Annotated Answers to the Multiple Choice Section of the 1984 AP...

Annotated Answers to the Multiple Choice Section of the 1984 AP Physics B Exam

1. C. The formula for the period of a pendulum does not involve mass; Tpendulum = 2π l

g

2. A. Conservation of momentum: m1v1 + m2v2 = (m1 + m2)v /. Being careful about the signs, we have

2,000(10) + 3,000(–5) = 20,000 – 15,000 = 5,000 = (2,000 + 3,000)v / ; v / = 1 m/sec.

3. C. The area under the velocity–time graph is the displacement, the change in position xf – xo. We are asked to findthe time at which the displacement is zero (i.e., when xf = xo.) That means, find the time when the area under thecurve adds up to zero (don’t forget that area under the graph counts as negative.) Here’s the diagram:

It is fairly obvious that the two areas are equal at a time between 1 and 2seconds (to be precise, at 1.707) 1

1 0

2

2

–1

4. D. The kinetic and potential energies trade off; when one is large, the other is small and vice-versa. This rules outboth A and B. At the endpoints, the energy is entirely potential since v = 0 momentarily; (this contradicts E.), andat the midpoint, the energy is entirely kinetic (since the stretch, x, equals zero there; which contradicts C.)

5. C. Choice D is nearly true if the quantities are taken as vectors, but even then, it isn’t quite right. The true vectorstatement is T1 + T2 + mg = 0, or T1 + T2 = – mg . The reason why T2 > T1 is because T2 is more nearlyvertical. (If T2 were vertical and T1 horizontal, then we’d have T2 = mg and T1 = 0.) The tensions would be equalonly if the angles were equal.

6. E. The normal force is always perpendicular to the surface, and the friction is always parallel to the surface.

7. C. A tricky problem. The first thought is, f = µkN = µkmg cos θ. Unfortunately, they didn’t give us the value of µk.They did tell us the value of a, though; a = 2 m/sec2. We also have the (memorized!) inclined plane equation

a = g sin θ – µkg cos θ.

Multiplying this equation by m gives

ma = mg sin θ – µkmg cos θ = mg sin θ – f.

But ma = 2*2 = 4 nt; mg sin θ = (2)(10) sin 30 = 10, or 4 = 10 – f ; and f = 6 nt.

8. B. Newton’s Third Law. A. violates this law. C. is wrong because the friction pushes the person towards the center(it is the only force that can provide the centripetal acceleration.) D. and E. are simply wrong; in E. we know thatthe frictional force is proportional to the normal force, which is proportional to the mass.

9. B. We can completely ignore the left hand column, Mass, because the mass of the satellite always cancels in orbitcalculations. Set the centripetal force, mv2/R, equal to the gravitational force, GMm/R2. Then

mv2/R = GMm/R2 or v2 = GM/R.

Clearly, as R decreases, v increases, so choose the answer B.

10. E. The kinetic energy to start off is 1/2 mvo2. At any time later, we know vx = vo (no change), and vy = at = gt.

Then v2 = vx2 + vy

2 = vo2 + a2t2, and

K = 1/2 m(vo2 + a2t2)

This is a parabola (since quadratic in time) which does not go through the origin (since vo ≠ 0).

11. E. F = ma = 3a; so a = F/3. The tension T is the net force on the 1 kg mass, so T = ma = 1*(F/3) = F/3.

12. E. As the mass on the spring is increased, the time of oscillation increases (since a larger mass is more reluctant to beaccelerated). The only answer with a time greater than 2 seconds is E. If you want to be exact, use the formula

Tspring = 2 π m

k

Doubling the mass increases the time by √2 = 1.414; 2*1.414 = 2.828. (Calculators were allowed in 1984.)

13. B. A torque problem (the only one on the test, and one of perhaps four on the five multiple choice tests released since1974.)

* P

x

500 nt

1000 nt

2.5 m

•

Take torques about the point P(•) as shown. The gravitational torque on theplank is found by considering all of its mass to be concentrated at its center ofmass, that is, at the center of the plank (denoted *.) This torque tries to turnthe plank counter-clockwise; (from * to the pivot P is 0.5 m)

τcounter-clock = (0.5 m)(100 kg)(10 m/sec2) sin 90 = 500 nt-m

The fool on the plank exerts a torque clockwise on the plank. Let his distance from P be x. Then

τclock = (x m)(50 kg)(10 m/sec2) sin 90 = 500 x nt-m

If the plank is not to rotate, these must be equal, or x = 1 meter.

14. E. The point P is called the perihelion; the point Q is called the aphelion. Recall Kepler’s Third Law, that equal areasare swept out in equal times. That means when r is large, v is small and vice versa. (In fact, Kepler’s Third Law isnothing but the conservation of angular momentum; mvr sin θ = constant, where θ is the angle between thevelocity and position vectors.) Going from P to Q, the speed decreases, the angular momentum stays the same,the total energy stays the same, and the kinetic energy decreases. It must be that if K decreases, U must increase.

Another, though trickier, way to see the same thing: Recall Ugrav = – GMm/R. Your first thought might be,“Hmm... U is proportional to 1/R, the distance from the sun. Since R at Q is greater than R at P, the potentialenergy must be smaller.” But be careful! U is negative. Just as –7 is less than –2, so too –GMm/(big) is larger than –GMn/(small). So U does increase as R increases, even though it depends on 1/R.

15. B. You should have converted 1 eV = 1.6 x 10–19 J enough times by now. Also, a volt is a joule per coulomb, so acharge (in coulombs, or a fraction of a coulomb) times a volt gives an energy (recall Work = qV.)

16. E. The charge moves in such a way as to minimize the potential energy. That is, charges want to maximize theirseparation from each other. If they are free to move, as they are on a conductor, they move to the surface. (This istrue for static charges only; if charges move in a wire, to make a current, the charges are equally distributed overthe entire volume of the wire.)

17. E. First off, the charge Q has nowhere to go (once the battery is disconnected) so the charge remains fixed. Next,recall that the electric field for a capacitor is given by E = 4πkσ = 4πkQ/A. This does not change as the plateseparation is increased. So the electric field stays the same. Finally, V = Ed; if E stays the same but d increases,then V must likewise increase. Put another way, V is the work per charge to move a charge from one plate to theother. The force stays the same since E stays the same, and F = qE. Then the work increases (Work = F∆x = qEd)and so V increases.

Another way to see the same thing: Recall Q = CV, and C = A/4πkd. If d increases, C decreases, but Q stays thesame, so V must increase.

18. B. Just work out the net resistance: I is 2 Ω , II is 4 Ω , III is 1 Ω and IV is 2 Ω .

19. D. For a point charge, E = kq/R2 = (9 x 109)(5 x 10–8 )/(0.5)2 = 1.8 x 103 nt/coul. (Remember, in 1984, calculatorswere allowed on the multiple choice part.)

Annotated Answers to the Multiple Choice Section of the 1984 AP Physics B Exam p.2 v.1.0 25 May 2000

20. C. For a point charge, φ = kq/R = (9 x 109)(5 x 10–8 )/(0.5)= 9.0 x 102 V.

21. A. Consider a short section of the wire, for example at the extreme right edge. To a first approximation a shortsection looks like a straight line. We know that the magnetic force on a straight wire is given by F = Il B sin θwith the direction given by the first right hand rule. On a short section at the extreme right of the circle, we find aforce to the right. Similarly treating the force on the extreme left hand edge of the circular wire gives a force to theleft, and on the extreme upper edge a force upward, and so on. That is, each section of the wire feels a pushoutwards. This tends to make the wire expand in radius.

If the wire had not been in the plane of the paper originally, the pulls would have tended to turn the loop backtoward the page; but since the wire is in the plane of the paper, there is no rotation.

22. B. You can go the full route. The net resistance is 6 Ω (the 3 and 6 in parallel become a 2; and the resultant 2 and 4are in series, which gives 6Ω .) The net current is 6V/6Ω = 1 A. The current splits up at point X; the 3 Ω gets 2/3of the current, and the 6Ω gets 1/3 of the current. That is, I3 = 4/3 A, and I6 = 2/3 A. Then V3 = VXY = 2 V.

Easier: once you find out that Inet = 1 A, then use the Loop Law (battery; 4Ω resistor, 3Ω resistor) to say

6V – 1A*4Ω – VXY = 0; VXY = 2 V.

23. C. This is the definition of the volt. If the work needed to bring a coulomb at constant speed from one point toanother is one joule, then the potential difference between these two points is one joule per coulomb, which is onevolt.

24. A. Recall that Felec = qE. We want to find the least magnitude of the force; that is, as close to zero as possible. Isthere a spot where the electric force on a charge would be zero? Yes: at A. There, the push from each charge isequal and opposite, and so the net force is zero. If the force is zero, the electric field is zero.

Some people no doubt thought about this in a different way: since E is proportional to 1/R2, the least E should bewhere R is the largest. This would lead them to choose point C. But there is still a push from both charges, in thesame direction, to the right. C is certainly the second best choice, but a small value is still larger than zero. If onlyone charge, to the left of the y axis, had been present, C would have been the best choice.

25. A. Now we don’t have to worry at all about direction (since electric potential is a scalar, not a vector.) By definition,

φ = Σ kqi/Ri where qi is the ith charge and Ri is the distance of the ith charge from the point of interest.

To maximize this, we need to minimize the distances. Clearly, the smallest distances from the charges are foundfor point A.

[I think that the examiners were actually trying to teach the candidates something here: that electric potential and electric field are reallyvery different. In this situation, where the field is least, there the potential is greatest. For the benefit of the calculus students, it turns out thatE = – dφ /dr. A function can have a value of zero where its derivative may be very large, and vice-versa.]

26. A. Think of each lamp as a resistor, R. Adding as many lamps in parallel as we like will not change the voltmeterreading (if the battery is ideal, i.e., if it has no internal resistance), because the battery’s job is to maintain aconstant voltage. On the other hand, if we add another resistor in parallel, we change the net resistance from R toR/2 (add two resistors in parallel to obtain 1/Rnet = 1/R + 1/R = 2/R.) That will double the net current. Theammeter is in position to record the net current, so its reading will increase (in fact, it should double.)

27. D. For a fixed voltage source, we have P = IV = V2/R. To maximize the power output, use the least resistance, whichis given by the two resistors in parallel, D.

28. B. Resistors cannot store energy; we need a capacitor for that. The only candidates are B., C. and E. But the capacitorsin C. and E. will have their opposite sides connected (with a wire or a resistor), so charge will move from one sideto another after the battery is disconnected. Only in circuit B will charge have no way of moving from one side tothe other, so it will stay charged, and the energy will be stored indefinitely.


29. C. Faraday’s Law and Lenz’ Law. The loop is losing X’s, so it wants to make up for the loss. That means the currentgoes clockwise. The size of the current is going to be V/R = Blv/R.

30. B. Amusingly enough, the mass of the block cancels. We have Ustart = mgh, so half of this is turned into heat. Thatis, Q = 1/2 mgh. This raises the temperature of the block according to the old familiar equation Q = MC∆T;

Q = mC∆T = 1/2 mgh, or ∆T = 1/2 gh/C = 1/2 (10 m/sec2)(100 m)/(100 J/kg-K) = 5 K.

31. C. “Maximum efficiency” is a code phrase, meaning, “Figure out what the Carnot efficiency would be.” This is

efficiency = 1 – (Tcold/Thot) = 1 – (600/1500) = 60%

32. B. The molecules have a range of speeds. It is true that there is an average speed, and that the temperature isproportional to the average kinetic energy, but the atoms’ energies are spread out over a range from nearly zero toquite large.

33. E. Unfortunately, the examiners use the extremely misleading symbol ∆U for the change in internal energy of a gas,whereas we’ve used ∆E. For dilute monatomic ideal gases, there isn’t any potential energy; the energy is entirelykinetic. So if you were to choose a symbol other than ∆E, you’d be much better off using ∆K than ∆U.

With that bit of editorializing out of the way, here’s the answer. We know ∆E = Q – W, or Q = ∆E + W, and weare told that ∆E is positive. Work is given by the area under the P–V curve, or what is the same thing for anisobaric (constant pressure) process like X→Y, Work = P∆V. This is positive for X→Y, since ∆V is positive.That means that W is positive. Q = ∆E + W = (positive) + (positive), so Q is also positive.

34. C. The choices are immediately narrowed to B. or C., because the process Y→Z is constant volume (isochoric), andW = 0. Then ∆E = Q. We are told Q > 0, so ∆E > 0 also.

35. B. Doppler effect. The waves are bunching up on the right, and so the object is moving towards the right.

no motionof source

v > vsource wave

v = vsource wave

v < vsource wave

Here arethe four possibilities:

F C

Evidently the answer is the second, that vwave > vsource.

36. E. The relevant formula is: sin θcritical = n2/n1. In this problem, θcritical = 30°, so sin θcritical = 1/2, and as n2 =nair = 1, it follows n1 = 2.

37. E. Recall for a circular mirror, R = 2f. The rule for a converging lens or mirror is that if do > 2f, the image isinverted and smaller. But rather than memorize yet another rule, why not just draw the relevant diagram?

38. A. To use a magnifying glass, the object or print to be enlarged is placed on the far side of the lens, and you lookthrough the lens to see the enlarged image. Since the print is not upside down, it must be a virtual image.


39. E. Young’s Patterns; d sin θ = jλ . For the first order maximum, j = 1, so we have

d = 1*(0.02 m)/sin 6° = 0.2 m.

40. B. This requires an educated guess. Recall that for a spring, Tspring = 2 π m

k so it seems at least reasonable that

the time for an oscillation should decrease as the tension increases (as k determines the tension in a spring.) Also,the square root is suggestive that we should have Tstring proportional to 1/√(tension). Since f = 1/T, we can hopethat the frequency of the vibrations is proportional to the square root of the tension. If the tension doubles, then fshould presumably increase by √2. (Anyone who plays a stringed instrument or a piano knows that as the key istightened, the pitch increases. So the only sensible choices would be A. or B.)

[You weren’t supposed to know this, but vwave = T µ

where T is the tension in the string, and µ is the mass per unit length of the string.

Then f = v/λ is likewise proportional to the square root of the tension.]

41. D. The energy lost by the atom is the energy of the light emitted; so choose the most energetic light of the answersgiven: violet.

42. A. This is just the standard graph of the photoelectric effect.

43. E. Intensity does not affect the energy of the individual electrons; it only affects the number of emitted electrons:more light means more electrons.

44. A. Maybe the easiest problem on the test. We just have to make sure that the top numbers add up on both sides of theequation, and the bottom numbers add up on both sides of the equation. That is, we need the unknown top numberto equal 236 – 145 = 91, and the bottom number to equal 92 – 56 = 36.

45. D. The deuteron is the nucleus of Deuterium, also known as “heavy hydrogen”, 2H1. The rule is that the pieces thatmake up a nucleus always weigh more separately than they do if bound in the nucleus. One has to put in a certainamount of work, called “binding energy”, to pull any nucleus apart. The mass equivalent to this energy is given bythe famous formula E = mc2. That is,

mass of individual pieces = mass of nucleus + (mass equivalent of binding energy) or mass of nucleus = mass of individual pieces – (mass equivalent of binding energy)

mdeuteron = mn + mp – (mass equivalent of binding energy)

46. D. Very few students got this one right. First off, you need to know what the de Broglie wavelength is; λ de B = h/pwhere h is Planck’s constant and p is the object’s momentum. For an electron, p = mv, so

λ de B = h/p = h/mv.

If the de Broglie wavelength is to be reduced by a factor of 2, it must be that the momentum increases by a factorof 2. The mass of the electron doesn’t change, so the speed must increase by a factor of 2. The kinetic energy thusincreases by a factor of 22, which is 4.

How does the voltage affect the charge’s energy or momentum? If accelerated from rest, to attain a kinetic energyof 1/2 mv2 requires an equal amount of energy in the form of work. The electrical form of work is qV. In thiscase, q = e, the charge of an electron. That is, we need

eV = 1/2 mv2

To increase the energy by a factor of 4, we need to increase the voltage by the same factor, from 150 V to 600 V. 47.* Relativity. No longer on the syllabus. (The answer is A., because moving meter sticks are measured to be shorter

by stationary people.)


48.* Relativity, again. C., because perpendicular lengths are measured the same by all observers.

49.* Relativity, yet again. A., because moving clocks run slowly.

50. C. P1V1/T1 = P2V2/T2; in this case P1V1/T1 = P2(4V1/2T1) and cancelling common factors we find

P1 = 2P2, or P2 = 1/2 P1.

51. D. Fcentrip = mv2/R; if the forces are the same before and after, we need mv12/R1 = mv2

2/R2 = mv22/4R1 ; and

cancelling common factors gives

v12 = v2

2/4 or v2 = 2v1.

52. C. We are told that the box moves with constant speed. This means that the net force is zero, and the push exactlymatches the friction. The power is

Work/time = F∆x/t = F(∆x/t) = Fv = fv = (µmg)v = (0.25)(40 kg)(10 m/sec2)(0.5 m/sec) = 50 W.

53. E. Faraday’s Law. If we change the flux, we get an induced voltage (and thus an induced current.) The flux changes inthe first four situations because the strength of the magnetic field depends on the distance from the end of themagnet. As the distance between the loop and the end of the magnet decreases, the field through the loop increases,and so the flux increases. This is a change, of course, so we get a voltage induced in the loop. However, in the lastsituation, there is no change of distance between the loop and the magnet. Then the flux does not change, and sothere is no induced voltage or current.

[Note that the only thing that matters is relative motion. It doesn’t matter if we move the magnet or the loop or both, as long as there isrelative motion towards or away. This was first pointed out by the great Albert Einstein in 1905, and marks the genesis of his famous Theoryof Relativity. There is no relative motion in E.]

54. C. The formula for the capacitance of a parallel–plate capacitor is C = A/4πkd, where A is the area and d theseparation between the plates. If we double A and at the same time double d, there is no change in C.

55. A. (A mean question.) We know v = λf, and f = 1/T. Then the distance ∆x travelled is just v∆ t = vT = v/f = λ.A lot of people probably confused T with “one second” and so said “Oh, about 340 meters,” answer B. It is also true that D. and E. are plausible, since the speed of sound is directly proportional to the square root of thepressure, and inversely proportional to the square root of the air density.

[You certainly do not need to know this formula, but for completeness,

vsound = γ P ρ

where P is the air pressure, ρ the density and γ the “adiabatic index”. Amusingly enough, the great Isaac Newton got this one wrong. Heassumed that the expansion and compression of sound was an isobaric process; in fact it is adiabatic. Newton’s formula lacked the factor ofγ. If you put in reasonable guesses for γ (1.4), P (1 x 105 nt/m2) and ρ (1.2 kg/m3), you get vsound = 341 m/sec at 20o C and 1 atm.]

56. D. The question is not well-worded, though it is not hard. Let m be the (common) mass of the two test objects, andlet the masses and radii of the two planets be M1, M2 and R1, R2 respectively. Let one mass m be on one planet,and let the other mass m be on the other planet. Then the parts of the questions are asking the following:

I. The first mass falls with acceleration g1 = GM1/R12 , the second with g2 = GM2/R2

2. Does g1 = g2 ? No.

II. The first mass is balanced against some new mass m/ on the first planet. The second mass is balancedagainst some new mass m// on the second planet. Does m/ = m// ? Yes.

III. The first mass is given an impulse ∆p, and its velocity increases by ∆v1 = ∆p/m. The second mass isgiven an identical impulse ∆p, and its velocity increases by ∆v2 = ∆p/m. Does ∆v1 = ∆v2 ? Yes.


57. B. The resistance is given by the fairly obscure formula R = ρl/A, where ρ is the resistivity, l the length and A thecross-sectional area. Given a variety of wires all made from the same material, The longer and thinner the wire, thegreater the resistivity. (C. has the least resistance.)

58. C. Accelerating charges, according to the work of James Clerk Maxwell, must radiate: they must glow with someform of light (although it might not be visible light; it could be anything from radio waves to gamma waves.)This means the electrons would lose energy. They’d have to slow down; and if they were to slow down, theywould pulled into the nucleus. The atom would fall in on itself. This doesn’t happen. We know today why itdoesn’t happen (the orbits must be a whole number of de Broglie wavelengths) but at the time, it was a bigmystery. (To be honest, it is still a mystery: why is quantum theory true?)

59. E. You have to work this out slowly. According to Coulomb’s Law, the net force is proportional to the product ofthe charges. Initially this is (–2Q)*(Q), or –2Q2, the magnitude of which is 2Q2. Let the two spheres touch, andthe net charge will be –2Q + Q = –Q. This charge will be shared equally (50-50) between the two identical spheres(by symmetry.) That means that each sphere now has –1/2 Q. When the spheres are removed to their originalposition, the force will be proportional to (–1/2 Q)*(–1/2 Q) = 1/4 Q2. The denominator of Coulomb’s Law, R2,is not important here since the charges are just as far apart in both situations. That is, the magnitude of the forcehas gone from 2Q2 to 1/4Q2, which is a factor of 8:1.

60. D. Charged particles in magnetic fields tend to move in circular or spiral paths. The spiral path would indicate a lossof energy (see 58., above.) So the only choices are B. or D. Use the Right Hand Rule, and find that the force on apositive charge would be B. But electrons are negative. Therefore, the answer is D.

61. A. Sliding friction is equal to µkN. The normal force is usually just mg, but due to the broom’s downward push ofT sin θ, the normal force increases to mg + T sin θ. That means f = µk(mg + T sin θ).

62. D. Recall Q = CV. Also, the capacitors are in parallel, so the V’s are equal. Then

Q1/Q2 = (C1V)/(C2V) = C1/C2 = 1.5/1 = 3/2.

63. D. Each wire carries a magnetic field B = 2k/I/R, where I is the current in the wire. Each current is acted on by theother current’s magnetic field according to the formula F = Il B. That is, we can say

F = I1l B2 = I2l B1 = 2k/I1I2l /R.

Since the force is proportional to the product of the currents, it follows that D. is the right answer.

64. C. Easy. P = IV, so for a P of 10 W and a current I of 4 A, we have V = 10 W/4 A = 2.5 V.

65. C. Slightly tricky, but a reasonably simple kinematic. At t = 1 sec, the object has a position of 24 m, a velocity of24 m/sec, and an acceleration of – 6 m/sec2. Then its position at any point is given by

x = xo + vot + 1/2 at2 = 24 + 24 t – 3t2.

Do we now put in t = 11? We do not. We have “reset the clock” to call 1 sec t = 0. So we put in t = 10 sec (!), and

x = 24 + 24*(10) – 3(10)2 = – 36 m.

66. A. The relevant formula is g = GM/R2 . If M is doubled, g doubles; if R is doubled, g goes to one-fourth its originalvalue. The new g will be 2/4 or 1/2 the old g; and the new force will be one half the old force, since F = mg.

67. A. In uniform circular motion, the acceleration is always directed towards the center. Theacceleration can be directed south only when the car is at the top of the circle. We are told thatwhen the acceleration is south, the velocity is east. The situation must be as shown: the motionis clockwise. The velocity satisfies

v2/R = 3 m/sec2 = v2/300 m; v = 30 m/sec.

a v


68. C. Remember Q = MC∆T. In this case, Q = P∆ t (power times time.) That means

P∆ t = MC∆T or ∆T = (P/MC) ∆ t

Where the graph is horizontal, the material is experiencing a phase transition. So that rules out A. Melting beginsat temperature T1 and boiling at temperature T2. This rules out B. If the heat of fusion and vaporization wereequal, since energy is poured in at a constant rate, the lengths of the horizontal pieces at T1 and T2 would be equal.(We’d need equal time to give the sample equal amounts of energy to boil and to melt.) That rules out D. Theanswer E. is false because the specific heat is inversely proportional to the slope of the T vs t graph; if the slope isconstant, then the specific heat is likewise constant. The answer is C. Note that the distance from To to T1 ispretty much the same as the distance from T1 to T2, but that the time needed to reach T2 is much greater than thetime to reach T1; the slope for the heating of the liquid is less than the slope for the heating of the solid. ThenCliquid > Csolid.

69. D. The giveaway here is the negative focal length. That means the lens is a diverging lens, and the image has to bevirtual. Virtual images formed by a single lens or mirror are always upright. We need only to find the distance. Butthis is easy: use the ol’ standby 1/f = 1/do + 1/di . Plugging in gives

– 1/0.15 = 1/0.3 + 1/di or 1/di = – 3/0.3 = – 1/0.1;

di = – 0.1 or 0.1 m on the same side of the lens as the object.

70. B. The electrons are initially in the ground state. That means they have an energy of –14 eV. They can accept photonsonly of 4 eV (to jump to the first excited state), 9 eV (to the second state), 11 eV (to the third), and so on. But thephotons we bombard the atoms with have energies in the range 7 to 10 eV. The only acceptable energy in thatrange is 9 eV. Consequently, a bunch of electrons will jump to the second excited state.

Now that we have electrons in the second excited state, what will happen? Some will drop directly back to theground state, providing light of energy 9 eV. Some will drop in a two step process, to the first energy level(giving off light of energy 5 eV) and then to the ground state (giving off light of energy 4 eV), or three colors inall. The relevant diagrams are shown below.

–1 eV –1 eV–3 eV –3 eV–5 eV –5 eV

–10 eV –10 eV

–14 eV –14 eV

0 0

~~~ γ<

~~~ γ > ~~~ γ >

~~~ γ > 9 eV

4 eV

9 eV 5 eV

Bombarding the atoms with light in therange 7 – 10 eV. Only the 9 eV light is absorbed.

Falling from the second excited state tothe ground state. Three spectral lines are observed.

Here’s a breakdown of the 1984 Multiple Choice by topic:

Unit Problems on test Actual % Target %Mechanics 1–14, 51, 56, 61, 65–67 20/70 = 28.6% 33%Electricity & Magnetism 15–29, 52–54, 57, 59, 60, 62–64 24/70 = 34.3% 25%Heat & Thermodynamics 30–34, 50, 68 7/70 = 10% 10%Optics, Sound & Waves 35–40, 55, 69 8/70 = 11.4% 16%Nuclear & Quantum Physics 41–46, 58, 70 8/70 = 11.4% 16%*Relativity *47–*49 3/70 = 4.3%

Note that the target percentages of mechanics vs. electricity and magnetism were inverted; on the other hand, the 1984 FreeResponse had two mechanics questions and only one electricity and magnetism question.


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