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[28.51s -> 40.78s] And the first one, to determine what the y-intercept of this graph means, we rewrite this formula by saying the energy of the photon is the product of Planck's constant and frequency. |
[40.78s -> 54.37s] and that's equal to the work function plus the kinetic energy. This is the y-intercept which means at this point the frequency must be equal to zero which means that zero is equal to |
[54.37s -> 68.29s] the sum of work function and kinetic energy which then can be rewritten as the work function is equal to negative kinetic energy and this is when frequency is zero so that tells us |
[68.29s -> 75.66s] that the y intercept on this graph is the negative work function. |
[77.17s -> 91.92s] The second intercept that we would want to find here is the intercept with the x-axis. We know that that is going to happen when the kinetic energy is equal to zero, and we get this by rewriting this equation. |
[91.98s -> 106.22s] where the energy of the photon is a product of Planck's constant and frequency. The work function is the product of Planck's constant and the threshold frequency, and in this case the kinetic energy is now zero. |
[106.54s -> 119.74s] And so we can then rewrite this or simplify it. Let me write that or leave that as kinetic energy, which leaves us with H times F is equal to H times F0. |
[119.74s -> 133.52s] which then simply tells us that the frequency at the point where kinetic energy is zero is our threshold frequency. So what this tells us is that we now have a graph that can tell us where the |
[133.52s -> 147.54s] work function is or what the work function is it can also tell us what the threshold frequency is and now this graph makes more sense because it tells us that when the photons have an energy or frequency less than the threshold frequency |
[147.54s -> 160.16s] There is no kinetic energy because there are no photoelectrons ejected. We know that once the threshold frequency is surpassed then the kinetic energy of those photoelectrons increases. |
[160.16s -> 173.78s] and it increases with a steady gradient and the gradient of that graph is kinetic energy over frequency which we know then is Planck's constant, a constant value. |
[173.78s -> 176.40s] for a constant gradient on that graph. |
[177.74s -> 190.26s] The second common use of the photoelectric effect or the way that we show it is in an electric circuit that's set up as follows. We have a battery connected to an ammeter connected to a... |
[190.67s -> 204.72s] photocell that we can see here essentially breaks the surface where we have a metal over here connected to the negative terminal of the battery and here metal connected to the positive terminal of the battery and what we can see here |
[204.72s -> 208.21s] is that by shining a light onto this metal |
[208.69s -> 222.99s] If the frequency of the light exceeds the threshold frequency, then electrons will be ejected from the surface of this metal and be attracted towards the positive terminal of the battery, which would then |
[222.99s -> 237.33s] register a current on this ammeter. So the first graph that we can draw here is a graph of current versus frequency and what this shows us is that no current would register until the point |
[237.33s -> 249.65s] where we reach our threshold frequency and from then onwards a current would register because now electrons are being allowed to bridge that gap which completes the circuit and allows a current to flow. |
[250.03s -> 259.60s] This circuit can also be used to demonstrate the fact that intensity changes the number of electrons that are ejected because what we will find here |
[259.95s -> 271.02s] is that if instead of plotting current versus frequency, we now plot current versus intensity, we'll find that as we increase the intensity of this light, |
[271.76s -> 286.38s] We increase the number of electrons that are ejected and more electrons being ejected means more charge moving across that gap. More charge we know would increase the amount of current because current is the charge. |
[286.54s -> 293.14s] over the amount of time. So what we find is that our current versus intensity graph |
[293.58s -> 303.70s] will increase for as long as the intensity increases. Again, as long as the frequency exceeds the threshold frequency. |
[303.98s -> 317.63s] So once again, these are two different ways in which we can see the photoelectric effect, but both of them show us the same thing. They say that no photoelectrons are ejected until the threshold frequency is surpassed. |
[317.63s -> 331.31s] They tell us that the kinetic energy increases once the threshold frequency is surpassed. And they tell us that the intensity affects only the number of electrons ejected. It does not affect... |
[331.31s -> 335.09s] whether or not an electron is ejected as a photoelectron. |
[7.57s -> 21.68s] Over the years of my teaching career I must have taught accounts of thousands of year 10 and 11 students and some of them were very bright and capable physicists but not once was I ever asked, hold on sir, how come the nucleus |
[21.68s -> 30.74s] With these positive things packed close together doesn't just go bang. Good question. And that's the subject of this video. |
[31.54s -> 41.58s] Imagine two protons which of course we know are both positive. We're going to consider the forces acting on these protons as we try to bring them closer together. |
[42.64s -> 55.39s] By convention, we say that repulsive forces are positive. There is an electrostatic repulsive force between these two protons then. If you've had the joy of doing electric fields in year 13, |
[55.39s -> 69.60s] You'll know that this follows a 1 over r squared pattern. Don't worry if that didn't mean anything to you, it soon will. But for now, just accept the fact that the force that is repulsive between these two protons grows large rapidly. |
[69.60s -> 79.58s] We can show this graphically. As one proton approaches its stationary friend you can see this in the green curve. Notice how quickly the green curve rises. |
[79.58s -> 87.60s] Considering how small the mass of a proton is, this repulsive force between these two protons is truly enormous. |
[88.05s -> 99.66s] The diameter of a proton is around about give or take roughly to a good approximation 10 to the minus 15 meters. Now we call 10 to the minus 15 a femto. |
[99.66s -> 111.98s] And this is easy to remember because both the femto and 15 begin with an F. This point here on the X axis is showing a distance of three femtometers then, which is roughly about three proton diameters. |
[111.98s -> 120.78s] Notice all the hand waving that's going on here. We're not really interested in exact measurements but approximations are going to be fine for us. |
[121.55s -> 135.89s] Given how large this repulsive force is between these protons, we can see that we've got a bit of a problem developing here. How can a nuclear stay together? Surely the protons should just literally blow each other apart. |
[139.25s -> 153.68s] In order to put a lid on the potential problem of the exploding nucleuses, there must be another force at play. Clearly this is the case because, well, we just wouldn't be here, would we? So, enter the strong nuclear force. |
[153.71s -> 161.39s] The nature of the strong nuclear force is quite different to any other force that you've met. For starters, it's a short-range force. |
[161.68s -> 172.35s] Let's picture the moving proton coming closer to its stationary friend. At large distances, whatever that means, there simply is no strong nuclear force. |
[172.35s -> 176.80s] There will, however, still be some electrostatic repulsion. |
[176.80s -> 189.25s] As we move closer we find that around about four or five centimeters the strong nuclear force is beginning to slowly increase but overall there is still a larger repulsive force between the protons. |
[189.25s -> 201.89s] Moving closer still and arriving at around 3 femtometers, we see that the strong nuclear force starts to increase rapidly in magnitude, peaking at a separation of around 1.5 femtometers. |
[201.89s -> 209.66s] Notice though that the strong nuclear force is negative which according to our convention means that it is an attractive force. |
[209.66s -> 218.42s] Therefore you can see that at around about 1.5 femtometers the resultant force between the particles is now strongly attracted. |
[218.80s -> 232.08s] As we go even closer, the strong nuclear force begins to become rapidly less attractive and at around half a femtometer flips over to being repulsive and continuing to increase rapidly as you get closer. |
[232.11s -> 238.67s] Like I said the electrostatic force and gravitational force are nothing like the strong nuclear force. |
[239.41s -> 252.74s] So you can see that if you try and push two protons closer than, well, shall we say, I don't know, one femtometer, very quickly the force is going to become extremely repulsive. In other words, you can't get them closer than that. |
[252.74s -> 265.52s] and there is a sweet spot round about one and a half femtometers one other important point is that the strong nuclear force applies to all hadrons and naturally that includes neutrons |
[265.78s -> 273.49s] And so a proton and a neutron will feel the strong nuclear force if they get close enough, as indeed will two neutrons. |
[274.93s -> 289.63s] The neutron is often described as the glue that holds the nucleus together. Let's see if we can see why. I want us to imagine a slice, one row, through a relatively large nucleus. We're going to go proton, neutron, proton, neutron, alternately, just to keep it simple. |
[289.63s -> 298.56s] let's pick any two particles that are separated well any two particles that are adjacent to each other obviously we have one proton one neutron so there's going to be a |
[298.56s -> 311.25s] strong attractive force between those because their distance between their centers is going to be around about one to one and a half centimeters. However, there'll be no repulsive force, because one of them is neutral, therefore strongly held together. |
[311.25s -> 325.81s] What if we take alternate protons? So we go proton, neutron, proton. Well, in this case, we have two positive particles separated at a distance of around about two to two and a half centimeters. And as I'm sure you can see from the graphs earlier, the strong force is beginning to... |
[325.81s -> 337.55s] of fall off a bit at this point but we also now have a repulsive force caused by the fact that they are both positively charged so now it's getting a bit more complex we're going to have to look for a resultant force |
[337.55s -> 349.14s] probably between those two protons is going to overall be attractive but what happens now if we move out to two protons separated by let's say around about four to five femtometers |
[349.26s -> 361.04s] Now in this case the strong force has fallen away basically down to zero but there is still a repulsive force between those particles because well as we know electrostatic force is not a short range force. |
[361.90s -> 376.34s] Although we are still firmly in hand-waving territory I think you can begin to see that the neutrons are acting like glue since they attract all of their nearby neighbours without experiencing any repulsive force. |
[3.70s -> 11.63s] Consider our planet Earth in space and these sample positions of an Earth satellite in circular orbit. |
[12.59s -> 24.11s] before the idea of inertia was understood people thought that a force in the direction of motion was responsible for motion they imagined angels pushing on the planets |
[24.59s -> 36.58s] newton changed all that and taught us that the only force acting on a satellite acts toward the body it orbits that's the gravitational force and what if this force disappeared |
[36.58s -> 48.75s] Then the satellite would fly off on a straight line path. No force, no orbital path. Note that each of the force vectors are the same size and point to Earth's center. |
[49.01s -> 58.00s] Same size because the satellite circles at the same distance from Earth. Note also that they're perpendicular to the orbital path. |
[58.48s -> 71.60s] the ninety degree angle means there is no component of force along the orbital path which further means no change in speed from an energy standpoint that means no change in kinetic energy |
[72.14s -> 83.02s] and since distance remains constant no change in potential energy so both kinetic energy and potential energy are constant all along the orbital path |
[83.76s -> 97.74s] So a satellite in circular orbit effectively coasts at an unchanging speed all the way around and around and around Now things are different for a satellite in an elliptical orbit |
[98.80s -> 113.36s] consider these sample positions for such a satellite since the distances from earth are different the forces are different weaker far from earth and stronger when closer in accord with the inverse square law |
[113.68s -> 116.05s] So the speeds are different also. |
[118.80s -> 132.02s] kepler was the first to discover the elliptical paths of planets about the sun early in the seventeenth century he discovered that planets travel fastest closest to the sun and slowest farther away |
[132.21s -> 143.47s] but he had no explanation as to why just as a projectile tossed upwards slows as it rises and speeds up as it returns so it is with any satellite |
[144.24s -> 158.67s] kepler never viewed a satellite as a projectile that planets are projectiles falling around the sun just as our moon is a projectile falling around earth this way of thinking escaped kepler |
[160.24s -> 170.35s] let's talk energy conservation the sum of the potential energy and kinetic energy at any point along the satellite path is the same as at any other point |
[170.54s -> 184.37s] hence where the kinetic energy is greatest potential energy is least and vice versa we can look at the changes in speed by considering the components of gravitational force along the satellite path |
[185.14s -> 196.78s] the component perpendicular to the satellite path shown as a white vector here doesn't affect speed but changes the direction of motion curving it away from a straight-line path |
[197.30s -> 209.81s] more interesting is the component along the path which we make purple here that component of force changes speed when in the same direction of motion speed is increased |
[210.32s -> 218.26s] But here on the other side, the purple component slows the satellite. That's because the satellite is going against gravity there. |
[218.90s -> 231.18s] so for our satellite we see it has the least speed farthest from earth and the most when closest it falls around and around indefinitely i want to leave you with a question |
[231.47s -> 246.38s] What becomes of the purple component of force along the satellite's path when the satellite is closest to and farthest from Earth? And more important, why? Until next time, good energy! |
[0.02s -> 8.75s] Professor Dave here, let's discuss angular motion. |
[9.84s -> 18.14s] We talked about uniform circular motion, but we need to make an important distinction between spin and orbital motion. |
[18.14s -> 31.47s] An object can spin around an internal axis that goes through its center of mass, and it can also orbit around some external axis. This means that the earth will spin on its axis |
[31.47s -> 41.78s] but it orbits around the sun. We have discussed spinning objects and the way that tangential speed will vary according to distance from the axis of rotation. |
[41.78s -> 56.13s] But this can be a tricky way to view the rotation of something like a ferris wheel, because there is a certain context in which we would like to say that every part of the wheel is spinning at the same speed, since it is one solid object. |
[56.13s -> 67.31s] For this reason, when we are looking at a rotating rigid object, rather than looking at translational motion, we will want to discuss angular motion, or rotational motion. |
[67.89s -> 72.46s] To do this we can begin to discuss certain angular quantities. |
[72.46s -> 83.14s] angular displacement represented by theta will be the angle swept out by any line passing through a rotating body that intersects the axis of rotation. |
[83.14s -> 89.28s] This value will be positive if the motion is counterclockwise and negative if clockwise. |
[89.28s -> 103.60s] you have a hard time remembering this just remember that in going counterclockwise around the xy plane we will cover quadrants one two three and then four so positive angular displacement correlates with a positive |
[103.60s -> 107.92s] increase in the number quadrant that is being traversed over time. |
[108.27s -> 122.86s] The SI unit for angular displacement will not be degrees, but rather radians. We will discuss the derivation and importance of the radian in the upcoming mathematics course, but for now, we simply need to know that one |
[122.86s -> 137.55s] revolution is equal to two pi radians, so if the rotation completed by the object is a counterclockwise quarter turn, the angular displacement is one half pi radians, or pi over two. |
[137.97s -> 148.66s] Angular velocity, as one might expect, is equal to the angular displacement over some time period, just the same way that linear velocity is equal to linear displacement over time. |
[148.66s -> 156.05s] average angular velocity, represented by the Greek letter omega, is therefore given as delta theta over delta t. |
[156.05s -> 168.19s] The SI unit for this value will be radians per second, although we will frequently encounter other units like revolutions per minute, or RPM. Again, counterclockwise rotation involves |
[168.19s -> 181.01s] positive angular velocity and clockwise will be negative. Lastly, angular acceleration, represented by the Greek letter alpha, is equal to the change in angular velocity over some time period. |
[181.01s -> 187.12s] This will be equal to delta omega over delta t, with units of radians per second squared. |
[187.22s -> 201.81s] Now that we have defined these terms, we can begin to understand that rotational kinematics will utilize equations that are direct analogs of those involved with linear kinematics. We just swap out linear components |
[201.81s -> 211.54s] for angular ones. Looking at these familiar equations, if we just exchange displacement for angular displacement, velocity for |
[211.54s -> 222.18s] angular velocity and acceleration for angular acceleration, we arrive at a new set of equations that allow us to calculate rotational motion. |
[222.18s -> 228.91s] they will apply to any system with rotational motion under constant angular acceleration. |
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