 to n=1, etc. The Balmer series, on the other hand, involves transitions from some n=s ton=2, such that s ≥ 3 and is an integer. The longest wavelength for both series involves the smallesttransition, i.e. n=2 going to n=1 for the Lyman Series and n=3 going to n=2 for the Balmer. TheRydberg formula can then be used to find the wavelength for each of the two transitions

1

λ= R

(1

n2f− 1

n2i

)(1.54)

For this problem we won’t need to compute anything, just compare λL and λB. Doing this forthe shortest Lyman transition gives

1

λL= R

(1

12− 1

22

)(1.55)

1

λL=

3

4R (1.56)

λL = 4/(3R) (1.57)

and for the Balmer transition

1

λB= R

(1

22− 1

32

)(1.58)

1

λB=

5

36R (1.59)

λB = 36/(5R) (1.60)

26


making the ratio

λL/λB =4/(3R)

36/(5R)= 5/27 (1.61)

Correct Answer

(B)

27


1.22 PGRE0177 #22


Recall our good friend Galileo, who demonstrated that an object in free fall in a vacuum will fall atthe same acceleration regardless of its mass. In the problem given, the moon is our object, space isour vacuum and we conclude that the mass of the moon is unknowable with the given information.

Correct Answer

(B)

28


1.23 PGRE0177 #23


In this problem, we have three vectors to consider. The first is the centripetal acceleration vectorgenerated by the rotation of the particle. The second two are both tangential to the circle but oneis a velocity vector, v = 10 m/s, and the other is a tangential acceleration, aT = 10 m/s2. First,get the net acceleration by adding aT and aC

Figure 1.4: Tangential and centripetal acceleration of a particle constrained to a circle

Since aT and aC are both 10 m/s2, the angle between them must be 45. This tells us that thenet acceleration is always 45 from any tangential vectors and since the velocity vector is also atangential vector, the angle between v and anet is 45.

Correct Answer

(C)

29


1.24 PGRE0177 #24


Start by realizing that without air resistance, the horizontal velocity will always be constant andpositive, so vx vs. t must be plot II and we can eliminate (A) and (E). Next, in the y-direction weknow that velocity must not be constant as it reaches some peak and its velocity becomes zero,before then changing direction and increasing its velocity in a negative direction. This descriptionperfectly describes plot III and so we choose (C).

Correct Answer

(C)

30


1.25 PGRE0177 #25


Start with the moment of inertia for a disk, I = 1/2mr2, which is given in the front of your testbooklet. Next, we can find the moment of inertia for the rest of the pennies using the parallel axistheorem,

I = Icom+ml2 (1.62)

again, the moment of inertia for a single penny is I = 1/2mr2 but they have a radius from thecenter of mass of l = 2r, so we get

Iop =1

2mr2 +ml2 (1.63)

=1

2mr2 +m(2r)2 (1.64)

=9

2mr2 (1.65)

however, since we have 6 pennies, the sum of all 6 is just

Iop,6 =54

2mr2 (1.66)

then add the moment of inertia of all 6 pennies with the inner penny to get,

31


Iip + Iop,6 =54

2mr2 +

1

2mr2 =

55

2mr2 (1.67)

Correct Answer

(E)

32


1.26 PGRE0177 #26


Before the rod has moved, it only has potential energy of

U =mgL

2(1.68)

where we use L/2 because that is the location of the center of mass. Next, recall the equationfor rotational kinetic energy, which all of the energy is converted to at the bottom of the rods fall,

mgL

2=

1

2Iω2 (1.69)

The moment of inertia for a rod rotating at its tip is

I =1

3mL2 (1.70)

substitute Equation 1.70 into Equation 1.69 and substitute ω = v/r to get

mgL

2=

Iω2

2(1.71)

33


=mL2v2

6L2(1.72)

gL =v2

3(1.73)

v =√

3gL (1.74)

Correct Answer

(C)

34


1.27 PGRE0177 #27


The Hermitian operator is defined as

〈A〉 =

∫ψ∗(r)Aψ(r)dr (1.75)

where A is the expectation value and is an observable. Therefore, A is, like all other observables,is real valued and we choose (A).

Correct Answer

(A)

35


1.28 PGRE0177 #28


Recall the condition for orthogonality,

〈x|x〉 = 1 (1.76)

from this, we can solve for x by taking

〈ψ1|ψ2〉 = 0 (1.77)

(5 · 1) + (−3 · −5) + (2 · x) = 0 (1.78)

20 + 2x = 0 (1.79)

x = −10 (1.80)

Correct Answer

(E)

36


1.29 PGRE0177 #29


Recall that the expectation value of any state is,

〈A〉ψ =∑j

aj |〈ψ|φj〉|2 (1.81)

where aj is the eigen value. Thus, we square each term, multiply it by its eigenvalue and sumeverything up to get

〈O〉 = −1

6+

1

2+

2

3(1.82)

= 1 (1.83)

Correct Answer

(C)

37


1.30 PGRE0177 #30


Start by realizing that the wave function should decrease as the radius goes off to infinity, whichis not true of option II and so we eliminate (B), (C) and (E). Next, as the radius goes to zero, thewave function should go to A, which is only true of I and we can choose (A).

Note that even if you don’t recognize that the wave function should go to A when r → 0, youcan at least be sure that it shouldn’t blow up to infinity like it does in option III.

Correct Answer

(A)

38


1.31 PGRE0177 #31


One exceedingly useful piece of information to memorize is that the ground state energy of positro-nium is half that of the hydrogen atom,

E0,pos =E0,H

2=−13.6 eV

2= −6.8 eV (1.84)

Then, using the Bohr model energy equation, substitute −13.6 eV with −6.8 eV and solve

E = −6.8 eV

(1

n21− 1

n22

)(1.85)

= −6.8 eV

(1− 1

9

)(1.86)

= 6.0 eV (1.87)

Correct Answer

(A)

39


1.32 PGRE0177 #32


Start by equating the total relativistic energy, E = γm0c2, to the rest energy, E = m0c

2, in theproportions given

Enet = γm0c2 = 2m0c

2 (1.88)

which clearly tells us that the lorentz factor is γ = 2. Next, solve for velocity in the Lorentzfactor

2 =1√

1− v2/c2(1.89)

2√

1− v2/c2 = 1 (1.90)

v2/c2 =3

4(1.91)

v =

√3

2c (1.92)

Finally, using our relativistic momentum and Lorentz factor, we can solve for p

p = γmov (1.93)

= 2m0

√3

2c (1.94)

=√

3m0c (1.95)

Correct Answer

(D)

40


1.33 PGRE0177 #33


First, immediately eliminate (E) because a pion can’t achieve light speed. Next, notice that if youtry to apply the classical concept of velocity (i.e. ∆x/∆T ), you get a ridiculously large number sorelativity must be in full effect and (A) isn’t nearly fast enough. Now, start with our equation forthe space time interval

∆s2 = ∆r2 − c2∆t2 (1.96)

in the pion’s frame, its change in position is 0 so we have

∆s2π = −c2∆t2 (1.97)

= −(3× 10−8)2(1× 108)2 (1.98)

= −9 m/s (1.99)

Now, in the lab frame, we get the space time interval,

∆s2lab = (30 m)2 − c2∆t2lab (1.100)

∆t2labc2 = 909 (1.101)

∆lab =

√909√c2

(1.102)

=√

101× 10−8 (1.103)

Finally, using our standard velocity equation, plug in our values to get

v =∆Xlab

∆tlab(1.104)

=30 m√

101× 10−8(1.105)

≈ 2.98× 108 (1.106)

Correct Answer

(D)

41


1.34 PGRE0177 #34


From the the displacement 4-vector, we get 3 general types of space-time intervals: time-like,space-like and light-like. These are typically drawn out on a space-time diagram as such

Figure 1.5: Relativistic space time diagram for time-like, space-like and light-like reference frames

each interval is characterized by the following criteria

time-like: if |∆x/∆t| < c, two events occur at the same location

space-like: if |∆x/∆t| > c, two events occur at the same time

light-like: if |∆x/∆t| = 0, two events are connected by a signal that moves at the speed of light.

Of the choices, the problem is describing a space-like interval which is (C).

Correct Answer

(C)

42


1.35 PGRE0177 #35


According to our models for simple blackbody radiation, a blackbody’s power is proportional tothe 4th power of its temperature

P = κT 4 (1.107)

From this, the increase in power from tripling the temperature is

P = κ(3T )4 = 81κT 4 (1.108)

which is (E).

Correct Answer

(E)

43


1.36 PGRE0177 #36


An adiabatic expansion is not the same thing as an isothermal expansion and an isothermal ex-pansion is the only type which maintains a constant temperature. Additionally, from the ideal gasequation we know that a change in volume should be accompanied by a change in temperature

P∆V = nR∆t (1.109)

Correct Answer

(E)

44


1.37 PGRE0177 #37


The quickest method to determine the thermodynamic work is to estimate the area under the P-Vcurve. The curve is roughly triangular shaped with base of 2 and height of 300. Calculate the areaunder the triangle as

1

2PV =

1

2(2)(300) ≈ 300 (1.110)

this eliminates all but (B) and (D). We can then determine whether the work is negative orpositive by the direction of the process. Recall that a clockwise process represents a heat engineand the work is positive. Alternatively, if the process is counter-clockwise, it’s a heat pump andthe work is negative. This process is counter-clockwise so we choose (D).

Correct Answer

(D)

45


1.38 PGRE0177 #38


Amplitude is maximized in an RLC circuit when we’ve reached the resonant frequency. This canbe calculated by

ω =1√LC

(1.111)

plug in your values to get

ω =1√LC

(1.112)

C =1

Lω2(1.113)

=1

(25 millihenries)(1× 103 rad)2(1.114)

=1

2.5× 107 millihenries-rad2 (1.115)

= 40 µF (1.116)

Correct Answer

(D)

46


1.39 PGRE0177 #39


At low frequencies and low energies, a capacitor and a resistor look like an impassible gap so thesewill act as a high pass filter (i.e. only high frequencies can pass) but an inductor will appear likejust another bit of wire. At high frequencies and high energies, capacitors and resistors can bepassed quite easily but passing through an inductor will generate a strong magnetic field and willimpede the flow, making this a low pass filter. Looking through the choices,

I. The inductor impedes high frequencies before it reaches the terminals so this can’t be a highpass filter

II. The resistor allows the high energy to pass and the low frequencies will drop via the resistor.We know this is a high pass filter.

III. The capacitor will allow high energies to pass through and low frequency voltage will drop.We know this is a high pass filter.

IV. The high energy passes the resistor and passes with the capacitor without a drop in voltage.Without a drop in voltage, this can’t be a high pass filter.

Correct Answer

(D)

47


1.40 PGRE0177 #40


After the switch is closed, we will start off with the maximum voltage and continually decrease asthe resistor and inductor eat away at the initial voltage. From this, we can eliminate all optionsbut (D) and (E). Next, compare the amount of time between the two remaining options to seethat only half of the voltage dropping after 200 seconds is a vastly unrealistic voltage drop for anycircuit, so we choose (D).

Correct Answer

(D)

48


1.41 PGRE0177 #41


The existence of magnetic charge would be, in essence, the same thing as saying that magneticmonopoles exist. If this were the case, then II would not have to be changed to allow magnetic fieldlines to diverge completely and we can eliminate (A), (C) and (D). Next, consider that a magneticmonopole will also allow magnetic field lines to exist without curling back to its opposite pole, soIII must change.

Correct Answer

(E)

49


1.42 PGRE0177 #42


From Lenz’s law, we know that anytime a current is generated by a change in the magnetic or electricfield, the generated emf will be such that it opposes this change in direction and magnitude. As thecenter ring moves towards ring A, this increases the field present and ring A will oppose this changewith a current of opposite current, i.e. clockwise. Ring B, on the other hand, will have its fielddecreased so a current will be induced to oppose this decrease, which requires an anti-clockwisemotion.

Correct Answer

(C)

50


1.43 PGRE0177 #43


Start with the commutator relation

[AB,C] = A[B,C] + [A,C]B (1.117)

and apply it to the given commutator

[LxLy, Lz] = Lx[Ly, Lz] + [Lx, Lz]Ly (1.118)

and replace the bracketed portions on the RHS with the commutation relations given in theproblem (while recalling that the have the commutation relation in the opposite order from left toright simply changes the sign), to get

[LxLy, Lz] = Lx[Ly, Lz] + [Lx, Lz]Ly (1.119)

= Lx(ihLx) + (−ihLy)Ly (1.120)

= ih(L2x − L2

y) (1.121)

Correct Answer

(D)

51


1.44 PGRE0177 #44


The problem gives us the energy equation as

En =n2π2h2

2mL2(1.122)

this tells us that the only difference between, say E1 and E2 will be the change in Energy fromthe squared quantum number, n = 1, 2, 3, . . .. So, we know that all allowed energies will be somesquared integer multiple of E1. Of those given, only (D) has a coefficient with a perfect squarevalue (i.e. 32 = 9).

Correct Answer

(D)

52


1.45 PGRE0177 #45


First, recall that the expectation value for any operator is the sum of its terms, each one individuallysquared and multiplied by it’s respective eigenvalue. First, find the ”ket” values for |1〉, |2〉 and 3〉,

|1〉 =3

2hω (1.123)

|2〉 =5

2hω (1.124)

|3〉 =7

2hω (1.125)

then multiply these by their respective coefficients, having been squared, to get

|ψ〉 =1√14|1〉 − 2√

14|2〉+

3√14|3〉 (1.126)

=

(1

14

)(3

2hω

)+

(4

14

)(5

2hω

)+

(9

14

)(7

2hω

)(1.127)

=86

28hω (1.128)

=43

14hω (1.129)

Correct Answer

(B)

53


1.46 PGRE0177 #46


Start with the de Broglie wavelength equation

λ =h

p(1.130)

and then recall the Schrdinger 1-D energy equation for a particle in a potential

E =p2

2m+ V (x) (1.131)

without doing much work, you can see that combining Equation 1.130 with Equation 1.131 willforce you to have an inverse square root to substitute the p2 in Equation 2 into the p in Equation1. This only matches option (E) so it must be the correct choice.

Correct Answer

(E)

54


1.47 PGRE0177 #47


First, eliminate choices (D) and (E) as both predict a negative change in entropy, which should notbe true of a gas expanding outward to occupy a new open space. Next, recall from thermodynamicsthat entropy is defined as the natural log of the total number of states of a system

S = kb ln(Ω) (1.132)

where Ω is the number of states. For a volume divided into two parts, for n particles the numberof states will be

Ω = 2n (1.133)

Plugging Equation 1.133 into Equation 1.132 and applying the rules of logarithms gives

S = kb ln(Ω) (1.134)

= kb ln (2n) (1.135)

= nkb ln(2) (1.136)

which is given by choice (B).

Correct Answer

(B)

55


1.48 PGRE0177 #48


You can initially eliminate all solutions that are less than 1, as we would expect the smaller andlighter N2 molecules to move faster than the larger and heavier O2 molecules. This step eliminates(A), (B), and (E). Between (C) and (D), you can either take a guess, or recall Graham’s law ofeffusion, which states

Rate1Rate2

=

√M2

M1(1.137)

at which point, (C) becomes the obvious choice.

Correct Answer

(C)

56


1.49 PGRE0177 #49


A Maxwell-Boltzmann system has the canonical partition function

Z =∑s

g e−Es/kbT (1.138)

Where g is the degeneracy. Since we have two energies, we will need at least two terms in ourenergy equation so we can eliminate (A), (B) and (C). Next, plug everything in to see our coefficientof 2 appear, as in (E).

Z =∑s

g e−Es/kbT (1.139)

= 2e−ε/kT + 2e−2ε/kT (1.140)

= 2[e−ε/kT + e−2ε/kT

](1.141)

Correct Answer

(E)

57


1.50 PGRE0177 #50


Recall our equation relating frequency to wavelength,

ν =v

λ(1.142)

if our velocity is 3

vf = 0.97vo (1.143)

so the initial velocity is

νo =ν0λ

(1.144)

440 hz =v0λ

(1.145)

v0 = (440 hz)(λ) (1.146)

then, compare this to the final velocity

νf =vfλ

(1.147)

=(0.97v0)

λ(1.148)

=(0.97)(440 hz)

λ(1.149)

=427 hz

λ(1.150)

Correct Answer

(B)

58


Note: if you have any experience with wind instruments, which fortunately I do, you knowfrom experience that a cold instrument always starts off a bit flat and the pitch of your instrumentcontinually increases as it warms up. From this little fact, you should at least be able to determinethat (D) and (E) can’t be correct.

59


1.51 PGRE0177 #51


Consider a beam of unpolarized light headed towards you and you have three ideal polarizers inhand.

In the figure above, grey vectors indicate individual directions of oscillation, black vectorsindicate the net polarization, greyed out areas indicate areas of absorbed polarization and whiteareas indicate unaffected polarization. If we wanted to completely eliminate all light using twoof the polarizers, we could place them in series with a rotation of exactly π/2 between them, forexample using the horizontal and vertical polarization. In this arrangement, anything that survivedthrough the horizontal polarizer would be caught by the vertical polarizer and no light would betransmitted on the other side. However, imagine we were to place another polarizer between thehorizontal and vertical polarizers such that this third polarizer is rotated π/4 or 45 with respect tothe other two. As the light first passes through the horizontal polarizer, only half of the photons thathit the polarizer will pass through. These photons will continue to the 45 polarizer where, again,half of the remaining photons get absorbed and the other half pass through. After the photonspass through the 45 polarizer, the remaining photons will spread out from -22.5 to +67.5. This

60


occurs because linearly polarized light will always completely fill a full 90 of angular spread, whichI didn’t mention previously because the previous instances came out with a 90 spread. Finally, theremaining photons will pass through the vertical polarizer giving a final total of 1/8 the originalnumber of photons

Correct Answer

(B)

61


1.52 PGRE0177 #52


Recall the three generic types of cubic crystals: Simple Cubic, Body Centered Cubic and FaceCentered Cubic.

Sum up the total area in each cubic to find

Simple Cubic 1 Atom

Body Centered Cubic 2 Atoms

Face Centered Cubic 4 Atoms

Thus the primitive unit cell (simple cubic) contains half as much area, a3/2

Correct Answer

(C)

62


(a) Simple Cubic (b) Body Centered Cubic (c) Face Centered Cubic

Figure 1.6: 3 generic forms of cubic crystals

1.53 PGRE0177 #53

63



Recall that semi-conductors, unlike regular conductors, conduct well at high temperatures andeffectively not at all at extremely low temperatures. From this, we know that the resistivity at0 should be very high and it should decrease as temperature increases. Only (B) matches thisdescription.

Correct Answer

(B)

64


1.54 PGRE0177 #54


Recall that impulse is

I =

∫ t2

t1F dt (1.151)

So we just want to find the area under this plot. If you don’t recall this equation, take notethat we are given a y-axis in newtons and an x-axis in seconds. The only way to get these twounits to end with kg ·m/s is to take the area. Using the equation for the area under a triangle,you can quickly get I = 2 kg ·m/s

Correct Answer

(C)

65


1.55 PGRE0177 #55


We can quickly eliminate choices (B) and (D) based on the fact that they both suggest an asym-metrical set of values for an exceptionally symmetric problem. We also know that (A) can’t beright because it would violate conservation of momentum for a particle of mass m and velocityv0 to generate two particles with mass m and velocity v0. Finally, since we know that horizontalmomentum must be conserved and each piece gets half the horizontal momentum of the initial,then the addition of a vertical component of velocity requires, by the Pythagorean theorem, thatthe net velocity be greater than just the horizontal component. In other words

v2net = v2x + v2y (1.152)

Correct Answer

(E)

66


1.56 PGRE0177 #56


Start by finding the difference between the air density and the helium density to get the averagedensity the balloon will experience

ρavg = ρair − ρHe = 1.29 kg/m3 − 0.18 kg/m3 = 1.11 kg/m3 (1.153)

Now, note that the only way to get units of m3 from kg and kg/m3 is to do the following

V =m

ρavg(1.154)

=300 kg

1.11 kg/m3 (1.155)

= 270 m3 (1.156)

Correct Answer

(D)

67


1.57 PGRE0177 #57


Consider that if the density of the water were to increase, so to would the force and we don’t seethis feature in (D) and (E). Next, eliminate (C) because the acceleration due to gravity,g, shouldnot be a part of these calculations. Finally, compare units on (A) and (B) to get

(A) ρv2A = (kg/m3)(m2/s2)(m2) = kg m/s2 = N

(B) ρvA/2 = (kg/m3)(m/s)(m2) = kg/s

and we choose (A).

Correct Answer

(A)

68


1.58 PGRE0177 #58


The problem tells us that the electric field is pushing the proton in the +x-direction and, fromFleming’s left hand rule, we can quickly deduce that the magnetic field is pushing the proton inthe -x-direction. Since we are told the proton isn’t deflected with a potential difference of V , weknow that the forces must have balanced out exactly. On the second pass, however, the potentialdifference is doubled and, from the Lorentz force F = qvB, this will increase the force from themagnetic field but not for the electric field. In the second pass, the magnetic field force will begreater and will exceed the electric field force and the overall direction of deflection will be in the-x-direction.

Correct Answer

(B)

69


1.59 PGRE0177 #59


The equation for a simple harmonic oscillator is

mx+ kx = 0 (1.157)

which we get when L = m, C = 1/k and Q = x.

Correct Answer

(B)

Alternate Solution

Start by eliminating any choices that suggest that Q = m, i.e. (C) and (E), as a change in massover time is not a necessary condition for a simple harmonic oscillator. Next, since Q has to bex, from our previous elimination, we can reasonably conclude that 1/C should be 1/k so that thesecond term gives us Hooke’s law, i.e. kx, and you can eliminate (A) and (D).

Correct Answer

(B)

70


1.60 PGRE0177 #60


Start by recognizing that the flux through the Gaussian surface should have some dependence onx, because if x blows up to infinity, so to should the flux. With (A) and (B) eliminated, nexteliminate (C) because the area under consideration should be the area cut out by the entire circleminus the area cut out by x, not the area of the difference of the two. Finally, note that if x = 0,the area under consideration should just be that of a circle, which is πR2 as opposed to 2πR2.

Correct Answer

(D)

71


1.61 PGRE0177 #61


Considering the electric field first, when an electric field orthogonal to a conductor interacts withthat conductor, the field (in a sense) disperses over it and the E field just to the left and to theright will be 0, from which we can eliminate (B), (D) and (E). Next, as the ~E interacts with theconductor, charges are moved around and this induces a magnetic field, so ~B 6= 0 and we choose(C).

Correct Answer

(C)

72


1.62 PGRE0177 #62


Start with our cyclotron frequency equation

f =Bq

2πm(1.158)

and then re-write Equation 1.158 to solve for m

m =Bq

2πf(1.159)

Now, plugging everything in gives us

m =(π/4 teslas)(2e−)

2π(1600 hz)(1.160)

=e−

(4)(1600 hz)(1.161)

=1.6× 10−19 kg/s

(4)(1600 hz)(1.162)

=1× 10−22 kg

4(1.163)

= 2.5 × 10−23 kg (1.164)

Correct Answer

(A)

73


1.63 PGRE0177 #63


Ideally, we would be able to recall Wien’s displacement law

λ =b

T(1.165)

in which b is the Wien displacement law constant. Alternatively, you could quickly get exactlythe same relationship with a bit of dimensional analysis by noticing that the only way to get afinal unit of Kelvins would be to take the quotient of Wien’s displacement law constant with themaximum wavelength. Either way you do it, convert the peak wavelength of 2 µm to 2× 10−6 mand solve

T =2.9× 10−3 m ·K

2× 10−6 m(1.166)

= 1, 500 K (1.167)

Correct Answer

(D)

74


1.64 PGRE0177 #64


Options (C) and (D) are very much true and are frequently mentioned in a lot of pop physicsbooks, television and other forms of media, so these two should be easily dismissed. (E) is also truebecause band spectrum, i.e. spectral lines so close together as to form a band, can’t appear whena single atom produces a single spectral line. Lastly, (B) is true because absorption spectroscopyand emission spectroscopy are essentially different ways of measuring the same thing, specificallywavelengths. Only (A) is left so that is our answer.

Correct Answer

(A)

75


1.65 PGRE0177 #65


Start with the Taylor Series expansion of ex to get an approximation

ex ≈ 1 + x (1.168)

ehν/kT = 1 + hν/kT (1.169)

Plug this approximation into the denominator of Einstein’s formula

C = 3kNA

(hν

kT

)2 ehν/kT

(1 + hν/kT − 1)2(1.170)

= 3kNA

(hν

kT

)2 ehν/kT

(hν/kT )2(1.171)

= 3kNAehν/kT (1.172)

Finally, as temperature blows up to infinity, ehν/kT = 1 and so we are left with

C = 3kNA (1.173)

Correct Answer

(D)

76


1.66 PGRE0177 #66


Ignore the fact that this problem mentions anything about physics and treat it like a standard rateproblem. If Jimmy can eat a whole apple pie in 24 minutes and Susie can eat a whole apple pie in36 minutes, how long will it take to eat that same pie if both Jimmy and Susie are sharing a singlepie?

24 ∗ 36

24 + 36= 14.4 minutes (1.174)

Correct Answer

(D)

77


1.67 PGRE0177 #67


First, eliminate (A) because we know that Uranium-238 can fission spontaneously. Next, eliminate(B) because, aside from it being a very vague claim that is not typical of a GRE answer, it is astrange conclusion to try and draw from the splitting of a Uranium atom that barely loses anymass/energy in the process. We can then eliminate (C) simply because it isn’t true that a nucleiwith roughly half the particles of Uranium-238 would be equally massive. Finally, between (D) and(E), recall that binding energy generally increases as we move towards the most stable element,iron, and decreases as we move away. A change from Uranium-238 to a nuclei with A=120 wouldbe a movement closer to iron, so we would likely so a binding energy of 8.5 MeV/nucleon ratherthan 6.7 MeV/nucleon.

Correct Answer

(E)

78


1.68 PGRE0177 #68


Since an alpha particle is a helium atom, specifically He2+ with two protons and two neutrons,if lithium lost an alpha particle it would have lost one to many protons to become beryllium and(A) is eliminated. Losing an electron, positron or neutron wouldn’t alter the actual type of atom,so emitting these wouldn’t change anything and we can eliminate (B), (C) and (D). This leaves us(E).

Correct Answer

(E)

79


1.69 PGRE0177 #69


From the description, we know that the light first interacts with the oil surface and the phase ofthe blue light is shifted 180. Next, the light interacts with the glass surface and, because of theprevious phase shift, we use the thin film interference equation

2nd = mλ (1.175)

where m = 1, 2, 3, . . .. The thinnest oil film will be when m = 1, so we solve for d in Equation1.175 and plug in our values

d =mλ

2n(1.176)

=480 nm

2(1.2)(1.177)

= 200 nm (1.178)

Correct Answer

(B)

80


1.70 PGRE0177 #70


Starting with the double slit equation

ω sin(θ) = mλ (1.179)

re-write Equation 1.179 to solve for the angle between fringes, θ, to get

sin(θ) =mλ

ω(1.180)

from Equation 1.180, we can easily see that doubling the frequency, ω, will result in a separationangle half as large as the initial 1.0 millimeter and we pick (B).

Correct Answer

(B)

81


1.71 PGRE0177 #71


We can eliminate (E) right away because the proposed velocity is faster than light. Next, eliminate(A) and (B) because we know, from the Doppler effect, that an increase in wavelength correspondsto an object moving away from the observer. Lastly, we can choose between (C) and (D) by utilizingthe doppler redshift equation

λ

λ0=

√1 + v/c

1− v/c(1.181)

607.5 nm

121.5 nm=

√1 + v/c

1− v/c(1.182)

25 =1 + v/c

1− v/c(1.183)

24− 26(v/c) = 0 (1.184)v

c= 12/13 (1.185)

v = (12/13)c (1.186)

v = 2.8× 108 m/s (1.187)

Correct Answer

(D)

82


1.72 PGRE0177 #72


Right as the string breaks the block will be accelerating due to gravity and it will also be acceleratingfrom the spring pulling down on the top block from the bottom block. Based on this, we knowthat the net acceleration of the top block must be larger than just that from acceleration, and weeliminate (A), (B), and (C). Lastly, summing the vertical forces on the top block, we get

−ma = −mg − k∆x (1.188)

2mg = k∆x (1.189)

which tells us that the acceleration should be 2g.

Correct Answer

(E)

83


1.73 PGRE0177 #73


In order for block B to not fall, the gravitational force downward must be in equilibrium with thefrictional force upwards. Adding up the forces and setting the net force to 0, we get

Fnet = f − FG (1.190)

0 = µFN −mg (1.191)

FN =mg

µ(1.192)

=(20 kg)(10 m/s2)

0.50(1.193)

= 400 N (1.194)

Correct Answer

(D)

84


1.74 PGRE0177 #74


The Lagrangian equation of motion can be calculated with

d

dt

(∂L

∂q

)− ∂L

∂q= 0 (1.195)

perform both derivatives to get

d

dt

(∂L

∂q

)= 2aq (1.196)

∂L

∂q= 4bq3 (1.197)

plugging our results from Equation 1.197 into Equation 1.195 gives

d

dt

(∂L

∂q

)− ∂L

∂q= 0 (1.198)

2aq − 4bq3 = 0 (1.199)

2aq = 4bq3 (1.200)

q =2b

aq3 (1.201)

Correct Answer

(D)

85


1.75 PGRE0177 #75


We could approach this problem theoretically, but it will be quicker to just give the matrix a vectorand see where it goes. to simplify things, use the vector

~v = 〈1, 0, 0〉 (1.202)

multiplying everything out, we get a transformed vector

~v′ = 〈1/2,√

3/2, 0〉 (1.203)

which represents a 60 rotation clockwise about the z-axis.

Correct Answer

(E)

86


1.76 PGRE0177 #76


The first of the five choices we can eliminate is (D), because electrons travel quite slowly in metals,making them exceptionally non-relativistic. We can also quickly eliminate (B) because there is nocondition specified in the problem that would take the metal out of equilibrium. Next, eliminate(A) because electrons in a metal move freely but are generally more constrained in their degrees offreedom than free atoms. Finally, between (C) and (E), choose (C) because interactions betweenphonons and electrons are more technical than we would likely ever see on the GRE.

Correct Answer

(C)

87


1.77 PGRE0177 #77


By Maxwell-Boltzmann statistics, we get the equation

Ni = Ngie−εi/kT

Z(1.204)

kT and the energy are both given so plug them both into Equation 1.204 and solve to get

Ni = Ngie−εi/kT

Z(1.205)

= Ngie

(−0.1 eV)/(0.025 eV)

Z(1.206)

= Ngie−4

Z(1.207)

which results in our ratio of e−4.

Correct Answer

(E)

88


1.78 PGRE0177 #78


Since we are talking about changing the arrangement of neutrinos and anti-neutrinos, both of whichare leptons, we would be wise to check lepton number first. Doing so, we find the typical leptonicdecay as

µ− → e− + νe + νeL : 1 = 1 − 1 + 1Le : 0 = 1 − 1 +0Lµ : 1 = 0 + 0 + 1

removing the anti-neutrino from our typical lepton decays results in

µ− → e− + νeL : 1 = 1 + 1Le : 0 = 1 + 0Lµ : 1 = 0 + 1

so lepton number is not conserved.

Correct Answer

(E)

89


1.79 PGRE0177 #79


Start with the relativistic energy equation

E2 − (pc2) = (mc2)2 (1.208)

we are given E = 10 GeV and p = 8 GeV/c so we plug these in and solve for m

(10 GeV)2 − (8 GeV/c)2c2 = m2c4 (1.209)

(100 GeV2)− (64 GeV2) = m2c4 (1.210)

36 GeV2 = m2c4 (1.211)

m2 = 36 GeV2/c4 (1.212)

m = 6 Gev/c2 (1.213)

Correct Answer

(D)

90


1.80 PGRE0177 #80


Start by eliminating option (E) because light will move slower than c when interacting with somemedium, relativistic or not. Next, consider that when the tube and water are moving at 1/2c andmoving in the same direction as the light, we would expect the light to pass through more quicklythan if the tube was stationary. Recalling our equation for light speed in a medium,

v =c

n=

3

4c (1.214)

so our answer should be larger than this and we eliminate (A) and (B). Finally, we can decidebetween (C) and (D) by considering the sum of our two relativistic velocities using

v′ =u+ v

1 + vuc2

(1.215)

=1/2c+ 3/4c

1 + (1/2c)(3/4c)c2

(1.216)

=5/4

11/8c (1.217)

=10

11c (1.218)

Correct Answer

(D)

91


1.81 PGRE0177 #81


Start with the equations for L2 and Lz

L2Ylm(θ, φ) = l(l + 1)h2Ylm(θ, φ) (1.219)

LzYlm(θ, φ) = mhYlm(θ, φ) (1.220)

applying both eigenvalues to their respective operators, we get

l(l + 1)h2 = 6h2 (1.221)

l = 2 (1.222)

and

mh = −h (1.223)

m = −1 (1.224)

at which point we plug in our values for m and l into the original momentum eigenfunction andget

Y ml (θ, φ) = Y −12 (θ, φ) (1.225)

Correct Answer

(B)

92


1.82 PGRE0177 #82


For a two electron system, there are three triplet state configurations and one singlet state config-uration. The single singlet configuration occurs when the total spin, S = S1 + S2 equals 0, whichis only true when S1 = −S2 and can be written as

1√2

(|α〉1|β〉2 − |α〉2|β〉1) (1.226)

we can then eliminate all choices with II as an option and, more to the point, all other config-urations must be one of the three possible triplet configurations.

Correct Answer

(D)

93


1.83 PGRE0177 #83


Depending on how well you recall your linear algebra, you might quickly recognize that the Paulispin matrix given in this problem is a relatively typical transformation/rotation matrix. To demon-strate this, you can pick any arbitrary up and down spin vectors (I’ll be using 1,2,3 & 4 just for thepurposes of distinguishing separate elements of the vectors). Start with spin up and down vectors

| ↑ 〉 = (1, 2) (1.227)

| ↓ 〉 = (3, 4) (1.228)

Now multiply each of those to the transformation matrix given in the problem to get thetransformed vectors

| ↑ 〉σx = (2, 1) (1.229)

| ↓ 〉σx = (4, 3) (1.230)

94


which shows that this Pauli matrix performs an orthogonal transformation on the vectors. Ofthe potential solutions, only (C) gives us the necessary result that swapping the order of the vectorswill also swap the sign.

Correct Answer

(C)

95


1.84 PGRE0177 #84


For a single electron transition, utilize our selection rules for the orbital and total angular quantumnumbers

∆l = ±1 (1.231)

∆j = 0,±1 (1.232)

we can eliminate transition A because ∆l = 0. For B and C, however, we have transitions

B: ∆l = −1 and ∆j = −1

C: ∆l = −1 and ∆j = 0

both of which are allowed and so we choose (D).

Correct Answer

(D)

96


1.85 PGRE0177 #85


The resistance of the wire is related to its length and area by

R =ρl

a(1.233)

this gives us two resistances

R1 =ρ2L

A(1.234)

R2 =ρL

2A(1.235)

which gives us a ratio for the resistances of R1/R2 = 4/1. This ratio also represents, fromV = IR, the ratio of voltages

V1V2

=4

1(1.236)

Finally, find the net voltage as

V1 + V2 = 7 volts (1.237)

and use Equation 1.236 and Equation 1.237 to get

V = 1 + IR (1.238)

= 1 V + 1.4 V (1.239)

= 2.4 V (1.240)

Correct Answer

(A)

97


1.86 PGRE0177 #86


Start with Ohm’s law and solve for current

I =V

R(1.241)

Then, we can find the induced emf from a changing magnetic field by

ε = N |dφBdt| (1.242)

φB =

∫~B · d ~A (1.243)

98


The area and, therefore, the magnetic flux changes during rotation so we find the magnetic fluxas

φB = Bπr2 sin(ωt) (1.244)

and, therefore, the induced EMF is

ε = NBπr2ω cos(ωt) (1.245)

Finally, substitute the EMF from Equation 1.245 into Equation 1.241 to get the final solution

I =NBπr2ω cos(ωt)

R(1.246)

=(15 turns)(0.5 tesla)(0.01 m)2(300 rad/sec) cos(ωt)

9 Ω(1.247)

= 25π cos(ωt) (1.248)

Correct Answer

(E)

99


1.87 PGRE0177 #87


For the left side of the diagram, the test charge falls inside the sphere meaning that the potentialis constant and the field is 0. This means that the only force on the test charge will be the resultof sphere Q pushing q to the left. Using Gauss’s law with a distance of 10d− d/2 = 19/2d, the netforce is

F =1

4πε0

qQ

r2(1.249)

=4qQ

4πε0(361d2)(1.250)

=qQ

361πε0d2(1.251)

Correct Answer

(A)

100


1.88 PGRE0177 #88


Eliminate (A) since a changing field must generate a magnetic field. Next, compare the units foreach of the potential solutions to get

(B): kgA·s2 = tesla

(C): kgA·s2 = tesla

(D): kgA·s2m 6= tesla

(E): kgA·s2m 6= tesla

We are left with (B) and (C) at which point you can guess or try to recall that the Biot-Savartlaw is proportional to 1/4π rather than 1/4π2

B =

∫µ04π

I dl × r|r|2

(1.252)

Correct Answer

(C)

101


1.89 PGRE0177 #89


From conservation of angular momentum,

Lf = L0 (1.253)

When the child is in the middle of the merry-go-round, he/she/it won’t contribute to the angularmomentum, so the final momentum will be

Lf = Iωf =

(MR2

2

)ωf (1.254)

however, for the initial momentum we must use both the child at a distance R and the merry-go-round,

102


Li = Iω0 =

(MR2

2+mR2

)ω0 (1.255)

Plug Equations 1.254 and 1.255 into Equation 1.253 to get

(MR2

2

)ωf =

(MR2

2+mR2

)ω0 (1.256)

ωf = ω0 +2m

Mω0 (1.257)

= 2.0 rad/sec +2

5(2.0 rad/sec) (1.258)

= 2.8 rad/sec (1.259)

Correct Answer

(E)

103


1.90 PGRE0177 #90


From simple common sense, with a shorter length in which oscillations can occur and a strongerspring constant in Figure 1, the period should be less than that of Figure 2. From this, we caneliminate all options that are 1 or greater, i.e. (C), (D) or (E). Between (A) and (B), recall thatthe period of a SHO can be written as

T = 2π

√m

k(1.260)

and

T = 2π

√l

g(1.261)

which tells us that doubling k and doubling l will both scale the period by√

2. However, sinceeach of these changes are happening in such a way as to give Figure 1 a longer period, the twoscalings combine to give a total change in period of twice as much for figure 1 or, equivalently,

T1T2

=1

2(1.262)

Correct Answer

(A)

104


1.91 PGRE0177 #91


At the top of the wedge, net energy is just the gravitational energy,

Enet = mgh (1.263)

all of this energy is converted to translational and rotational at the bottom of the wedge,

mgh =1

2mv2 +

1

2Iω2 (1.264)

solving for I in Equation 1.264 gives,

I =2mghR2

v2−R2m (1.265)

and substituting the value given for v2

105


I =2R2mgh

v2−R2m (1.266)

=7

4R2m−R2m (1.267)

=3

4R2m (1.268)

Correct Answer

(B)

106


1.92 PGRE0177 #92


The Hamiltonian is the sum of kinetic and potential energy terms, as opposed to the Lagrangianwhich is the difference. The energy of a harmonic oscillator is

107


V =1

2k(∆l)2 (1.269)

and the kinetic energy term is

T =1

2mv2 =

p2

2m(1.270)

adding the two of these gives us

H = T + V (1.271)

=p212m

+p222m

+1

2k(l − l0)2 (1.272)

=1

2

[p21m

+p22m

+ k(l − l0)2]

(1.273)

Correct Answer

(E)

108


1.93 PGRE0177 #93


If you do well with your physics history, you may recall that Bohr radius is defined as the smallestpossible orbital distance for the hydrogen atom in its ground state, which is sufficient to choose a0.

Correct Answer

(C)

Alternate Solution

If you don’t recall the fact in the recommended solution, start by taking the squared wavefunctionand multiplying it by a spherical shell

dP =

[1

√πa

3/20

e−r/a0

]24πr2 dr (1.274)

=4

a30r2e−2r/a0 dr (1.275)

differentiating dP in Equation 1.275 with respect to r and setting it equal to 0 allows us tosolve for the minimum radius

2re−2r/a0 − 2

a0r2e−2r/a0 = 0 (1.276)

2re−2r/a0[1− r

a0

]= 0 (1.277)

109


at which point, it is easy to solve for the radius as

r = a0 (1.278)

Correct Answer

(C)

110


1.94 PGRE0177 #94


From quantum mechanics, the first order energy shift for our specific hamiltonian is

E(1)n = 〈n0|V (a+ a†)|n0〉 (1.279)

multiply the Hamiltonian out to get(a+ a†

)2= a2 + aa† + a†a+ a†2 (1.280)

at which point we can eliminate a2 and a†2 by orthogonality. Finally, solve for the energy withour remaining terms

E = 〈n|(aa† + a†a|n〉 (1.281)

= (2n+ 1)V (1.282)

= 5V (1.283)

Correct Answer

(E)

111


1.95 PGRE0177 #95


Start by recalling the relative permitivity equation

κ =ε(ω)

ε0(1.284)

where ε(ω) and ε0 are the absolute permitivity and electric constant, respectively. Next, re-calling that the Electric field is inversely proportional to ε(ω), which you could potentially realizefrom Coulomb’s law, we get

E ∝ 1

ε(ω)(1.285)

112


∝ 1

ε0κ(1.286)

Then applying E0 ∝ 1/ε0, we get

E =E0

κ(1.287)

Correct Answer

(A)

113


1.96 PGRE0177 #96


Recall the Larmor formula, which gives the power radiated by a charged object,

P =e2a2

6πε0c3(1.288)

since the sphere is not moving, just expanding at a stationary location, we get an acceleration,a = 0. Plugging this into Equation 1 clearly results in a total radiated power of zero as well.

Correct Answer

(E)

114


1.97 PGRE0177 #97

115



Consider the limiting case of θ = 0. In this case, there will be no difference between δθ′ and θ′,which tells us that the solution should have some dependence on θ and, furthermore, that δθ′ = 0when θ = 0. This eliminates (A), (B) and (C) due to their lack of theta dependence and (D) basedon the its failure to have δθ′ = 0 when θ = 0. Note that taking the limit in (D) requires you toutilize L’hospitals rule to deal with the division of 0/0, at which point you get a pair of cosineswhich go to 1 at θ′ = 0 and θ = 0 and keeps δθ′ from going to 0.

Correct Answer

(E)

Alternate Solution

Rather than applying a limiting case to the angle, θ, we can apply the limiting case n = 1. Whenn = 1, there is no transition to a new index of refraction and δθ′ should be equivalent to θ′. We canimmediately eliminate (D) because it lacks the dependence on n and we can next eliminate (A),(B) and (C) because their lack of θ′ dependence means we could never get δθ′ = θ′ when n = 1.

Correct Answer

(E)

116


1.98 PGRE0177 #98


Examining both sums in the expression, it should become quite clear that whatever it represents,it will have units of energy. Looking at the units for the potential solutions,

(A) Average energy is an energy!

(B) The denominator of this particular expression is the partition function. You don’t really needto know that, however, provided you recognize that it is not an energy.

(C) Absolutely not an energy.

(D) Probability of a certain energy value is not, in itself, an energy value.

(E) Entropy has units of J/K so it is not an energy.

Correct Answer

(A)

117


1.99 PGRE0177 #99


Considering the problem qualitatively, after the collision occurs, we have three particles each witha mass m and some non-zero velocity. This means that our final energy, which by conservationof momentum must equal our initial energy, must be the sum of three particles with rest energyof mc2 (i.e. net rest energy will be 3mc2) and some kinetic energy. This tells us that the initialphoton energy must be greater than just the final rest energies so we can eliminate (A), (B) and(C). Finally, equating the coefficients 4 and 5 in choices (D) and (E), respectively, to the Lorentzfactor, γ, gives us a strong hint that higher values of γ will give us less realistic particle velocities(i.e. far too high) and we can choose (D).

Correct Answer

(D)

118


1.100 PGRE0177 #100


Recalling the acronym for the visible light spectrum, ROYGBIV, it should be clear that red lighthas a longer wavelength and lower energy than green light. Red light is given to us in this instanceas λred = 632.82 nm so we know that the wavelength should be less than this and we can eliminate(D) and (E). Next, recall width of the visible spectrum is roughly between 380 nm and 750 nmand that green light is exactly in the middle of the spectrum. Thus, averaging the two wavelengthsshould give us a good approximation for the wavelength of green light

380 nm + 750 nm

2=

1130 nm

2= 565 nm (1.289)

and this average is closest to (B).

Correct Answer

(B)

119

complete solutions to the physics gre pgre0177

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