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Lecture Comments (14)

2 answers

Last reply by: Sai A.
Sun Sep 25, 2016 12:56 AM

Post by Anh Dang on September 2, 2015

Regarding the AP Practice Exam, why are we going over the 1998 AP Physics B exam?  Shouldn't we be reviewing a more recent AP Physics 1 & 2 exam which would reflect more of what will actually be on the new exam?
Thank you

1 answer

Last reply by: Professor Dan Fullerton
Thu Jul 31, 2014 11:19 AM

Post by Timothy Holmes on July 30, 2014

Hi Dan, just joined. I do not see any "quick link" to the AP exams

3 answers

Last reply by: Professor Dan Fullerton
Mon May 12, 2014 8:06 PM

Post by Milan Ray on May 12, 2014

angular momentum is not on the new ap tests right?

0 answers

Post by Torrey Poon on April 16, 2014


3 answers

Last reply by: Professor Dan Fullerton
Thu Jul 31, 2014 11:19 AM

Post by Torrey Poon on April 16, 2014

Hey Dan, I don't see the link

AP Practice Exam: Multiple Choice, Part 1

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Problem 1 1:33
  • Problem 2 1:57
  • Problem 3 2:50
  • Problem 4 3:46
  • Problem 5 4:13
  • Problem 6 4:41
  • Problem 7 6:12
  • Problem 8 6:49
  • Problem 9 7:49
  • Problem 10 9:31
  • Problem 11 10:08
  • Problem 12 11:03
  • Problem 13 11:30
  • Problem 14 12:28
  • Problem 15 14:04
  • Problem 16 15:05
  • Problem 17 15:55
  • Problem 18 17:06
  • Problem 19 18:43
  • Problem 20 19:58
  • Problem 21 22:03
  • Problem 22 22:49
  • Problem 23 23:28
  • Problem 24 24:04
  • Problem 25 25:07
  • Problem 26 26:46
  • Problem 27 28:03
  • Problem 28 28:49
  • Problem 29 30:20
  • Problem 30 31:10
  • Problem 31 33:03
  • Problem 32 33:46
  • Problem 33 34:47
  • Problem 34 36:07
  • Problem 35 36:44

Transcription: AP Practice Exam: Multiple Choice, Part 1

Hi everyone and welcome back to

I am Dan Fullerton and today we are going to start our work by going through the 1998 AP Physics B Examination, and you will find the link below where you can go print out a copy of the examination.0003

Before we get into the test itself, I would like to talk a little bit about how the test works.0014

You have two parts to the test: a multiple choice and a free response portion.0019

For the multiple choice part, you will get 90 minutes to do roughly 70 questions, so quite a few questions.0023

If you target about a minute a question you will be in pretty good shape with a lot of time to go back.0029

The second part of the test, the free response, is somewhere between 6 and 8 questions usually and you will have 90 minutes for that as well.0034

So what we are going to do is we are going to start with the multiple choice, we will split it in halves, we will do 35 questions first, then we will stop, take a break, do another 35 and then we will come to the free response.0042

Now as you do these, a couple of things: do not leave anything blank.0053

You do not lose any points for a wrong answer so you might as well guess and if you can narrow down, eliminate a couple of the wrong answers, the odds of getting that right answer go way up.0057

But also do not feel like you have to answer every question correctly.0068

You can miss quite a few questions on this test and still get a very, very good score.0071

With that, why not dive into the multiple choice part of the test?0077

I would recommend instead of just watching me as you answer the questions, why not take something, like 45 minutes or so, try to go through the test on your own and then come back and we will see how you did as we go through it.0081

Taking a look at Number 1 -- We are talking about a solid metal ball and a hollow plastic ball and they are released from rest.0093

Assuming that we have no air resistance because we are in a vacuum chamber, they are going to fall with the same acceleration, the same rate, and therefore they will have the same (B) speed because there is no air resistance.0103

In Number 2, we are talking about a student climbing at a constant speed -- the moment you see constant speed right away think acceleration must be 0 -- to the top of an 8 m rope in a certain amount of time. 0117

Find the power exerted by the student.0131

Power is going to be the work done divided by the time, the rate at which work is done.0133

And the work is going to be force times that displacement divided by time or 700 N × 8 m and it took a time of 10 s, so that is going to be 700 × 8 = 5600/10 or 560 and the units of power are watts.0140

So the correct answer for number 2 would be C.0154

Taking a look at number 3, we have a railroad car moving and it is going to collide with another railroad car.0160

Right away I am starting to think conservation of momentum -- the initial momentum must equal the final momentum.0177

Let us write that down -- initial momentum must equal final momentum and if our initial momentum is the mass times the velocity -- because only one of those is moving -- that must equal the final momentum.0182

Now when these two come together, they stick together now, they are treated as if they are one object, so that must be their combined mass times some new velocity which we will call V prime.0195

We are trying for that new velocity so I would rewrite this as V prime = M/total mass (m + M) × initial velocity (V), and that is going to give us our answer E.0206

Moving on to Number 4 -- A cart moving down the surface is collecting rain and as it does that it's mass must be increasing, but because its mass is increasing and we have that Law of Conservation of Momentum...0226 its mass is increasing its velocity must be going down -- so our correct answer for 4 must be C: it is decreasing because of conservation of momentum.0239

Taking a look at Number 5 where we are talking about units of power.0253

Well, power we know is measured in watts (W), which a watt by the way is a joule per second (J/s), so those would both be correct answers, and our other choice is a kilowatt-hour and if you recall a kilowatt-hour is actually a unit of energy, not power.0257

So both 1 and 2 are correct which gives us the correct multiple choice answer of C: both 1 and 2.0272

Let us take a look now at number 6 -- We have a 2 kg object that is moving in a circle.0281

Let us draw our circle and we will put our object on it. 0288

It has some velocity 3 m/s and the circle has a radius of 4 m.0292

Now as I look at this we are looking for the angular momentum with respect to the axis perpendicular to the circle and through its center.0304

Let us start with our formula for angular momentum: L = MVr sin(θ), but in this case we know that θ is going to be 90 degrees and the sine of 90 degrees then is 1, so L = MVr.0311

Well with that I can substitute in my values and this implies then that (L) is going to be equal to my mass (2 kg) × velocity (3 m/s) × radius (4 m), which is going to be 24 kg-m2/s.0334

That would be answer E, and as you take a look at that one there, notice as you look at all the answers you can eliminate a couple of the answers just by looking at the units of angular momentum.0356

On to Number 7 -- Three forces are acting on an object. 0372

If the object is in translational equilibrium, what do we know?0376

Well translational equilibrium means that all of the forces must sum to 0, the net force must be 0, and also the acceleration then is going to be 0.0380

So what are our correct answers?0388

Well Number 1 -- Absolutely that is correct.0390

Number 2 -- No, the magnitudes of the forces do not have to be equal, they just have to sum to 0, so that is not it, and Number 3 -- The forces must be parallel -- that is just silly.0394

So our correct answer must be A, 1 only.0404

Number 8 looks extremely similar to some of the graphs we drew when we were talking about energy back in our energy lectures, and what it tells us is we are given a graph of the potential energy and we need to figure out the kinetic energy.0413

And remember the kinetic energy is just the inverse of the potential energy as long as you have a constant total energy, so a very familiar graph in physics.0427

Here we have our potential energy, which looks kind of like that, and we know in this closed system we must have a constant total energy.0438

So what is going to happen? 0450

Well when it is not potential it must be kinetic and vice versa, so our kinetic energy graph is going to look like we have here in purple.0451

That is our kinetic energy graph, so the correct answer there must be D.0460

Moving on to number 9 -- We have a child pushing a box -- we really do not know why -- and they are doing that with a constant speed and we are given a coefficient of friction. 0469

We need to find the rate at which the child does work on the box.0481

The rate at which you do work means we are looking for power.0485

First thing is let us draw a free body diagram (FBD) for our box.0490

We have the normal force acting up, the weight of the box with gravitational force down, we have some applied force to the right -- the force of the child pushing -- and we have some frictional force that we will call (f).0493

We know since it is at constant speed that right away that tells us acceleration is 0, therefore it must be in translational equilibrium; there is no net force on the box, so all the forces must sum to 0.0509

So if we look in just the x-direction, it should be pretty obvious if those are balanced out, if we are at a constant velocity that our applied force must be equal to the frictional force.0522

But we know the frictional force is μ times the normal force, and the normal force here must be equal to the gravitational force, again because it is in equilibrium.0533

So I can replace normal force with (mg), so the applied force is μmg.0543

If we want to know the power then, remember that power is force times velocity, so that is going to be our force: μmg times the velocity, and there is our answer, choice A.0549

All right, we are on a roll here.0569

Number 10 -- We are talking about quantum transitions that leave you with x-rays. 0571

Well if you remember from our nuclear lectures, we talked about how x-rays are formed. 0578

You knock an electron out of the lowest level and an electron from something like a Molybdenum plate falls down to the first level and as it does that it emits a photon that has enough energy to be an x-ray.0583

The way you do that is you knock an electron out of the lowest or the closest, the innermost shell, so the answer, therefore, must be A.0595

Number 11 -- We are looking for an experiment that provided evidence that electrons exhibit wave properties.0608

All right, electrons -- matter behaving as a wave.0614

Well the Millikan oil drop experiment -- what that was all about was determining the elementary charge -- That is not it.0617

Davisson-Germer, electron diffraction experiment -- Yes, absolutely.0624

If you recall in the Davisson-Germer experiment, we had two slits, we shot an electron or electrons toward it and then we looked on a screen behind it and we observed a diffraction pattern, which was evidence that electrons behave as waves, so that must be one of our correct answers.0628

And 3, Thomson's measurement of the charge-to-mass ratio -- well important work but really did not have anything to do with proving electrons or waves, so the correct answer is going to be 2 only or answer B.0645

Number 12 -- What quantities are conserved in nuclear reactions?0663

Well let us think. We have all of our conservation laws, so Number 1 is going to be charge.0668

A charge must be conserved, so 1 is a correct answer.0674

Number of nuclei and protons? -- I do not recall any laws about conservation of nuclei or protons, so our correct answer here must be 1, A.0678

And 13 -- We are looking at an uncharged conducting sphere in a uniform electric field.0690

Let me draw our electric field and we will say it is pointing in that direction and in it we will put an uncharged conducting sphere.0697

Well because it is a conductor, the electrons in that conductor are free to move, they are going to wind up along with the electric field.0707

So we are going to have the electrons wanting to go opposite the direction of the electric field and they are going to try and hang out over there. 0715

That is going to leave positive charges on this side.0721

You are going to have a force from the electrons in the electric field and the protons in the electric field, but because it is a uniform electric field and the entire conducting sphere is in it, those are going to balance out by conservation of charge.0724

The net force you are going to have here is going to be 0, so your correct answer must be A.0739

Let us take a look here at 14 -- We have two parallel conducting plates connected to a constant voltage source.0750

So I always like to draw these when I can.0757

All right. We know the magnitude of the field between the plates is 2000 N/C, and what we are going to do is we are going to double the voltage and take the distance down to 1/5, the distance between the plates. 0765

How would you do that?0784

Well I am going to start by writing the relation between electric field, voltage and distance between the plates.0785

If you recall, that electric field is the potential divided by (d) and now what happens here if we double the voltage, I have to put a 2 there and if I do anything on the right side I must do it on the left.0790

And then for the distance between them, that is going to be 1/5 or .2 -- what it was initially in the denominator -- so I have to divide by .2 over here.0805

Really, what I have is if we focus on this section, .2/2, I have multiplied the electric field strength by 10, so 10 times the initial electric field strength, 2000 N/C from right up there, is going to give us 20,000 N/C -- Answer E.0814

Moving on, Number 15 -- 15 and 16 both refer to the circuit diagram, and we are asked to find the electrical resistance of the part shown between point X and Y.0841

Well to do that, I am going to simplify my circuit a little bit. 0853

Instead of drawing it that way, I can see that I have a 1 and a 3 in series, so I am going to replace that with its equivalent resistance, so now my circuit is going to look like this with a 4-ohm resistor here and let us put a 2-ohm resistor down here.0857

Now I have two resistors in parallel.0874

Before I even do the math to find the equivalent resistance between the two, I know because they are in parallel that my equivalent resistance must be less than the smallest resistor in that parallel configuration.0877

So R-equivalent, my equivalent resistance, must be less than 2 ohms.0888

Looking at my answer choices, I only have one answer that is less than 2 ohms -- Answer A. 0894

We did not even have to go calculate.0900

In Number 16, we are working with the same circuit and asks if there is a steady current in the circuit, the amount of charge passing a point per unit time.0905

Well that is current flow. Current flow is what we are looking for.0913

What is going to happen? Well we are going to get the most current through the branch with the smallest resistance. 0918

That is going to be that side that is going to get the most current flow, so we are going to have more current in this resistor than we are in this network of resistors, which is actually broken up into a 1 and a 3-ohm resistor, but they will have the same current because they are on the series path.0923

Without a doubt, we are going to have greater current through the branch with the smallest resistance, so E must be our correct answer; it is going to be greater in the 2-ohm resistor than the 3-ohm resistor.0940

Number 17 and 18 have to do with that diagram where we have two positive charges at corners of a square and the lengths are side (D) and we want to find the direction of the net electric field at point (P).0956

Well let me just re-draw this a little bit. 0970

For number 17, we have some charge (+q), some distance (d) away, our point (P) is our point of interest and over here we have another charge (+q).0972

If we want to find the net electric field at (P), well let us figure out the electric field due to this charge (+q). 0988

That is going to give us an electric field away, so at (P) it will be pointing this way, and over here on our (+q) charge to the right, that is a positive charge -- the electric field points away from positive charges -- so at (P) that is going to point that direction.0994

When I add those together to get the net electric field, I have one to the left, one down, and so my net electric field is going to have that direction, so the correct answer for 17 must be C.1010

For 18, we are trying to find potential energy if we put a charge of (+q) over here at point (P).1026

So over here at point (P) now we are going to put a test charge (+q). 1034

We need its potential energy and the way I would start that is realizing that the electrical potential energy is just going to be that charge (+q) times the voltage.1037

If we can figure out the voltage, we will be all set.1047

Well the voltage is just going to be the sum of Kq/R for both of these, so that is going to be (q) times -- well this charge, we will have (K) times (Q) divided by our distance (d) plus another one of those K × Q/d or that will be 2q × 2KQ/d.1057

And I am just going to rearrange this a little bit and I am going to pull the 2 and the (K) outside, so I am going to say that this is 2K × qQ/d, which is starting to look like my answer but it has answers that are in terms of 4 π ε0 in the denominator.1080

Well recall that K = 1/4 π ε0, so I can replace this or rewrite it as 2/4 π ε0 × q × Q/d, and that is going to give me answer choice D.1098

All right, let us take a look here at Number 19.1122

We have a loop of wire with a magnetic field running through it, it gives us some dimensions there, the strength of the magnetic field and we are looking for magnetic flux.1125

All right, magnetic flux (φ) is BA cos(θ), but in this case, our angle θ is 0 degrees.1133

The cos(0 degrees) is 1, so this just becomes BA, and our magnetic field strength (B), well that is just going to be 2 T. 1147

Our area is the length times width of our rectangle of wire, so that is going to be 0.05 m × .08 m, and therefore my flux is going to be 2 × 0.05 × 0.08 = 0.008 T-m2.1160

Or if I want to write that in scientific notation, that is 8 × 10-3 T-m2 or answer choice C.1185

Let us take a look here at Number 20 -- Talking about a coffee pot tells us the current that it draws, its potential difference that it requires, and it also tells us about electrical energy costs.1199

We have 10 cents per kilowatt-hour (kWh) and if you recall kilowatt-hour is a unit of energy.1210

So we need to figure out how much it is going to cost us to run this device for 2 hours.1220

Well the way I would do that is I would start by taking a look at the power that we have here.1226

If power is IV, which is also equal to work/time, I could rearrange that to find the work or the energy is going to be equal to IVT, which is going to be the current flow through our device (4 amps) times the potential difference, the voltage (120 V), times the time.1233

I am going to leave this in hours since we are after energy in kilowatt-hours.1256

This implies then that the work or the energy, 4 × 120 = 480 × 2 = 960 watt-hours (Wh), but we want kilowatt-hours.1261

All right, well kilowatt is 1000 watts, so that is pretty easy to convert and say that that is going to be 0.96 kWh.1274

Now for our answer, we need to know what the cost is going to be to operate it.1283

So let us take our 0.96 kWh and we are going to convert that into a monetary unit because we know that 1 kWh is 10 cents...1287 my units here, kilowatt-hours are going to make a ratio of 1 or cancel out, so 0.96 × 10 is going to give me an answer of 9.6 cents.1304

Therefore the correct answer here would be D.1316

In 21 -- We have an electron in a uniform magnetic field and as it is moving in the plane of the page, we need to find the force on the electron.1323

This is a right-hand rule problem, however it is a negative charge so we are going to use our left hand.1333

Point the fingers of your left hand in the direction of the electron's velocity and then bend them in the direction of the magnetic field.1339

You should see then the direction of your thumb pointing in the direction of the force.1347

Point the fingers of your left hand in the direction of the electron's velocity, bend your fingers in the direction the magnetic field points, your thumb should point (A), toward the right.1354

All right. For 22 and 23, it looks like we have a PV diagram from our thermodynamics study.1368

We are talking about a couple of different processes here and asks which of the following is true of the work done on the gas.1375

Well remember the work done from a PV diagram is the area under the graph, and if it is done on the gas we must be going to a smaller volume, to the left of that PV diagram.1385

Which one of those has the greatest area under it as we move to the left on the PV diagram? 1396

That answer must be Number 1, or A.1401

And for 23, we are asking about the final temperature of the gas.1407

Tell us about the temperatures of the gas. 1412

Well remember on a PV diagram, as you go up and to the right, you get to higher temperatures.1414

The point there that ends up at the highest, at the furthest up and most to the right -- well they are all at the same point left and right, so the highest point is going to be 1. 1422

That is higher than 2 and 3 and it is the most up and to the right, so the correct answer must be A (1): get higher temperatures on graphs as you go up and to the right.1432

Number 24 is another thermal question.1446

In a certain process we are going to add heat to a system and we are also going to do work.1450

We are trying to find a change in internal energy, so right away δU = Q + W, our First Law of Thermodynamics.1456

Now our key here is remembering what is positive: 'heat added to' is a +Q and 'work done on a gas is +W.1464

So δU is going to be -- well we added 400 J of heat so that is going to be 400 J -- and 'the work done on' is positive.1474

In this case, the system does 100 J of work.1484

So it is not 'work done on' -- the system is doing work, so that must be -100 J for a total of 300 J.1488

Really answer C -- really testing whether you understand the sign conventions of that First Law of Thermodynamics.1497

Number 25 -- This one is a question that is not specifically on the test anymore.1507

The AP curriculum no longer covers this specific topic, but if you are interested we will go through it really quick, and if you are not, just go ahead and skip right through this question.1513

We are talking about how much energy is required to convert an ice cube to water at temperature (T2).1521

Well we are going to break this up into three parts.1528

First, you need the temperature to heat it to 273, and that is going to be the mass times the specific heat of ice times the difference in temperature (273 - T1), and then you have to convert it from ice to water.1530

That is just going to be the mass times L (latent heat of fusion of water), and finally to convert from water to our final temperature (T2) -- well that is going to be M × the specific heat of water × T2 - 273.1549

So I need to add all three of those up in order to get the total amount of energy required, and when I do that, add them all up in a little bit of Algebra, I am going to find that I get M × CI (273 - T1) + L + CW (T2 - 273).1569

The correct answer for all of that is B -- not specifically in the AP course anymore even though it is on this old released exam, so not really one to worry about.1594

Number 26 -- Now we are into concave mirrors and it gives us the radius of curvature.1606

The moment I see the radius of curvature I can find out the focal length.1611

Focal length is the radius of curvature divided by 2, so that will be 1 m/2 or 0.5 m.1616

We want to know the distance between the mirror and the image of a star, and if it is a distant star, we can assume that the object distance is closing in on an infinite distance away.1623

We know it is not really infinity but it is a long, long, long ways away, and we will assume it is close to D0 being infinity.1634

So when we write our equation, if 'I do, I die' -- 1/F = 1/object distance + 1/image distance, then 1/.5 = 1/infinity + 1/di.1642

And 1/infinity, 1 over a really, really, really, really, really big number is going to be 0, so that term goes away and it is easy to see that 1/0.5 = 1/0.5. 1659

Therefore our image distance must be 0.5 m, and the correct answer there multiple choice-wise would be B.1671

Number 27 -- We have light passing from air into water and we know of course when that happens the frequency remains the same. 1683

We need to find the speed in the wavelength of the light as it crosses that boundary.1691

All right, well we know V = F(λ), and the frequency remains the same.1695

Now as you go from one material into another -- the different index of refraction -- the speed of that electromagnetic wave is going to change and as you go from air to water, it is going to slow down.1701

If our velocity is decreasing, and frequency is constant, that means wavelength must be decreasing.1714

The answer that gives us a decreasing speed and a decreasing wavelength is answer E.1720

Number 28 -- We have a student doing another optics experiment here.1738

We are going to place an object 6 cm from a converging lens, so the object distance is 6 cm, and our focal length is 9 cm.1744

Find the magnification of the image or the magnitude of the magnification.1753

Well we will use our equation again: 1/F = 1/do + 1/di, to find that 1/9 = 1/6 + 1/di, and I am going to rewrite that as 1/9 (2/18) = 1/6 (3/18) + 1/di.1758

All right, that makes it pretty easy to see then that 1/di must be -1/18, so 1/di = -1/18, therefore di = -18 cm.1783

Once we have that, the magnification is -di/do or -(-18 cm)/6cm (do) = 3, so our correct answer here would be choice D.1797

Number 29 -- Now we have a string fixed to a wall, and that is a hard boundary.1819

We are going to send a wave toward it and we want to know what happens when it comes reflecting back off the wall.1825

Well the first choice: having a greater speed? -- No, that speed depends on the medium, it cannot be that.1831

Choice 2 has a greater amplitude -- well that would violate the Law of Conservation of Energy, so it cannot be that answer either. 1838

Choice 3 -- It is on the opposite side on the string -- Absolutely. When a wave goes in and it hits a hard boundary and reflects, it comes back inverted, so 3 would be our correct answer, 3 only or choice B.1846

If it were to hit a soft boundary, it would come back on the same side of the string as before, but since it hits a hard boundary, it is reflected inverted.1858

On to Number 30.1869

We have an object placed at some distance from a converging lens of focal length (F). 1875

Tell us about the image formed and its size relative to the object.1880

Well the best way to do this is probably to go ray tracing right on the test sheet itself.1884

I will try and draw this by hand as best I can here just to give you an idea of what that would look like.1889

There is our lens and we will call that (F), and it looks like right about here is our object and we will try and space these out at about the same spacing on the other side of our lens, where this is (F).1896

All right. So if I were to start drawing rays, the easiest one to draw I think is the one that is parallel to the principal axis and then it bends down through a focal point.1915

The other one that is really simple is the one that goes right through the middle of the lens, so that is going to do something like that, and I am going to end up with an image way over there.1929

I would recommend on your test you do that with a straight edge to make sure that you are as accurate as possible.1943

But if you look, right away we can see we have a real image because the light rays are actually converging where we have our image; it is inverted and it is magnified, so the correct answer here would be D -- It is real and magnified larger.1948

All right. Number 31 -- We have a light ray passing through a few substances and it shows the index of refraction of those and it tells you ray segments 1 and 3 are parallel.1973

Well if it is parallel at 1 and 3, right away with Snell's Law we know 1 and 3 must have the same index of refraction, so N1 = N3, and we are pretty sure of that.1984

Now as it goes from 1 to 2, it bends toward the normal, and that means that N2 must be greater than N1.1996

I do not see that as any of my choices.2002

And then as it goes from N2 to N3 -- well it bends away from the normal, so that means N2 must be greater than N3, which of course makes sense if N1 and N3 are equal.2005

But as I look at all those choices for answers, the only one that works out, looks like it is going to be A -- N3 is equal to N1.2014

A couple of more here. Number 32 -- We are talking about a radioactive sample and it is decaying at a rate of 4000 counts per minute.2026

At 12:30 PM, that rate has decreased to half of what it was before, so find the predicted rate at 1:30.2037

Well what we are really talking about is a half-life here and it looks like our half-life must be 30 minutes because the decay rate cuts in half every 30 minutes based on the information that is given there.2044

So I would say that at noon, we had 4000 counts per minute, at 12:30 we had half of that (2000), which means at 1, you must have had 1000 and then at 1:30, you probably had 500.2055

It asks you what the decay rate is at 1:30 so the answer there must be B -- 500 counts per minute.2076

Number 33 -- A little bit more of nuclear decay.2087

We have a negative beta particle and a gamma ray emitted during a radioactive decay of some lead.2091

Let us see what this is going to look like.2098

We have 214-82 lead and it is going to decay by -beta decay plus a gamma ray and we are going to be left with something leftover X.2100

Let us see if we can fill these in in a little more detail so we know exactly what is happening.2115

214-82 lead is going to be -- well, our beta decay is going to give us -1, 0, e, and we have our gamma ray and then we have this (X), but what do we know about the (X)?2120

Well if we have 82 here, (-1), that must be an 83 and we still have 214 nucleons there.2136

So I really do not know what element that is off hand, but as I look at my answer choices, I only have one answer that has 214 as the mass number and 83 as the atomic number and it looks like that is bismuth, which is answer choice D.2146

All right. Number 34 -- We are talking about the momentum of an electron and figuring out what happens to its De Broglie wavelength.2166

Well let us start by writing the equation λ = H/P for the De Broglie wavelength, and it is saying if the momentum doubles -- well if we multiply that by 2, then over here 2 in the denominator means we have to have 2 in the denominator over here.2175

So what happened? We have half the wavelength we had before -- answer B.2191

And one more here. Number 35 -- We are talking about quantum concepts being critical in explaining these different things.2199

Well Rutherford scattering experiments -- those were all about showing that we had a small positive nucleus, we had orbiting electrons, and the atom itself was mostly empty space.2210

Nothing in there really relates to quantum phenomena, so right away I am betting the answer is A.2221

Bohr's theory of the hydrogen atom? -- Well that is all about quantum theory and energy levels.2228

Compton's scattering -- that is about quantum theory and momentum of a photon.2232

The blackbody spectrum -- that was evidence of the particle nature and had to do with quantum effects.2236

And the photoelectric effect -- absolutely about energy levels and quantum effects again, so our answer there must be A -- Rutherford's scattering experiments have nothing to do with these quantum concepts.2241

All right, I think we are going to stop there for the first half of the test, give you a second to catch up, recommend going through making sure you understand these answers. 2255

And if you do not as you go through them, go back to the previous portions of the lectures, the videos and see if you can pick up on it so when you come back to these, you go "Yeah, I have that down pat."2263

We will do part 2 of the test here, in our next part 2 of the multiple choice in our next session.2273

Thanks so much for your time everyone, and make it a great day.2278