physics comprehensive review · 2010. 9. 22. · physics comprehensive . review . by larry, jeff...

Fields Test Example Questions Guide

PHYSICS COMPREHENSIVE

REVIEW By Larry, Jeff and Jeff

Not for circulation outside USNA

Physics Handout Series – fields.tank with jeff and jeff page 1


CAUTION: This guide covers 35 problems that will not appear on your Fields Test. Also under

stand that the guide makes statements that are usually true without adding qualifying remarks or

apologies. Advanced or special cases are not treated. In some cases, one or two steps are omitted.

Solve each problem as you read through this guide.

STRATEGY: The best approach is to recognize the problem and solve it. If you are unsure, find

physical reasons to eliminate as many of the five options as possible. Make an educated guess as you

pick from among the remaining options. Mark the problem so that you can return to it later at which

time you can attempt a more thorough analysis. As you work through the guide, pay special attention

to the examples for which options are found not to be viable.

DO NOT MEMORIZE: Memorizing values and facts from this guide is not recommended. Do

think about the concepts and orders of magnitude as you read this guide. Think and read just

carefully enough to be left with a hazy recollection of the experience. THINK, think, think while you

take the test.

The sample questions appear boxed. Search for “Ex.” to find the next problem.

MECHANICS AND RELATIVITY:

Newton’s Laws are best written for particle mechanics. The laws are written in forms appropriate for

inertia reference observes. A particle is a body for which every point in the body has the same

velocity. If one considers a rotating billiard ball, there is no one velocity and hence no one

acceleration for all the mass points comprising the ball. The center-of-mass theorems are the bridge

that allows one to generalize Newton’s laws to forms that apply to more general bodies.

I.) A particle maintains its state of uniform motion unless acted upon by an (external) force.

Uniform motion is motion at a constant velocity v .



II.) The net force that acts on a particle causes it to accelerate. The net force is the vector sum of all

the forces that are acting on the particle, and that net force is equal to the mass of that particle times

its acceleration. (The word causes is of cosmic importance in physics.)

III.) If particle A exerts a force on particle B then particle B exerts a force of equal magnitude and

opposite direction on particle A. AB BF F A

Newton’s Third Law deals with forces and particles.

General Reciprocity (NOT Newton’s Law, an extension of Newton’s third law): If a physical

entity A exerts an influence on entity B, then B exerts a proportionate counter influence on A. For

example, an electric charge causes an electric field and an electric field exerts a force on electric

charges. A moving charge creates a magnetic field, and a magnetic field exerts forces on moving

charges. Note that the entities in these examples were not equivalent so the counter influences were

not equivalent. The counter nature is observed when a time-varying magnetic field induces a current

in a conductor. The magnetic field caused by the induced current fights (counters) the variation in

the magnetic field. The magnetic field in the interior of a perfect conductor must be constant in time

as the currents induced in such a conductor are as large as required to completely counter the

disturbing influence. Newton was only setting down laws for forces and particles, not for these

more complex situations.

Newton’s Laws: Use when asked about forces and acceleration at a single position and time.

Circular Motion : 22 ˆ ˆˆ ˆcircle v dvr dta r t r r r t

The centripetal part is always in play. The tangential part applies if the speed is changing.

Friction: Static friction acts with a magnitude from 0 to s N to keep the contacting surfaces from

slipping relative to one another. Once the surfaces are slipping, the friction force on each surface has

magnitude equal to k N directed to oppose the relative motion. Factoid: k s.

Use Newton’s laws to solve Here & Now problems.



Work-Energy: Work is W = suggesting that work-energy by used when a change is

associated with a change in position.

F dr

Mechanical energy: Ei = K1 i + K2 i + … + UA i + UB i + …..

K = = ½ McmVcm2 + ½ Icm2 = ½ McmVcm2 + ½ mvrel 2 + ... = .... 2½ i iparticles i

m v

U = ½ k (stretch)2 + {mgh; - GMm/r} + {q V; kQq/r} choose near-earth or Universal form choose appropriate form

Ei + Wnon-conserve = Ef Wnc-frintion = - k N *(distance that surfaces slip relative to one another)

A conservative is any force for which the work integral can be evaluated as the difference in the

values of a scalar function at the integration endpoints. (Can do if the work integral around any

closed path is zero; curl of the force is zero).

Use Work-Energy to solve Here to There problems.

Momentum and Collisions: Impulse is 2

1

t

tF dt

suggesting that impulse-momentum methods be

tried for a change that occurs as time changes – even for the time change from before to after.

, ,

tf

k k initial net external k k finaltim v F m v

, Internal forces do not change the total momentum.

Use Impulse-Momentum to solve Now to Then (Before to After) problems.

(1) Collisions: Before to after or now to then problems

Rule 1: Conserve momentum vectorially (component by component).

Rule 2: Kinetic energy is unchanged for an elastic collision. Some kinetic energy is converted to

other forms if the collision is partially or totally inelastic. As the total momentum is conserved, only

the kinetic energy associated with motion relative to the center of mass is available to be lost

(converted).

2 2 2,½ ½ ½total i i i relative i CM relative

i iK m v M V m v K K ;

2

2CMtotalP

MK

M: total mass; V: speed of the center of mass (CM); vrelative, i : speed of mass i relative to CM.



In a totally inelastic collision, the particles stick together so Krelative is lost. KCM remains as required

by conservation of momentum.

Ex. 1) The figure to the left shows two

particles with masses and velocities as

indicated. The objects are moving on a flat,

frictionless surface. When they collide, the

objects stick together. Their speed after the

collection is most nearly:

(A) 0.67vo

(B) 0.87vo

(C) vo

(D) 1.15 vo

(E) 1.73vo

m, 2 vo

300

300

2 m, vo

Type: Before to After impulse-momentum method collision sub-type

Rule 1 is to conserve momentum vectorially so a coordinate system is adopted.



2 m, vo

m, 2 vo

300

300 x

y

The momentum of the particles before the collision are 1 0 3 2 ˆ ˆ(2 )[ ]½p m v i j and

2 03

2 ˆ ˆ(2 ) [ ]½p m v i j . The total mass is 3 m and the total momentum is 0 3 2 ˆ4 [totP mv i

] .

03

2 ˆ4 [ ] 2 ˆ3 3

totalfinal

total

P mv i vvM m

0 i (inelastic; stick together) D

Was kinetic energy converted to other forms as a result of the collision?

The collision was totally inelastic. Was all the kinetic energy lost?

(2) Motion in a uniform gravitational field:

Ex. 2) Ball 1 is dropped from a height h and ball 2 is dropped form a height ½ h. Which of the

following gives the ratio of the speed of ball 1 to that of ball 2 just before they impact? (Assume that

air resistance is negligible.)

(A) 2 (B) 2 ½ (C) 1 (D) 2-½ (E) ½

First, the problem is a constant acceleration problem. Write down the master equations.

x(t) = xo + vo t + ½ a t 2; v(t) = vo + a t ; v 2 = vo2 + 2 a (x – xo); vave = ½(v +vo)



The balls fall from rest for a distance H. Using v 2 = vo2 + 2 a (x – xo), v 2 = 2 g H or 2v g H .

11

2 2

22 ½

g Hv hv g H h

= 2 B

Sanity Check. Three answers can be discarded immediately. Which three? Why?

If the answer were (A), ball 1 would have twice the average speed of ball 2. What would be the ratio

of the fall times in that case given that the first ball falls twice as far? Would you expect them to hit

at the same time if dropped at the same time?

(3) Law of Universal gravitation:

Ex. 3) Two planets of mass m and M respectively have center-to-center spacing R. At what distance

from the planet of mass M do the gravitational forces of the planets cancel each other.

(A) mM R (B) 1 mM R (C) mM R (D) 1 mM R (E) 1 mMR

One need only consider points along the line joining the planets. Why? What must be true about the

directions of two vectors that sum to zero?

Prepare a sketch: A large well-drawn sketch provides the greatest benefit. Assume M > m.

R

d R - d

M m

The distance from M is sought. Assign the symbol d to that distance.

Thought 1: Gravitational forces are attractive. Therefore, the ‘zero – force’ point must be between

the planets and along the line joining them. Compare/contrast this situation with that in Ex. 10.

Thought 2: The force is an inverse square (with distance) law force. 2

2 2 2

( )( )

G m G M R d m R d mMR d d d M d



Think about the result. At the point where their influences balance, the distance from the smaller

planet is smaller than that from the large planet. (Include sanity checks as you proceed through your

solution.) Assume m < M as you reason through the possibilities,

11

Rm mR d d R d dM M mM

E

Since m M is less than one, d > ½ R. The point is farther from the more massive planet. Reread

the question. The distance d of the point from the more massive M planet is requested. The unknown

label d is assigned to the value requested. Always review the question to ensure that you have

answered the question that was asked.

Assume m


Examine your options. The new reading should be less than the original reading. Answer (B) could

be chosen without much more as the one third density means that the mass will be buoyed up by 1/3

of its weight if completely immersed, but one should solve the problem before choosing an answer.

Prepare drawings

T = Wdry

mVg

wVg

T = Wwet

mVg

Wdry = m V g = 3 w V g Wwet =m V g - w V g = (m - w) V g

* The ratio of the density of a material to that of water is called the specific gravity of the material.

Wwet = Wdry ( - w/m) = Wdry ( - /) B

(5) Simple Harmonic Motion

Equation of Motion: x(t) = A cos[ t + ] = C cos[ t ] + D sin[ t ]

Take derivatives to find expressions for v(t) and a(t).

v(t) = - A sin[ t + ]; a(t) = - 2 A cos[ t + ]

Energy Conservation: ½ m v2 + ½ k x2 = ½ m (vmax)2 = ½ k A2 = Etotal mechanical

With the convention that U = 0 when the particle is at the equilibrium position, the energy is all

kinetic as the particle passes through the equilibrium position and all potential at the turning points.

= stiffness inertial propertygk

m I

A mass hanging from a linear spring executes simple harmonic motion about its equilibrium

position. Potential energies are associated with systems of interacting entities. Potential energies are

associated with pairs of things (at least) or the system while a kinetic energy can be owned by a



single entity. The potential energy can be shifted by an additive constant. For SHM, it is often set to

zero for a particle at the equilibrium position.

Ex. 5) A 1-kilogram particle is attached to a spring and exhibits one-dimensional simple harmonic

motion. The particle’s distance (!/* displacement) from the equilibrium position is given by the

expression: y(t) = A sin[ t + /2], where A = 1 meter and = 0.5 rad/s. If the potential energy of the

particle (system) at its equilibrium position is null (or zero), which of the following gives the total

energy of the particle (system)?

(A) 2 J (B) 1 J (C) ½ J (D) 1/8 J (E) 0 J

v(t) = - A sin[ t + ]; a(t) = - 2 A cos[ t + ]

vmax = A; amax = 2 A

½ m v2 + ½ k x2 = ½ m (vmax)2 = ½ k A2 = Etotal mechanical The total energy is kinetic at equilibrium as the potential is set to zero at that point.

½ m (vmax)2 = ½ k A2 = Etotal mechanical

The relation v(t) = - A sin[ t + ] shows that vmax = A = 0.5 m/s. Using m = 1 kg,

Etotal = ½ m (vmax)2 = ½ (1 kg) (0.5 m/s)2 = 1/8 J. D

Alternative: Etotal = ½ k A2 = ½ (m 2) A2 = ½ (1 kg) (0.5 rad/s)2 (1 m)2 = 1/8 J.

Write down the equation that is to be used. Substitute numerical values with units for each symbol

writing the values in the same geometric pattern that was used for the symbols. Adopt procedures

that reduce the chance of making a careless error.

(6) Equilibrium with torques

In equilibrium, the sum of the forces acting on a rigid body is zero, and the sum of the torques about

any axis ( free to choose axis where you wish) acting on the body is zero.

; sr F r F r F r F in

The displacement from the axis to the point of application of the force is r , CCW is positive, and

is the angle that takes you from the direction of r to the direction of F

.



Vector cross product: Extend the fingers of your right hand in along the direction of . Orient your

hand so that the fingers can curl toward the direction of

r

F

. The thumb of the right hand will be in

the direction of .r F

axis r F

line of action of F

r

500

Ex. 6) The figure shows a uniform rod of mass 4 kilograms that is pivoted at one end and supported

by a string at the other end. If the rod is at rest (in equilibrium) the tension in the string is most nearly

(A) 20 N (B) 26 N (C) 31 N (D) 40 N (E) 62 N

Use the figure The rod does not have an assigned length. Assign a length of , 1 m or even 4 m. If

the value is not given, the answer must be independent of its value.

Choose an axis: The pivot point is a natural choice for the axis. Picture the situation. If the tension

were changed, about what point would the rod rotate?

The tension force tends to cause a CCW (+)

rotation so the associated torque about the

pivot is T = + T sin(400). To compute

torque, the weight of the boom can be

assumed to be applied at its center of mass.

The weight has a lever arm of ½ and tends

to cause a CW rotation. W = - W (½ ).

The sum of the torques must be zero so:

0 0 0

½ (4 ) (9.8 )½ ½sin(40 ) sin(40 ) sin(40 )

NkgkgW m gT = 30.5 N C

500

400

T

Tr

W

½400



½ (4 ) (9.8 ) 20 28.3.707 .707

Nkgkg N N

lost calculator approximate using sine of 450; 40 < 45 so r

smaller and T larger for same = r

T

We expect an answer larger than 28 N so response (C) is chosen. The actual value is about 30.5 N.

(7) Relativity Relative speed v

Lengths are contracted along the direction of relative motion, but lengths running transverse to the

motion are unchanged. 21 ( ) ;vc

A clock in motion relative to the observer runs slowly. 21 ( )vc

tt

If a time t = 10 s elapses as measured by a moving clock, and it appears to be running slowly as

viewed by the primed observer, then the corresponding interval t is longer than 10 s.

Consider a primed observer moving at relative to an unprimed observer with the initial

conditions that their respective axes are parallel and the origins coincided at t = t = 0.

ˆv i

Adopt the notations: 2 2and 1 ( ) 1v vc c .

1

[ ]] [c t c t x x x cty yy y z z t t

Ex. 7) A stick of length L lies in the x-y plane as shown. An observer moving at 0.8 c in the x

direction measures the length of the stick. Which of the following gives the components of the

length as measured by the moving observer?

Lx

Ly

(A)

L cos 0.60 L sin

(B)

0.6 L cos 0.60 L sin

(C)

0.6 L cos L sin

(D)

0.64 L cos 0.64 L sin

(E)

0.78 L cos 0.78 L sin

O

y

x

L



The component along the direction of motion should be contracted by a factor of = 0.6 while the

transverse length should be unchanged.

Lx = 0.6 (L cos) ; Ly = Ly = L sin

Note that only answer (C) has an unchanged transverse component. C

The following relativity material is under development.

There are four vectors, and they all transform according to the same transformation rules (Assuming

the same relative velocity and with ˆv i 2and 1vc .).

0 0 1 2 2

1 1 0 3 3

[ ][ ]

V V V V VV V V V V

For the position vector, x0 = ct, x1 = x, x2 = y and x3 = z. The four-momentum is: (p0, p1, p2, p3) =

(E/c, px, py, pz). Note the manner in which factors of c are used to ensure that all components have the same dimensions.

The transformations ensure that there is an invariant metric product. 2 2 2 2 2 2 2

0 1 2 3 0 1 2 3V V V V V V V V2

Invariant interval: s2 = (c t)2 - (x)2 - (y)2 - (z)2 = (c t)2 - (x)2 - (y)2 - (z)2

This result is useful when one considers the interval between two events. If s2 = (c t)2 - (x)2 -

(y)2 - (z)2 is positive, the interval is called time-like and the two events can be causally related. If

s2 is negative the events are separated by a distance greater than that which light can travel in a

time |t|, and the interval is called space-like and the two events cannot be causally related. Points at

http://en.wikipedia.org/wiki/Light_cone

a time-like interval from the current event are in the future and past light

cones of that event and can linked causally to events inside the light cone.

Points at a space-like interval form the event are outside the light cone

and can not be causally related to the event at the apex of the cones.

Similarly: p2 = (E/c)2 - (px)2 - (py)2 - (pz)2 = (E/c)2 - (px)2 - (py)2 - (pz)2

When one considers the momentum of a single particle, the invariant p2 can be evaluated in the rest

frame of the particle with the result that 2particle2 2p m c in all frames where m is the mass of the

particle. The general expression for the moment of a particle is p = (v mc, v m ) where v v is the

standard three velocity of the particle and 21 (v v v )c

.


http://en.wikipedia.org/wiki/File:World_line.svg�


Using the invariant p2 = (E/c)2 - (px)2 - (py)2 - (pz)2 =m c2, E2 = m2c4 + c2 p p = Eo2 + c2 p p

E2 = Eo2 + c2 (v m v)2 = Eo2 + c2 (v Eo v/c2)

2 = (v Eo)2 E = v Eo. The kinetic energy of the

particle is the increase over the rest energy. K = E = E Eo = (v – 1) Eo.

Total energy: E = v Eo.

Kinetic energy: E = (v – 1) Eo 2 3 84

2½ ...v cmv m (small v expansion; binomial)

Sample calculations:

In one frame, event B occurs 4 s after event A and event B is 1 km from A. In a second frame,

event B occurs 5 s after event A. a.) Could event A have caused event B? b.) What is the observed

distance d between the events in the second frame?

a.)s2 = (c t)2 - (x)2 - (y)2 - (z)2 = (3 x 108 m/s * 4 x 10-6 s)2 – (1000 m)2 = 440000 m2 > 0

The interval is time-like so event A could have caused event B.

b.) The interval is invariant sos2 = 440000 m2 = (3 x 108 m/s * 5 x 10-6 s)2 – d2 d = 1345 m.

An electron has a rest momentum (0.511 MeV/c, 0, 0, 0). a.) What is the electron’s energy when it is if

it is moving at 0.8 c relative to the observer? b.) Assuming that the electron is moving in the positive

x direction relative to the observer, what is its velocity if its energy is 1.25 * 0.511 MeV?

Using the invariant p2 = (E/c)2 - (px)2 - (py)2 - (pz)2 =m c2, E2 = m2c4 + c2 p p = Eo2 + c2 p p

E2 = Eo2 + c2 (v m v)2 = Eo2 + c2 (v Eo v/c2)

2 = (v Eo)2 E = v Eo

a.) E = v mc = v Eo = [1 – 0.82]-½ = 0.852 MeV.

b.) E = v mc = Eo [1 – 2]-½ = 1.25 MeV. 1 – 2 = 1/(1.25)2. = 0.6 so v = 0.6 c.

With the other assumptions, the particle is traveling at 1.8 x 108 m/s in the positive x direction.

Alternative solution: p = (v mc, v m v ) (0.511 MeV/c, 0, 0, 0) in the rest frame. Apply the

transformations: 0 0 1 1 0 2 2 3' [ [ ], ' ; '1]; 3p p p p p p p p p p

a.) pc = [0.511 MeV/c - (0)] = c-1 Eo E = Eo ()

Attempt to solve b.) using the transformation method.

ELECTROMAGNETISM AND CIRCUITS:



(8) Basic Circuits and Circuit Elements

The three basic passive elements are the capacitor, resistor and inductor. Each is characterized by its

voltage rule. VC =( 1/C) Q ; VR = I R ; VL = L dI/dt (Note 1/C in the VC relation.)

Series Combos: 1 2

1 1

seriesC C

1C

Rseires= R1 + R2 Lseries= L1 + L2

Parallel Combos: Cparallel = C1 + C2 1 2

1 1

parallel

1R R R

1 2

1 1

parallelL L

1L

Study the location C relative to those of R and L in the voltage relations. Note the inverse location

of C in the voltage relations and hence in the rules for series and parallel combinations.

The voltage across a capacitor is a continuous function of time. Why?

The current through an inductor is a continuous function of time. Why?

Kirchhoff’s voltage rule: The sum of the voltages (potential changes) around a closed path is zero.

Kirchhoff’s current rule: The algebraic sum of the currents into a node is zero.

Elements are in series if the wiring of the circuit requires that the current through the elements is

the same. There can be no branch points between them.

The voltage across a series combination is the sum of the voltages across the individual elements.

Elements are in parallel if the wiring of the circuit requires that the voltage across the elements is

the same. One lead from each goes to a common point and the other two leads from the elements

join at another common point (connection path to join of high conductivity).

The current through a parallel combination is the sum of the currents through the individual

elements.



SERIES PARALLEL

http://en.wikipedia.org/wiki/File:Parallel_circuit.svg

Elements are in parallel if the wiring of the circuit requires that the voltage across the elements is the

same. One lead from each goes to a common point and the other two leads from the elements join at

another common point (connection path to join of high conductivity).

Ex. 8) A capacitor of capacitance 125 microfarads is initially charged such that the voltage across

the plates is found to be 100 volts. If the capacitor is the connected in series to (with) a pure inductor

of inductance 0.2 Henry, what is the maximum value of the current that is observed in the inductor?

(A) 1.0 A (B) 2.5 A (C) 4.5 A (D) 5.0 A (E) 10.0 A

Prepare a sketch:

When the switch is closed, the two elements will be connected in series and parallel. The capacitor initially has a charge Q = C VC = (125 F)(100 V) = 12500 C.

KVR: Q/C + L dI/dt = 0. Using I = - dQ/dt, 2 2

22 2

1 0 compare 0d Q d xQ xdt LC dt

L

C 100 V

125 F 0.2 H

A simple harmonic oscillation at = (LC)- ½.

Q(t) = Qi cos[t + ]; I(t) = - Qi sin[t + ]

The current through an inductor is a continuous function of time. It is zero just before the switch is

closed so it is zero just after it is closed. I(t) = - Qi sin[] = 0 = 0 and Qi is just Q(t = 0) or

12500 F. The frequency is (LC)- ½ = (125 x 0.2 x 10-6 )- ½ = (25 x 10-6 )- ½ = 200 rad/s.


http://upload.wikimedia.org/wikipedia/commons/0/0d/Series_circuit.svg�http://upload.wikimedia.org/wikipedia/commons/5/5a/Parallel_circuit.svg�


I(t) = - Qi sin[t + ] Imax = Qi = (12500 C) (200 rad/s) = 2,500,000 A. B

The potential drops as one moves down in both

elements. Note the pattern in which the static

charges accumulate on the inductor to cause the dI/dt . Starting at the lower left and proceeding CW,

the sum of the potential changes is Q/C – (L dI/dt) =

0.When the current is in the direction chosen as

positive for dI/dt, the charge Q decreases so I = - dQ

/dt. Better: Recall that an LC circuit is an

oscillator. 2

2

1 0d Q Qdt LC

Alternative: maxmax max maxmax

1 100 500 2.50.2

dI V V AI I Isdt L HLC A

The total electric field is the sum of the electrostatic field and the circulating field S CE E E

.

ˆ0S S c B tE d E V E d

ndA Kirchhoff’s law states that the sum of the potential drops is zero. The potential drop between A and

B is . The electric field in the integral is the electrostatic field only! A terminal to 0B

SAE d

terminal integration path can be adopted to avoid the regions with circulating field contributions.

C 100 V

+Q

-Q

+++ 2

2d QdI

dt dt C

100 V

+Q

0.2 H -Q - - -

Switch closed at t = 0.



0b c d a

S S S Sa b c d SE d E d E d E d E

d

0 ( 0) ( 0)S QCE d LdI dt 0

ALTRNATIVE: Ex. 8) A capacitor of capacitance 125 microfarads is initially charged such that the

voltage across the plates is found to be 100 volts. If the capacitor is the connected in series to (with)

a pure inductor of inductance 0.2 Henry, what is the maximum value of the current that is observed

in the inductor? (A) 1.0 A (B) 2.5 A (C) 4.5 A (D) 5.0 A (E) 10.0 A

The energy initially stored initially in the capacitor is all stored in the inductor when the current is a

maximum. ½ L Imax2 = Qmax2/2C. Imax = LC]-½ Qmax =(200 s-1) (1.25 x 10-2 C) = 2.50 A

Energy methods are more efficient when they work!.

Ex. 9) If the V is the potential difference between points I

and II in the diagram above and all three resistance have the

same resistance R, what is the total current between I and II?

(A) V/3R (B) 3 VR (C) 2V/3R (D) 3VR/2 (E) 3V/2R

R R

R

I II

What is not said? The lower conductor continues out of the field of view. The components illustrated

are embedded as a unit in some larger circuit.

R R

R

I II

Put the sub-assembly in a circuit and use Kirchhoff’s Laws.



Solution: Start with the voltage versus current relation: VR = I R. The voltage across a resistor is

equal to the current through it times its resistance. The first conclusion is that I = V/R for an

individual resistor with each symbol representing its value for that resistor.

1st Conclusion: Responses (B) and (D) have incorrect dimensions and so are eliminated.

2nd Conclusion: There is a current V/R in the lowest resistor. A smaller current exists in the upper

branch of two resistors. The currents are to be summed. V/R < I < 2 V/R. They make it rather easy

as only response (E) corresponds to a current greater than V/R. E

Related Techniques: When you encounter a network of passive elements, you should make series

and parallel combinations in sequence to reduce the network. The upper branch resistors are

combined in series (What is required for two elements to be in series?) to yield an effective

resistance of 2R. Then 2R and the R in the lower branch are then combined in parallel to yield a net

effective resistance of 2/3 R. If one is asked about the current, voltages, … for individual elements,

one should start with the fully collapsed network and find all the values. Next, back out one step (in

our case to R and 2R in parallel. Analyze to find all the values. Next, back out one more step and

analyze …. .

(10) Coulomb’s Law and Electrostatics

A physics major should know Coulomb’s law, the Biot-Savart law and Maxwell’s equations.

equation integral form differential form

Gauss’s Law 00

ˆV V

insideQE n dA dV

0E

Faraday’s Law ˆA A BtE d n dA

BtE

Gauss’s Law - Magnetism ˆ 0V B n dA

0B Ampere (Maxwell's 4th) 0 0 0 ˆA A EtB d J n dA

o o o EtB J

Charge Conservation (Continuity) ˆV V

insidet

dQdtJ n dA dV

tJ

Lorentz Force Law q others q othersF q E v B q others q othersF q E v B



/

3 3

2 20 00

( ) ( ) ˆ4 44

( )s

p ss s s s sp sp sp

sp spall r p s all rp s

r rr d r r d r qrr rr rr r

E r

2 r̂ Coulomb for E

30 0 03 3

( )4 4 4

( )s

3s s p S s p S s s p S

pall r p S p S p S

J r d r r r I d r r q v r r

r r r r r rB r

Biot-Savart B

Maxwell’s Equations Explicated

The Helmholtz theorem of vector calculus states that vector fields have two basic behaviors,

diverging and circulating. Maxwell has four equations.

' 4electric flux out divergence

Maxwell s ormagnetic circulation curl

The flux of a field through a surface is ˆF n dA

and in, ˆA

F n dA

is the net flux out of the field out of a closed surface. In the field line picture, this is the number of field lines that start in the

enclosed volume. The net flux out of a small (infinitesimal) volume divided by the volume is the

divergence of the field.

Divergence = Flux out density

The first of Maxwell’s equations 00

ˆV

insideQE n dA dV V

states that electric charge density is a divergence source for the electric field. Equivalently, electric charge starts electric field lines.

The notation V directs that the flux is to be that through the surface V that bounds the volume V.

The circulation integral of a field is C

F d , the integral of the tangential component with respect

to path length around a closed path. The component of the curl of the field in the (RHR) normal

direction to the area enclosed by the path is the circulation divided by the area enclosed by the path

for a small path.

curl = circulation/area or circulation density

The second of Maxwell’s equations ˆA A BtE d n dA

states that the electric field circulates in a region of space in which the magnetic field varies in time. This equation represents

the induction part of the Faraday’s flux rule. The notation A directs that the circulation path is the

path (curve) A that bounds the surface patch A.



The third of Maxwell’s equation ˆ 0V

B n dA

states that lines of the magnetic lines do not start or stop. (They may close on themselves.) Equivalently there is no such thing as magnetic charge.

The magnetic field does not have a divergence source.

The fourth of Maxwell’s equation 0 0 0 ˆA A EtB d J n dA

contains the fact that the

magnetic field circulates around its source which is moving charge. 0steady state

ˆA AB d J

n dA

.

The full equation states that the magnetic field circulates in a region of space in which the electric

field is time dependent. This behavior is the reciprocity action to that described in the second

(Faraday’s) law.

In the case of general time dependence with associated charge acceleration and time dependent

charge densities, electromagnetic radiation is generated.

charge electric field moving charge magnetic field

accelerated charge E-M radiation

Ex. 10) A charge of +Q is placed on the x-axis at x = - 1

meters, and a charge of -2Q is placed at x = + 1 meters,

as shown to the left. At what position on the x-axis will a

test charge of +q experience zero net force?

(A) (3 8) m (B) – 1/3 m (C) 0 m

(D) 1/3 m (E) (3 8) m x

Q -2Q

y

1 m 1 m

One should think about a problem before launching into a solution. The net force on the test charge

due to the two existing charges is to be zero. Two vectors can sum to zero only if they are anti-

parallel. A little reflection makes it clear that the solution points must lie on the (extended) line

joining the two charges. That is: The solution point is somewhere on the x axis.



x

y

-2Q Q

q

x

III

-2Q Q

II I

y

1 m 1 m

In order to sum to zero, the forces must have equal magnitude and opposite direction. For points on

the x-axis, the test charge must lie to the right of both charges or to the left of both charge to meet

the opposite direction requirement. To meet the equal magnitude requirement, the test charge must

be closer to the smaller magnitude charge. That places the solution point in region I, x < - 1 m. The

particular value of x is found using the inverse square law nature of the Coulomb force.

Look for equal magnitudes for points in region three. The distance from Q is | x – (-1)| and the

distance from -2Q is | x – 1|.

22 2

(2 ) 2( 1) ( 1)( 1) ( 1)

k Q k Q x xx x

2 or x2 + 6 x + 1 = 0

This equation has a root (3 8) m in region I and a root (3 8) m in region II. Our previous

arguments have identified region I (x < - 1 m) as only region in which both the equal magnitude and

opposite direction requirements can be met. Recall that quadratic equations often return one

physical root (answer) and one not-so-physical root.

Review your options. Which of the proposed answers corresponds to a point to the left of both

charges? Test them, simplest value first, to see if the equal magnitude condition is met.

You need x < -1 m; only A is possible.

(11) Gauss’s Law, Ampere’s Law and E&M Knowledge Points:

Gauss: 1

0 0ˆ ( )

V VencQE n dA dV

0ˆ

VencQE n dA EA



(a wa y)enco

QEA

directed away

A : The enclosing surface through the point where the field is to be determined is to be made of

portions perpendicular to the field and parallel to it. A : net area that is perpendicular to the field.

A plane = 2 A ; A cyl = 2 r L ; A sph = 4 r2

For applications to conductors, make a sketch. (For equilibrium,) The net charge is on the surface(s).

The field at points in the conducting material is zero. A small patch on the surface is essentially

planar, and the field only pokes outside. Qenc = local A; A = A; localo

E

away.

Ampere:

ˆenc encA AB d I J n dA I

0i i i encAi i

B d B I

o encIB

directed to circulate (RHR)

The encircling path through the point where the field is to be determined is to be made of portions

perpendicular to the field and parallel to it. : the net parallel length along which the field has a

constant non-zero magnitude. long st. wire = 2 r ; solenoid = ; toroid = 2 r

Applied to arbitrary paths: Note the direction of each integration path. Give the value of

for each path illustrated. d B s

CCW: out is positive; CW: in is positive C1: CCW; C2: CW; C3: CW

Cd B s

= o Ienc

1Cd B s

= o Ienc = o [5 A - 2 A] = m

2Cd B s

= o [- 1 A - 5 A] = - 7.54 T m

3Cd B s

= o [- 1 A + 2 A] = 1.26 T m

0 0 ˆenc RHRC A Ad I J n dA B s



The notation C = A means that C is the curve (path) than bounds the area A. The RHR subscript means that the normal to the area is in the direction dictated by the RHR. In the cases above, an

arrow indicating a CCW path means that the normal and hence the positive sense for currents is out

of the page. For CW paths, the positive direction is into the page.

Knowledge points:

(a) Field and potential of a point charge: 0

( )4sp sp

qV rr

; 3 20 0

ˆ( )

4 4sp sp

spsp s

q r q rE r

r r

p

spr : displacement from source point to field point

(b) The field of a uniform spherical shell of charge is zero inside and the same as a point charge with

qtotal located at the center outside.

(c) The field of a spherical ball of charge increases linearly with r inside and is the same as a point

charge with qtotal located at the center for points outside.

(d) The field of a long uniform line of charge is the charge per length divided by 2 times the

distance from the wire. The field is directed away from a uniform line of positive charge.

(e) much, much more …

Ex. 11) A uniform insulating sphere of radius R has a total charge Q and a uniform charge density.

The electric field at a point R/3 from the center of the sphere is given by which of the following?

(A) 2012

QR

(B) 208

QR

(C) 206

QR

(D) 204

QR

(E) 20

34

QR

By knowledge: The field grows linearly inside a uniformly charge sphere so the field has 1/3 of the

strength at r = R or 20

13 4

QR

. The direction is radially away by symmetry. A

By Gauss: enco

QEA

directed away. For the point R/3 from the center, the Gaussian surface is a

sphere surface of radius R/3 concentric with the sphere. (The point at which the field is to be found

must lie on the Gaussian surface. The surface has the symmetry of the problem.)

Qenc = (1/3)3 Q; A = (1/3)2 (4 R2)



(12) Faraday’s Law

ˆA A

BE d n dAt

Faraday’s Flux Rule

= Emf = - Bddt ˆB A B n dA B A

Schematically:

cos

coscos cos

cos cos sin[ ]

Bd d BAdt dt

dB dA dA B BAdt dt dt

dB dAA B BAdt dt

t

pure induction motional motional

transformer slidewire gen. rotating generator

In some cases, one needs to multiply by the number of turns.

Ex. 12) Four meters of wire form a square that is placed perpendicular to a uniform magnetic field of

strength 0.1 Tesla. The wire is reduced in length by 4.0 centimeters per second while still

maintaining its square shape. Which one of the following gives the initial induced emf across the

ends of the wire?

(A) 1 mV (B) 2 mV (C) 4 mV (D) 8 mV (E) 16 mV

The magnetic field is not varying in time, so we will use the flux rule form: Emf = - Bddt .

The easy way: The side s is the perimeter divided by 4. The side at time zero is s(0) = 1 m. At one

second, s(1 sec) = 0.99 m. Using B B A , B(0 s) = (0.1 T) (1 m2) and B(1 s) = (0.1 T) (.992 m2)

Emf = - Bddt

2 2(1 ) (0 ) (0.1 )(1 ) (0.1 )((.99)(.99) ) (0.1 )[1 0.98]1 1

B Bs s T m T m T Vs s

= 2 mV

As a function of time. Use: Area = side2 = (perimeter/4)2



B B A = B side2 = B (perimeter/4)2 = (0.1 ) (4.0 – 0.04 t )

2/16 Tm2

d/dt B =(0.1 ) ( [2 (4.0 – 0.04 t) (- 0.04)]/16) Tm2s-1

At time zero, the magnitude of the emf is |(0.1) ( [8 (- 0.04)]/16)| Tm2s-1 = 2 mV. B

The final technique to compute a motional emf is as ( )v B d B v . In this problem you

have 4 meters of wire moving inward at ½ cm/s at t = 0. Bv (0.1T)(4m)(0.005 m/s) = 2 mV.

OPTIC, WAVES, THERMO AND STAT:

(13) Properties of an E&M plane wave and the Poynting Vector

EM waves are transverse which means that the electric and magnetic fields are perpendicular to the

direction of propagation of the wave. The electric and magnetic fields are mutually perpendicular.

The wave propagates in the direction of the Poynting vector, 10( )S E H E B

. The E and

B fields oscillate in phase with one another and B/E = v, the wave speed (= c in vacuum). The

Poynting vector has units W/m2 and represents the power flow of the wave.

Energy Densities: E = ½ o E2 ; B = ½ 2/o

Ex. 13) A linearly polarized electromagnetic plane

wave carries energy in the positive z direction. At

some positions and time t, the magnetic field points

along the positive x-axis, as shown in the figure to the

right.

r

At that position and time , the electric field points

along the

(A) positive y-axis (B) negative y-axis

(C) positive z-axis

(D) negative z-axis

(E) negative x-axis

x B

y

z



EM waves are transverse which means that the electric and magnetic fields are perpendicular to the

direction of propagation of the wave. The electric and magnetic fields are mutually perpendicular.

As the wave propagates in the z direction and at that point and time has the magnetic field in the x

direction, the electric field must be in the y direction. The Poynting vector represents the

intensity of the wave times its propagation direction. Find the direction of .

S

S

10 ( )S E H E B

k

10

ˆ ˆˆ ˆ ˆ ˆ( ) and and ( )S E B j i k j i

The electric field is in the negative y direction at that point and time. B

(14) Diffraction and Interference Standing waves

As a first guess, diffraction limits angular resolution to min = /d where d is the width of the aperture.

(If the aperture is circular, use min = 1.22 /D where D is the diameter of the aperture.)

The N slit pattern constructive interference condition: d sinbright = m where m is the order of

the interference.

Array of equally spaced finite slits. A array of 100 identical slits spaced by 32 m and of width a

yields a diffraction pattern in which the first three orders of the multiple slit pattern appear, but the

fourth m = 4 peaks are missing (dark). What is the width a of the individual slits?

Intensity pattern = single slit * multi-slit pattern

The single slit pattern zeros: a sin = m m 0

The multi-slit bright: d sin = m m = integer

The fourth bright is missing: d sin = 4 anda sin = m

m = 1, because this is the first missing peak.

432sin 8.0

4 4 4d a d ma m



The figures below show the intensity pattern of six narrow (width


Reflection: The fraction of the light of each of the two polarizations that is reflected depends on the

angle of incidence. At Brewster’s angle, the refracted ray make a 900 angle with respect to the

reflected ray, and the reflected light is 100% polarized perpendicular to the plane of incidence. n1

sinp = n2 sin2 and p + 2 = 900 so sinp = cos2 tanp = n2/n1.

Birefringence (double refraction): Certain materials have different indices of refraction for the two

polarization so the two polarizations are refracted at different angles leading to a separation of the

incident beam into two beams of pure linear polarization (that are polarized perpendicular to one

another).

Scattering: The electric field accelerates the charges in material along its direction. The scattered

light has electric field (polarization) components in the direction of the charges acceleration that are

perpendicular to the direction of propagation.



Polazation, polarizers: The interaction of E-M waves with matter is described by the Lorentz force

law, . For a wave, |B| c-1 E so Fmagnetic (vq/c) Felectric. The electric

interaction dominates so the polarization of the wave is associated with the direction of the electric

field. That field is transverse to the direction of propagation and there are two independent transverse

directions so there are two independent polarizations for a given propagation direction. These

polarizations might be chosen to be vertical and horizontal linear polarizations or left- and right-

handed circular polarizations. Circular polarizations carry angular momentum along the direction of

propagation. Linear polarizations carry zero net angular momentum. Natural (unpolarized light) can

be assumed to be an equal mix of the two independent polarizations.

q others q othersF q E v B

A linear polarizer absorbs the component of the polarization perpendicular to its pass axis. It does

this by adding a wave polarized perpendicular to the pass axis that is out of phase with the

incident wave and of equal amplitude.



The incident wave has an E field that makes an

angle w.r.t. the pass axis. The charges in the

polarizer move perpendicular to the pass axis and

absorb energy effectively adding a field

perpendicular to the pass axis that is out of

phase. Superposing, the perpendicular part is

eliminated on only the part along the axis passes.

Incident

components

added by polarizer

transmitted

pass axis

The wave immediately after a polarizer has a polarization parallel to the pass axis and an average

electric field strength of Eincident cos. As the intensity of a wave is proportional to the square of its

amplitude, Itransmitted = Iincident cos2. (The Law of Malus.) Natural (unpolarized) light is an equal

mix of the two polarizations. If natural light is incident on a polarizer, light of one-half the intensity

that is 100% polarized along the pass axis is transmitted.

Io natural ½ Io vertical ½ Io cos2 along last pass axis ½ Io cos2cos2 along last pass axis

One half the intensity of the incident intensity of natural light passes through the first polarizer. All

the light that passes is polarized along the pass axis direction. Incident polarized light obeys the Law

of Malus, and the transmitted light is 100% polarized along the pass axis of the last polarizer.

(15) Lens Problems



(Ex. 15) Two thin converging lenses A and B, each having a focal length of 6 centimeters. are

placed 10 centimeters apart, as shown in the figure above. If an object is placed 10 centimeters to

the left of the lens A, the final image is

(A) 30 cm to the right of lens B (B) 30/11 cm to the right of lens B

(C) 30/10 cm to the right of lens B

(D) 30/11 cm to the left of lens B

(E) 30/10 cm to the left of lens B

A positive lens increases the convergence of the light that passes through it. Light diverges from a

real object. Real objects have positive object distances. Light converges to a real image with a

positive image distance.

The focal length of the lenses in this problem is f = 6 cm. The object is 10 cm before the left lens

and so it has an object distance + 10 cm. Using the lens equation,

Obj Img

1 1 1d d f

Img Img Img

1 1 1 3 1 5 1 210 6 30 30 30d d d

Lens A would form an image 15 cm to its right if there were no second lens. That would be 5 cm to

the right of lens B. Lens B is converging so the rays should converge closer to lens B than 5 cm.

(Lens B is converging so we are focusing on responses (B) and (C).) The appearance of 5 cm and 6

cm makes the 30/11 cm value look attractive.

The light arriving at lens B is converging. That case is called a virtual object and corresponds to a negative object distance. (Light diverges from a real object and converges to form real image. These cases correspond to positive object and image distances. Converging light incident on a lens and diverging light leaving the lens corresponds to virtual objects and virtual images negative distances for the lens formula.) Rays are converging as they reach the second optic lens two has a virtual object negative object distance. The rays as converging to a point 5 cm after the lens dObj,2 = - 5 cm. Applying the formula to lens B,

Obj Img

1 1 1d d f

Img Img Img

1 1 1 6 1 5 1 15 6 30 30 30d d d

1

The image distance is positive so the light converges to a point on the side of the lens opposite to

that from which the light was incident. right side at 30/11 cm. B



(1)

(3)

(2)

Rays actually converge atThe position of a REAL image

First lens. Three rays: (1) in parallel, out through focus; (2) in through focus, out parallel; (3) straight

o trace the second lens, you generate construction rays that are directed to converge at the position

through the lens center. Shows the image point if there were no second lens.

T

of the lens that would be formed by the first lens in the absence of the second.

The blue rays shows the final image after the action of the second lens. point.

s, out through

final

converging as they reach the second optic lens two has a virtual object negative

Real image: the rays actually converge to and pass through a real image

Second lens: Three rays: (1) in parallel directed at the image point of the first lenfocus; (2) in through focus directed at the image point of the first lens, out parallel; (3) straight through the lens center directed at the image point of the first lens. Shows the intermediate andimages. Rays areobject distance. The rays as converging to a point 5 cm after the lens dobj,2 = - 5 cm.



Virtual image: The rays appear

to diverge from an image point

that the rays did not converge to

and pass through. Rays are

diverging immediately after the

last optic so the image is virtual.

(16) Basic Wave Properties

Basic Wave Plug*: v = f The wave-speed is the frequency of the wave times its wavelength.

THE BASIC WAVE PLUG (BWP) is the most important wave property relation.

Wave Speed: v = /

Tension stiffnessmass length inertial density

Traveling Wave: y(x,t) = A sin[ k x - t + ] k = 2; T = 2

2 radians per cyle k *(cycle distance ) = 2 * (cycle time T) = 2



Standing Wave: y(x,t) = A cos[t + ] sin[ k x + ]

node to node spacing: ½ node to anti-node spacing: ¼

Wave types: longitudinal and transverse

There are two transverse directions so transverse waves have polarizations

Waves reflect with a sign change when they reflect from a region with lower wave speed.

Waves reflect with a sign change when they reflect from a fixed end.

Waves reflect with no sign change when they reflect from a region with higher wave speed.

Waves reflect with no sign change when they reflect from a free end.

(Ex. 16) Two identical sinusoidal waves travel in opposite directions in a wire 15 meters long and

produce a standing wave in the wire. The traveling waves have a speed of 12 meters per second and

the standing wave has 6 nodes, including those at the two ends. Which of the following gives the

wavelength and frequency of the standing wave?

Wavelength

Frequency

(A)

3 m 2 Hz

(B)

3 m 4 Hz

(C)

6 m 2 Hz

(D)

6 m 3 Hz

(E)

12 m 2 Hz

Prepare a sketch! The four interior nodes divide the wire into five equal length segments, node-to-

node spacings. Thus, 5 (/2) = 15 m or = 30 m/5 = 6 m. From the BWP, f = v/ = 12/6 s -1 = 2 Hz. C

(17) Kinetic Theory of Gases

Gas molecules have an average translational kinetic energy of 3/2 k T (three-halves times

Boltzmann’s constant times the absolute temperature). The pressure arises due to the many



molecular collisions per second with the walls so it increases for higher speeds v (v increases if

temperature increases) and for higher density. The molecules have a cross sectional area for

interaction and so sweep out a volume v t in time t. The mean time between collisions is found

as the time for a molecule to sweep out the average volume per molecule.

v tcollision free = V/N so tfree = V/Nv

The mean free path free is the distance traveled to sweep out V/N. free = V/N.

Ex. 17) Cubical tanks X and Y have the same volumes and

share a common wall. There is 1 gram of helium in tank X

and 2 grams of helium in tank Y, and both samples are held

at the same temperature. Which of the following is the same

for both samples?

(A) the number of molecular collisions per second on the common wall

(B) the average speed of the molecules

(C) the pressure exerted by the helium

(D) the density of the helium

(E) the mean free path of the molecules

The number of collisions per second, pressure and density for sample Y are double the X values. The

mean free path for sample Y is half that for sample X because there are twice as many things to hit in

the same volume. The average speed depends on the temperature and molecular mass. It is identical

for samples X and Y. B

(18) Ideal Gas Law

isothermal: process at constant temperature

isobaric: process at constant pressure

iso-entropic: process at constant entropy

adiabatic: process in which there is no heat transfer; insulated from the environment

isochoric, isometric, iso-volumetric -------- process at constant volume

The Ideal Gas Law: P V = n R T P V = N k T



number of moles: n

ideal gas constant R = 8.31 (J/ mol-K)

number of molecules: N

Boltzmann’s constant: k = 1.38 x 10-23 J/K

LaChatlier’s Principle: A stable system acts to counter any change in its parameters. If a the

volume of a sample of gas is reduced, its pressure increases to fight additional decreases in volume.

(Ex. 18) If one mole of an ideal gas doubles its volume as it undergoes an isothermal expansion, it

pressure is

(A) quadrupled (B) doubled

(C) unchanged

(D) halved

(E) quartered

The process is isothermal (same temperature). P1 V1 = n R T = P2 V2

12 1

2½VP P PV

1

D

Use ratios whenever possible.

(19) Thermodynamics and efficiency

Entropy Change: revQST

Ideal Heat Engine Efficiency: high low

high

T Te

T

Use absolute temperatures. 0 0C = 273 K

Ex 19.) A power plant takes in steam at 527 0C to power turbines and then exhausts the steam at

127 0C. In any given time, it consumes 100 megawatts of heat energy from the steam. The maximum

output power of the plane is

(A) 10 MW (B) 20 MW (C) 50 MW (D) 75 MW (E) 100 MW

The problem statement is a little fuzzy with regard to its use of the terms power and energy!



Using 0 0C = 273 K, Thigh = 800 K and Tlow = 400 K. The ideal thermal efficiency for a reversible

engine operating between these temperatures is 50% so the maximum possible mechanical/electrical

power out is 50 MW. C

QUANTUM MECHANICS AND ATOMIC PHYSICS:

Particles have associated wave properties. As a crude characterization, particles propagate like

waves and interact as (point) particles. It may be helpful to assume that each particle has a guide

wave that feels out the space to generate the probability distribution for the particle’s location.

Waves such as optical radiation also display particle-like character. When they interact, energy

transferred into and out of the wave in quanta (photons) each with energy hv (and momentum hv/c.) 34 34; ; 6.626 10 ; 1.0546 10h p k E hv h Js J s

The information that can be known about a system is encoded in a wavefunction (x,t) that satisfies

the Schrödinger equation. The probability to find the particle between x and x + dx is

(x,t)(x,t) dx (Born interpretation). Additional information can be extracted from the wavefunction

using the operator for the dynamical quantity of interest. The operator for x is x. The operator for px

is - i x. Dynamic quantities in Hamiltonian mechanics are functions of position, momentum and

time. The quantum operator for the quantity is formed by keeping the coordinates and the time while

replacing each momentum pq with –i/q. px - i x; p- i ; …..

Q(x,y,z, px,py,pz,t) (x,y,z, –i x, –i y, –i z, t); K Q̂2

2

2m ; ( ) ( )V r V r

Lz = xpy – ypx ˆ ( ) (z y xL x i y i )

Shrödinger’s equation identifies two operators for energy, the Hamiltonian and i t. In other

words, the Hamiltonian represents the total energy (kinetic + potential), and it is the time-

development operator for wavefunction (multiplied by –i/).

22ˆ ( , ) ( ) ( , ) ( , )

2H r t V r r t i r t

m t

(Use separation.)



22( , ) ( ) ( ); ( ) ( ) ( ); ( ) ( )

2 n n n n n nr t u r T t V r u r E u r E T t i T t

m t

The values En are the energy eigenvalues, the total energies, a set of real values that are

characteristic of the problem and the associated temporal dependence is: Tn(t) = nE

n i ti te e .

A general solution has the form, ( , ) ( )n nall states n

ni tr t a u r e .

( ,0) ( ) ( , ) ( )n n n nall states n all states n

ni tr a u r r t a u r e

Operators operate on wavefunctions to return other (or perhaps the same) wavefunctions.

Operators operate on their eigenfunctions to return a scalar multiplying the same wavefunction.

Eigenvalue equation: px u(x) = –i x u(x) = (momentum eigenvalue) u(x) = p u(x)

–i x u(x) = p u(x) u(x) = A ( )i xe p

A e i (kx - t) is a momentum eigenfunction with eigenvalue xp k

H n(x,t) = En n (x,t)

Operators for physical observables are Hermitian (or self-adjoint) an, as such, they have real

eigenvalues and expectation values.

Wavefunction Behavior: 2 22( , ) ( ) ( , )n

mr t E V r r t

In a classically allowed region, the

kinetic energy K, E – V, is greater than zero so the net curvature of has a sign opposite to that of .

The function value oscillates back and forth across zero. In the case that V > E or K < 0 (which is

forbidden classically; regions for which V > E are classically forbidden), the curvature and function

have the same sign behavior like growing and decaying exponentials.

All the functions oscillate in the classically allowed region and decay in the forbidden regions. The

decay is more rapid for larger energy deficits. Each higher state has an additional node.

Notice that the spatial rate of oscillation increases as the kinetic energy increases (or the ‘local

wavelength decreases).



-10 -5 5 10-0.2

0.2

0.4

2 22 2

2(2 )

2kK m 2m

The n = 36 QHO wavefunction is plotted for t = 0. Note the more rapid oscillation around x = 0 where the

kinetic energy is greatest.

Free Particle States for piecewise constant potentials are of the form (i kx te ) where

2 ( )mk E V x . At any abrupt change in V(x), the wave is partially transmitted and partially reflected. The wave function ca tunnel through classically forbidden regions (V > E k = i and the

wave is like ex).

Fails to show the reflected wave!

Energy eigenstates are stationary the associated probability density * is time-independent.

2( , ) ( ) ( ) ; ( , ) ( , ) ( )n

n n n n n nn

Ei ti tr t u r e u r e r t r t u r



Energy Level Spacing: Energy levels in an infinite well with width a have the form 2 2

222

nm a

so

the n to n + 1 spacing is 2 2

2(2 1) 2n

m a

; the level to level spacing grows with n and is larger for

smaller a (tighter confinement). The finite-well states penetrate into the forbidden region and so are

less tightly confined leading to lower energies as compared to the infinite well with the same inside

width. For the hydrogen atom problem, the electron’s range approaches infinitely wide as the energy

approaches zero from below. The level spacing approaches zero in this limit. For positive energies,

the particle is not confined and the allowed energies run continuously (zero spacing).

Transitions between levels: A quantum system absorbs or emits energy in chunks or quanta equal

to the energy difference between the initial and final state of the system.

A hydrogen atom in an n = 4 state can make a spontaneous transition to an n =2 state emitting a

photon of energy 2.55 eV.

- 13.6 eV/42 – (- 13.6 eV/22) = 2.25 eV = 486 nm a fantastic shade of blue or blue-green!

Commutation: If the operators for two quantities do not commute, then those quantities have a

minimum uncertainty product. [x,px] = i x px ½ .

General Uncertainty Relation: 2 2 12ˆ ˆ( ) ( ) [ , ]A B iA B A B

If operators commute, they can have simultaneous eigenvalues. For the hydrogen problem, H, L2 and

Lz (plus the operator for electron spin) commute. The hydrogen atom states are labeled by nm

representing the eigenvalues of energy ( - 13.6 eV/n2), orbital angular momentum squared ([+1]2)

and z component of orbital angular momentum (m) (plus spin – up or down).

Generalized Uncertainty Principle. Experimental measurements are performed on systems in

order to extract average values and the corresponding uncertainties. In quantum systems, constraints

can arise between two uncertainties due to the nature of the corresponding operators. Given two



operators A and B, and knowledge of the commutator [A,B] = iC, then there exists a hard lower limit

on the product of uncertainties A B >= ½ ||. Example: [x,px] = i x px ½ .

If two operators commute, then there is no lower-level uncertainty constraint on knowledge

of the simultaneous values of operators (the outcomes of measurements), and the expectation values

of these operators serve as a very useful label to identify and characterize a system. For the bare

Coulomb potential (simple H-atom problem), the operators H, L2, and Lz commute, so the values of

the three can be known exactly at the same time. Therefore we identify individual states by the

eigenvalues (energy, a.m. magnitude, a.m. z-comp) or their corresponding labels n,,m (quantum

numbers).

Expectation value: 3ˆ*( , ) ( , )everywherethatis non zero

Q r t Q

r t d r

Mixed State: ( , ) ( )n nall states n

ni tr t a u r e more than one an 0. Here, an is the amplitude to be

found in state n and | an |2 is the probability for the system to be found in state n. As energy

eigenstates have been used above, E 2n nn

a Eall states . A precise measurement of the energy will

yield one of the energy eigenvalues. For a set of identically prepared systems, the eigenvalue En will

be found with probability | an |2. Making a precise measurement and finding the value En places the

system in a state that has that energy. If a measurement on the state:

100 100 210 200 211 210 211 211 21 1 21 1 310 310hydrogen a a a a a a

returns the energy, - 3.40 eV, the energy for the n = 2 states, then the atom is left in a state of the

form by the measurement: 210 200 211 210 211 211 21 1 21 1after b b b b immediately after the

measurement. (All the n = 2 parts remain; the parts for other n’s are removed.) The act of making a

precise measurement of a physical observable collapses the state to a state that is consistent with the

measured value. (Removes the parts that are not consistent; preserves remainder with relative phases.)

After the collapse, the state will time develop as directed by the Hamiltonian, and the state may not

be consistent with the measured eigenvalue after some time has passed. If the operator for the

eigenvalue commutes with Hamiltonian, the expectation value for the quantity will be time

independent.



ˆˆ ,d Q i H Q

dt tQ

time development of expectation values

ˆˆ ,d Q i H Q

dt

for operators Q without explicit time dependence

Time independent operators that commute with the hamiltonian have time independent expectation values.

Example with non-zero commutator: p i ˆ , (

d p i H p V rdt

)

The time rate of change is momentum is the ‘force’. Ehrenfest’s Theorem: Quantum mechanical

expectation values obey the corresponding classical equations of motion.

Every Hermitian operator has a complete set of eigenstates (Q m = qm m) which can be used as an

basis set for representing a general solution. ( , ) ( , )m meigenstates m

r t c r t . It follows that:

2m m

eigenstates mQ c q . The value |ck|2 is the probability that the measurement returns the eigenvalue

qk for the state k. A precise measurement of Q will return an eigenvalue and place the system in a

state with that eigenvalue immediately after the measurement. The eigenvalues of the measurement

operator are the only possible outcomes of a good measurement.

(20) Ionization and the Periodic Chart

According to L. Pauling, “the power of an atom in a molecule to attract electrons to itself.”

Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons.

Electro-negativity increases from bottom to top of a column.

Electro-negativity increases from left to right across a row.



→ Atomic radius decreases → Ionization energy increases → Electronegativity increases →

The inner electrons screen the nuclear charge seen by the outer electrons. The apparent unscreened charge increases as you move across the table filling an electron shell. It is a maximum for group VIII and a minimum for the single electron outside a closed shell in group I. The effective unscreened charge increases as you move down the table to the higher Z atoms. Group I has one electron outside a closed shell and are the easiest to ionize (easiest at top after H; lowest screened charge). In group VII, the effective charge gets higher as you move down the table most electron hungry (highest electron affinity). Electron affinity increases as you move down and across from I to VII. VIII is the closed shell nobles; they wish neither to give nor receive electrons. Outer valence shell has two S plus 6 P states.


http://en.wikipedia.org/wiki/Atomic_radiushttp://en.wikipedia.org/wiki/Ionization_energy


Electron Shell Filling Sequence: Valence electrons appear rightmost

Z Atom Configuration Comments 1 H (1s) 2S½ Filling the 1s shell ; single active electron 2 He (1s)2 1S0 Shell filled spherical, zero angular momentum 1S03 Li (2s) 2S½ Filling 2s; single active electron outside closed shell 4 Be (2s)2 1S0 Sub-shell filled spherical, zero angular momentum 1S05 B (He)(2s)2(3p)1 2P½ Filling the 2p sub-shell 6 C (He)(2s)2(3p)2 3P0 Maximize unpaired electrons to minimize exchange energy 7 N (He)(2s)2(3p)3 4S3/2 Half filled sub-shell; max net spin J = |L – S| 8 O (He)(2s)2(3p)4 3P2 Spin decreases as electrons added pairing previous ones 9 F (He)(2s)2(3p)5 2P3/2 J = |L + S| 10 Ne (He)(2s)2(3p)6 1S0 Sub-shell filled spherical, zero angular momentum 1S0 11 Na (Ne)(3s)1 2S½ single active electron outside closed shell 12 Mg (Ne)(3s)2 1S0 Shell filled spherical, zero angular momentum 1S0 13 Al (Ne)(3s)2(3p)1 2P½ Filling the 2p sub-shell 14 Si (Ne)(3s)2(3p)2 3P0 Maximize unpaired electrons to minimize exchange energy 15 P (Ne)(3s)2(3p)3 4S3/2 Half filled sub-shell; max net spin J = |L – S| 16 S (Ne)(3s)2(3p)4 3P2 Spin decreases as electrons added pairing previous ones 17 Cl (Ne)(3s)2(3p)5 2P3/2 J = |L + S| 18 Ar (Ne)(3s)2(3p)6 1S0 Sub-shell filled spherical, zero angular momentum 1S0 19 K (Ar)(4s)1 2S½ single active electron outside closed shell 20 Ca (Ar)(4s)2 1S0 Shell filled spherical, zero angular momentum 1S0 21 Sc (Ar)(4s)2(3d)1 2D3/2 Filling inner sub-shell hidden inside the 4s electrons 22 Ti (Ar)(4s)2(3d)2 3F2 The full set has similar chemical character. 23 V (Ar)(4s)2(3d)3 4F3/2 Transition metals filling d sub-shell* 24 Cr (Ar)(4s)2(3d)4 7S3 25 Mn (Ar)(4s)2(3d)5 6S5/2 26 Fe (Ar)(4s)2(3d)6 5D4 27 Co (Ar)(4s)2(3d)7 4F9/2 28 Ni (Ar)(4s)2(3d)8 2F4 29 Cu (Ar)(4s)2(3d)9 2S½ 30 Zn (Ar)(4s)2(3d)10 1S0 Sub-shell filled spherical, zero angular momentum 1S0 31 Ga (Ar)(4s)2(3d)10(4p)1 2P½ Filling the 2p sub-shell 32 Ge (Ar)(4s)2(3d)10(4p)2 3P0 Maximize unpaired electrons to minimize exchange energy 33 AS (Ar)(4s)2(3d)10(4p)3 4S3/2 Half filled sub-shell; max net spin J = |L – S| 34 Se (Ar)(4s)2(3d)10(4p)4 3P2 Spin decreases as electrons added pairing previous ones 35 Br (Ar)(4s)2(3d)10(4p)5 2P3/2 J = |L + S| 36 Kr (Ar)(4s)2(3d)10(4p)6 1S0 Sub-shell filled spherical, zero angular momentum 1S0 The fine structure energies complicate the filling sequence post Kr so we stop. *The sequences of elements arising while an inner f sub-shell is filling are called rare earths.


Fields Test Example Questions Guide Fields Test Example Questions Guide




Electron shell filling sequence. Except for (H, He), the first electron in an S shell leads to an alkali metal (strong electron donor). Two s electrons yield a metal. The nth valence shell consists of the two ns and six np states. Beginning with period n = 4, inner shell (d and f electrons of lower n) fill after the nS, but before the nP. Shells fill first with spins parallel (say up) and add spins anti-parallel only after the sub-shell is half filled. Elements with approximately half-filled d and f sub-shells exhibit magnetic properties associated with the unpaired spins. The D, F, shells fill inside the valence electrons and have little impact on bounding. The transition metals and rare earths are the large group of elements differing by the number of electrons in the partially filled D and F sub-shells.

Ex 20.) For which of the following elements is the ionization energy of a neutral atom the lowest?

(Z) is the atomic number.

(A) Oxygen (Z = 8) (B) Fluorine (Z = 9) (C) Neon (Z = 10)

(D) Sodium (Z = 11) (E) Magnesium (Z = 12)

The first period of elements includes hydrogen and helium with one and two electrons. In the next

period (3 , there are up to eight electrons in the valance shell. Oxygen and fluorine need

electrons to close the shell and so are hard to ionize. Neon is the very stable closed shell

configuration. Sodium has just one electron outside a closed shell and so is easy to ionize. For

Magnesium, the nucleus plus inner electron shells combined have an effective charge of plus 2 and

so the screened nuclear charge is larger and it binds the outer electrons more tightly than does the

sodium nucleus plus inner shell electrons (net charge +1). D

Sodium is a light alkali metal and so is an electron donor. In compounds such as NaCl, it is strongly

ionic with the sodium being positive which means its electron has transferred to the chlorine. After

hydrogen, the Group I atoms are willing electron donors and easy to ionize. Group 7 is the group of

the strongest electron acceptors. Top Left strong donor; Bottom Right (grp.7) strong acceptor.

(21) Hydrogen Atom spectrum levels, …



2 2 213.6 1 113.6upper lowerupper lower

neV cE E h h E E eV

n n

n

20

0 2

4aZe

, ao = 0.0529 nm for hydrogen, reduced mass, Z nuclear charge

The radius varies as (the mass of the orbiting particle)-1 and as (the nuclear charge)-1.

H-atom Transition Table

Lyman Series (n 1) Series Limit: nupper = to nseries

Transition Series Limit Designation L L L L L L: 13.6 eV Wavelength nm 102 nm 97.2 nm 94.9 nm 93.8 nm 91.2 nmColor VUV VUV VUV VUV VUV VUV

Balmer Series (n 2)

Transition Series Limit Designation H H H H H H: 3.4 eV Wavelength nm 486.1 nm 434.1 nm 410.2 nm 397.0 nm 364.6 nmColor red blue-green violet violet violet UV

Paschen Series (n 3)

Transition Series Limit Designation P P P P B P: 1.51 eV Wavelength 1876 nm 1282 nm 1094 nm 1005 nm 955 nm 820.6 nmColor IR IR IR IR IR near IR

Brackett (4) limit 1458.03 nm; Pfund (5) limit 2278.17 nm; Humphreys (6) 3280.56 nm

Ex 21.) The Paschen series for hydrogen corresponds to transitions that end in states with quantum

number n = 3. The shortest wavelength line in the Paschen series is closest to which of the

following? (The ionization energy of hydrogen is 13.6 eV and hc = 1240 eV nm)

(A) 125 nm (B) 250 nm (C) 400 nm (D) 800 nm (E) 1800 nm

The shortest wavelength would correspond to the transition with the largest E from the highest

bound level down to the n = 3 level. You are given 13.6 eV; You must deduce En = -13.6 n-2.

2 21 1 13.613.6 3 9eVcE h eV

; 9 9 1240 800

13.6 13.6hc eV nm nmeV eV

(22) Blackbody radiation



Total power per area = Stephan Boltzmann constant times T to the fourth power = T 4

Wien’s Displacement Law: peak T = constant = 2.898 x 106 nm K (or vmax = constant * T)

Planck’s Law: 3

2

2 1( )1

hvkT

hvI dv dvc e

2

5

2 1( )1

hckT

c hI d de

Wien: I() is a function of T so the peak occurs

for a set value of T peakT = constant.

Stephan: 0

( ) ?I dv

change of variable 33 3

2 2

2 1 2 111

hv ukT

hv kT hu kTdv duc h c ee h

4 34 4

3 20 0

2( )1u

k u duI dv T Th c e

4 3

3 2 0

21u

k u dh c e

u

Ex 22.) Which of the following describes the effect of doubling the absolute temperature of a

blackbody on its power output per square meter and on the wavelength where the radiation

distribution is a maximum?

(A) The output power is increased by a factor of 16 and the maximum of the distribution shifts to

twice its original wavelength. (B) The output power is increased by a factor of 16 and the maximum of the distribution shifts to

half its original wavelength.

C) The output power is increased by a factor of 8 and the maximum of the distribution shifts to twice

its original wavelength.

(D) The output power is increased by a factor of 8 and the maximum of the distribution shifts to half


(E) The output power is increased by a factor of 2 and the maximum of the distribution shifts to four


Higher temperature suggests higher energy and hence shorter wavelength. Answers (B) and (D) are


http://en.wikipedia.org/wiki/Image:Wiens_law.svg�


viable! The power increase factors of 24 or 23 for (B) and (D). The fourth power should be familiar

as it is Stephan’s law, Power T 4. Wien: peakT = constant. Doubling the temperature moves the

peak to one half of the wavelength. B

hcE hv 1 eV 1240 nm

1 eV photon = 1240 nm 2 eV photon = 620 nm; 4 eV photon = 310 nm

hc = 197 eV·nm (atomic physics) = 197 MeV·fm (nuclear physics)

color (nm) v (1012 Hz) E(eV)

IR > 700 < 428 < 1.77

red 625-700 405-480 1.77-2.0

orange 585-620 484-513

yellow 585-570 513-526

green 570-505 526-594

blue 500-440 600-682

violet 440-400 682-750

UV < 400 > 750 > 3.10

ranges are approximate and arbitrary no universal agreement on the ranges

Ex 23.) Let H denote a Hermitian operator and suppose that H |= a |, where | is an

eigenvector of H. Which of the following is true of the eigenvalue a? (The symbols Re and Im

denote the real and imaginary parts, respectively.)

(A) Re(a) = Im(a) (B) Re(a) = - Im(a)

C) Re(a) = 0

(D) Im(a) = 0

(E) H = a SP352 Knowledge points: The eigenvalues of a Hermitian operator are real. The eigenvectors associated with

distinct eigenvalues are orthogonal. A Hermitian operator can be diagonalized by a unitary transformation. D

Ex 24.) The quantum numbers used to label the radial wave function solutions to the Shroedinger



equation for hydrogen atom are the principal quantum number n and the angular momentum

quantum number . If the principal quantum number is n = 2, which of the following gives the

possible values for the angular momentum quantum number ?

(A) 1, 0 (B) (C) 2, 1, 0 (D) (E) 3/2, 1/2

The states of the hydrogen atom are labeled by the quantum numbers n, and m plus the intrinsic

spin. The allowed values are n = 1, 2, 3, ….; = 0, 1, … , (n -1); m = -, - +1, … , 0, 1, …, ).

That is: m assumes the 2+ 1 values spaced by 1 running from - to . Note: n and are

For n = 2, can be 0 or 1 and m can be -1, 0, +1. A Each combination can be combined with spin

up or down. The value is the orbital angular momentum which assumes integer values. The total angular momentum is the orbital plus spin, and it assumes half-integer values for an electron in hydrogen.

Ex 25.) The three operators (Lx, Ly, Lz) for the components of the angular momentum commute with

the Hamiltonian of a particular particle. Therefore, the angular momentum of the particle is

(A) equal to zero (B) equal to the energy in magnitude

(C

physics comprehensive review · 2010. 9. 22. · physics comprehensive . review . by larry, jeff...

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