Edexcel IAL Physics Unit 2 January 2026 Full Paper Walkthrough

hello all. I'm back with another video and this is the most recent unit two from January 2026. And as is the case with the other units, for some reason since the October November 2025 season, there are two available versions for each unit. So, at some point might upload one more, which is going to be called unit two also from January 2026. But, anyway, let's dive in. The general remarks always say is that you should spend about minute per mark. So, if it's if if question awards one mark, it should take you minute. If it awards 12 minutes sorry, 12 marks, it should take you 12 minutes and so on. Right. Question one, the graph shows how current varies with potential difference for an electrical component. Which component is represented by this graph? So, this is the classic example of diode. And we can tell because there is cut-off, say not, voltage, after which current flows in the circuit and below that it doesn't. So, that's diode. Okay. Question two, phone is connected to charger. The current in the charger is 5 milliamps. Which of the following gives the charge in coulombs entering the phone in 8 minutes? okay, to do this we use the definition of the current. The definition of current is charge per time. So, charge is current times time. 5 * 10 to the minus three times eight times 60, which is 8 * 60 is 480. So, that's option Okay. Okay, question three. The diagram shows how sorry, the diagram shows some of the energy levels of hydrogen atom. an electron in the ground state absorbs photon. Which of the following could not be the energy of the photon? So, basically, you have to use your calculator and find the energy differences like -3.4 minus of minus, which becomes plus, so plus 13.6. So, 13.6 minus 3.4, for example, is 10.2, which is option So, that's possible transition, okay? So, this transition is possible because it's 10.2 electron volts. And similarly, you can play with all other transitions, but you can easily tell from the beginning that is wrong because it's large energy difference, okay? So, the only possible transition would be from the minus 13.6 eV energy level. And if we add 9.8, we don't get minus 3.4. So, that's that's wrong. So, it's kind of like calculator problem, okay? So, let's move on. light-dependent resistor is connected in circuit as shown. Okay. The internal resistance of the battery is negligible. The light intensity incident on the LDR increases. Which row of the table describes what happens to the readings on the voltmeter and the ammeter? So, So, okay, let's think about this. So, the resistance the resistance decreases for the LDR. Okay, and this is because light provides extra carriers, okay? Therefore, the resistance of that element decreases, okay? So, the current increases actually because there are more electrons in the circuit. Okay. And and current flows easier. So, current increases as decreases and there is an easier flow of current. Okay. And how about the voltage? Will it change? Okay, pause the video for minute. Sorry, for second. Think about it and try to think of the answer. So, so so the answer is that the voltage doesn't change because there's only one element in the circuit, which is the LDR. So, it has to have 12 volts of voltage pretty much, okay? So, it stays the same. So, stays the same as we only have one element in this circuit. Okay, so what did we say? Okay, the ammeter should increase, so is the correct answer as the voltage stays the same. Question five, light from two sources S1 and S2 is in phase. The light has wavelength lambda and meets at point as shown. The path difference for the light is lambda over four. Which of the following is the phase difference in radians at So, for this kind of problem you will need this formula, delta phi, the phase change the phase the phase difference, excuse me, is equal to 2 pi over lambda times the path difference, okay? So, that's 2 pi over lambda times lambda over four. Okay, so that's pi over two, which is option Okay. let's move on. Question six, student investigated stationary waves on string using the apparatus shown. We have 2.5 string and 20 Newtons of force are applied to it, okay? The vibrating length of the string has mass of 0.005 kg and length of 2.5 Which of the following gives the speed of transverse wave on this string in meters per second? Okay, so So, the formula is tension over the mass per unit length, mu, of this string. Okay? So, the tension is 20, right? 20 Newtons. And then the mass per unit length is 0.005 kg per 2.5 okay? Or in other words, mu is 0.005 over 2.5, which we plug in there, okay? So, the answer is which one is it? Option Okay. Question seven. Which of the following graphs represents the relationship between the frequency and the wavelength lambda of photons? Okay, so this is little bit tricky. mean, it's easy to mess it up because okay, remember that photons and any type of electromagnetic radiation, X-rays, gamma rays, radio waves, etc. For all of them, the speed all of them travel with the speed of light and the speed of light is therefore equal to their wavelength times their frequency. So, we need to solve for the frequency. So, the frequency is times 1 over lambda, okay? And from your math lessons, you know that this kind of graph looks like okay? So, it's it's the function the function of equals to 1 over which looks like this, okay? And you can if you don't remember it, you can calculate it by using, for example, if lambda is small, you can use lambda equals to, don't know, 0.1, say, okay? So, 1 over 0.1 is 10. So, is going to be large. Whereas down here, you can use the value of lambda of 100, say. So, if lambda is 100, is 1 over 100. So, it becomes small. So, you can take points like that and plot it yourself in case in case you don't remember the form, okay? Hope this makes sense. If it doesn't, ask me in the comments. And also, guys, mean, need some feedback about how all this thing is going. the channel is growing little bit, but not as much as would hope or wanted, but yeah, keep going. So, we'll see. But, mean, feedback is welcomed. So, question eight. The circuit diagram shows potential difference applied across the ends of uniform wire of length Okay, which of the following gives the potential difference across length of the wire? Okay, this one gave me little bit of trouble, would say. So, the way thought about it, let me use clean page, was the following. So, for piece of wire, the resistance is given by the resisti- its resistivity times its length over its cross-sectional area, right? For So, for the piece of length this resistance is going to be raw times over Okay? For the piece that has length of minus minus the resistance is going to be raw, the resistivity, times minus over Okay? And now recall that the current is going to be over right? So, it's going to be for the whole, mean, the current in the circuit, okay? The whole piece has length right? So, the current is going to be equal to the voltage over the resistance of the whole piece of wire of length say, okay? So, that's easier. So, maybe can write it here that the resistance of the whole piece of wire, is going to be raw times over So, if we plug in this result there, by the way, this is subscript, okay? So, raw So, if we plug it in, we will find times over raw times is the current in the circuit, and then the voltage across the piece of length is times RX, so that's times over raw times okay, which is the current, times RX, which is this result over there, which is raw over So, raw cancels, cancels, so it's simply times over Sorry, my and my raw look the same, so this is Okay? So, this is All right. So, the correct answer is option Okay? If you can think of an easier way to solve this, please let me know in the comments, okay? So, let's move on. So, question nine, ray of light is traveling through material of of refractive index 1.75. The ray is incident at the boundary with material of refractive index 1.4 as shown. Which of the following expression gives the condition for total internal reflection? So, so, 1.75, using Snell's law, okay? So, if we are in this material, which has an index of refraction of 1.74, times the sign of the critical angle angle will give us 1.4 times the sign of 90°, right? That's the condition, which is one, okay? And remember that to have our total internal reflection, we need to travel from an optically dense material to something that is optically less dense, meaning that the index of refraction of the first material has to be larger than the index of refraction of the second material. And anyway, sign then of the critical angle is 1.4 over 1.75. Therefore, the critical angle has to be larger than the inverse sign of 1.4 over 1.75, which is option okay? Is there anything else needed to be mentioned here? No. So, let's move on. Okay, question 10, when longitudinal waves pass through material, compressions and rarefactions are formed. Which of the table is correct? So, the pressure at rarefaction points is minimum, right? Because if this is the center of this oscillation, the particles will be away from it. Okay? So, the pressure is minimum at the rarefaction. Rarefaction, excuse me, point, okay? And the displacement at compression point is The compression point is where all particles come close to the equilibrium group to the equilibrium position, right? So, the displacement is zero. So, the correct answer is therefore Okay? Hope this makes sense. Let's move on to section now. Okay. Question Okay, question 11, there is current of 1.25 amps in wire. Calculate the drift velocity of the conduction electrons in the wire, and they give you the number of conduction electrons per cubic meter, which is 9 * 10^28 charge carriers or electrons per cubic meter, and the cross-sectional area of the wire. So, that's all the information we need because the current is given by the number of conduction electrons per cubic meter times the charge. By the way, No, it's not It's not necessary to So, let's move on. Times the charge times the drift velocity times So, the drift velocity is current over NQA. So, let's plug in the numbers. So, we have 1.25, and then we have 9 * 10^28, and by the way, we have Let me use the units as well. So, amps, and then cubic meters due to the carriers, and then the charge is 1.6 * 10^-19 coulombs, and then the cross-section is 2.7 * 10^-7, and that's square meters, okay? So, the meter squared cancel, okay? And then we can work this out. Let me show you. So, 1 amp is 1 coulomb per second, okay? 1 amp is 1 coulomb per second, so we have coulombs per seconds times meters over coulombs. So, you see how this actually works out to velocity, and the velocity is 3.22 times 10^-4 meters per second, which is the answer, okay? So, was trying to say earlier that the this charge over here in this formula is the charge of the charge carriers, okay? So, there are materials, don't know if this ever appears at this level, at the AS level or the level or in general, year 12 or year 13 schools, but this charge might be different than the charge of one electron. This This is okay? But, they will tell you when that is the case, okay? So, for now, you can use the charge of the electron there, and you're fine, okay? Let's move Question 12 is writing problem, okay? So, the filament in lamp is metal wire. current in the filament causes heating. Explain how the resistance of the filament changes as the temperature of the filament changes. So, the first thing to notice here is that the increase Okay? The increase in temperature Don't know why used capital, but anyway, the increase in temperature causes greater lattice vibrations, okay? So, it causes greater or maybe larger is proper word, larger lattice vibrations. Okay? So, electrons collide more frequently with the lattice, more frequently with the lattice. Okay? this op- opposes the flow of charge, okay? So, this opposes the flow of charge Okay? more than when the filament is cooler. Okay? And so, the resistance increases. Okay? And if you want to have picture of what's going on, so let's let's try to draw drawing. So, in cool filament, what's going on internally is that you have your lattice points. So, when we say lattice, we mean for example, if the filament is made of don't know, tungsten, okay? So, these positive ions are the mean, you can imagine it as one-dimensional material for simplicity. So, you can picture it as chain of positive ions that are slow and in the center, right? And around these ions, there are electrons flowing, okay? So, there are free electrons. This is This is property of metals. And they flow. Okay? If you heat it, okay, so this is cool filament case. Let's do the warm filament, okay? So, this is warm filament. So, you still have the ions. But they might you know, oscillate like this. They vibrate. Okay? And because they vibrate, the electrons cannot flow as easily as before because, you know, maybe an electron is flowing because and then because the other lattice ion is moving to the left, for example, the electron will want to flow to the left little bit and then go to the right again and then go back and then maybe collide with the ion and stuff like that. So, the motion becomes difficult. Okay? So, this is what is going on internally. mean, very, very roughly. For those of you who are really interested in physics and they want to become physicists or even engineers who study materials, this is something you can study by studying material science or solid-state physics, even quantum materials, you know, this kind of this kind of subject is taught in those kind of studies. So, hope this is helps. So, let's move on. Okay? Question 13. student investigated how the potential difference across battery varied with current in the battery. The student plotted her results on graph as shown. Determine the EMF and internal resistance of the battery. Okay, so the formula is equals to the EMF minus the current times the internal resistance. I'm wondering if they actually give you this formula. Let me check really quick. Let's go to the back of the paper. And we need unit two. Okay, do they No, this is something you have to remember because don't see it over here. So, let's move Let's go back. to question 13 over here. All right, so So, this is of the form equals some constant plus MX. So, the intercept is equal to the intercept, which is 1.5. mean, you can see very clearly. So, 1.5 volts. And then the gradient is going to be equal to minus the internal resistance, right? But the gradient is delta over delta Okay? You can choose any two points of your liking. chose the one circled and this one, okay? So, So, up here the voltage is 1.5. And down there the voltage is 0.5. So, we we subtract. And then the current in the first point is zero, right? And the current in the second point is minus is Sorry, two. So, we have to subtract two. So, we find minus 0.5. So, the internal resistance is 0.5 ohms, okay? This is straightforward problem, guys. All right, let's move on. Question 14. An electron is accelerated from rest through potential difference of 4.8 kV in cathode ray tube. show that the kinetic energy gained by the electron is about 7.7 * 10 to the negative 16 joules. And for the record, this is how quantum mechanics kind of started because they developed these cathode ray tubes. And because the voltage was high and there was gas and collisions with gas, etc. At some point they started observing some strange radiation coming out of these tubes, which was the spectrum of the atoms in the tubes, okay? So, this This could not be explained with electromagnetism that they knew at the time. So, yeah, mean, it's one of the events in the history of of quantum mechanics. So, question The energy is simply equal to the charge times the voltage of the tube, okay? So, the charge, since we're talking about electrons, 1.6 * 10 to the negative 19. And the voltage is 4.8 kilovolts. So, three. And then if we multiply coulombs and volts, we get joules, okay? So, this is in joules, okay? So, we find 7.68 times 10 to the negative 16 joules, which is about 7.7 * 10 Wait, they ask Yeah. 7 times 7.7 * 10 to the negative 16 joules. Sorry, hesitated because for moment thought they that they asked for the energy in electron volts, but no, they ask for the energy in joules, which matches the energy they ask. So, that's good. Okay. Calculate the de Broglie wavelength of the electron. So, our friend again, Prince de Broglie, excuse me. So, lambda is over And the energy as we know, the kinetic energy is squared over 2m. So, we can solve for to find is equal to the square root of 2m times the energy. So, lambda is over the square root of 2mE. Okay, so lambda is 6.63 times 10 to the negative 34 over the square root. No, that this doesn't work. tried to make it pretty, but okay. Okay. Now, maybe it will work. Okay, there you go. Okay, times the square root of Sorry, over the square root of 2 times 9.11 times 10 to the negative 31, which is the mass of one electron, right? Times the energy, which we will plug as 7.7 * 10 to the negative 16. Okay. This yields 1.77 times 10 to the negative 11 meters, which is about 1.8 * 10 to the negative 11 meters, which is our answer. All right, let's move on. This was also straightforward, think. So, yeah, let's move on. 15. The photograph shows rear window heater in car. okay, we see the metal strips. You're all familiar, hope, with this kind of device. So, let's read the question. Each metal strip in the heater has rectangular cross section with 1.64 and height 0.2 Show that the resistance of each metal strip is about 5 ohms. And they give you the resistivity of metal in the strip, okay? And the length. All right. So, this is when where we can use the formula we saw earlier, okay? So, the resistance is given by the resistivity times the length over the cross sectional area. So, the resistivity is 1.7 * 10 to the negative six. And this is ohms times meters. The length is 0.95 meters. Okay, so we get meter squared. And then the cross sectional area is equal to 1.64 times 10 to the negative three times 0.2 times 10 to the negative three. And this is in meters squared. And we convert it from millimeters to meters as usual. So, this yields 4.92 ohms, which is about 5 ohms, okay? Which is what they ask. Okay, good. Good, good. Let's move to the second part. The heater has 11 identical metal strips connected in parallel as shown. Determine the total resistance of the metal strips. Okay, so since they are connected in parallel, one over the total resistance is going to be one over plus one over plus dot dot dot plus 1 over 11 times, right? So, this is 11 times. Okay, so that's 11 over So, the total resistance total is going to be over 11, which is 0.45 ohms or about 0.5 ohms. Okay. question three. The potential difference across the heater is 12 volts and the internal resistance is negligible. Determine the current entering the heater. okay, so so total is voltage over current, right? So, current is 12 volts over 0.45, which we calculated earlier. So, this is 26 .8 amps. don't know why was writing ohms. Okay, that's the answer. Now, part An alternative design is to connect the metal strips in series. Explain why this would not result in the same power being dissipated by the heating circuit. You do not need to include calculations in your answer. Okay. Now, the resistance will be higher, right? Because we add them in series. Okay, so the resistance will be higher. Okay, and since current is voltage over resistance the current will be lower. Okay. So, the power dissipated Okay. And the and the power dissipated is given by squared times okay? Will also be lower. Also be lower. Okay? And you can see this because will be lower. And if you square something something small, it's going to get even smaller, right? If it's less than one, of course. So, we will have less power dissipated. Okay, hope this makes sense. If it doesn't, ask me in the comments. Question 16. student connects the circuit shown in the in this diagram, okay? We have 3-V power supply and two resistors in series. Determine the reading in the voltmeter. Okay, so let's find the total resistance so we can find the current first. So, the total resistance is 6.8 + 4.7 and that's kiloohms, which yields 11.5 kiloohms. And then the current is then the voltage over the total resistance. So, that's 3 over 11.5 * 10 to the 3. Sorry, was ready to write -3 for some reason. Okay. Which is 2.6 * 10 to the -4 amps. And now the voltage, this voltage over there, is going to be times the resistance of that element. So, 2.6 * 10 to the -4 * 6.8 times 10 to the third. And we get 1.77 volts, which is about 1.8 volts. Determine the power dissipated by the circuit. So, the power is going to be squared times the total resistance. So, we did all the work earlier, so we are rewarded now. So, 2.6 times 10 to the -4 squared times 11.5 * 10 to the third. So, this is 7.77, so 7.8 * 10 to the -4 watts. Okay. Now, question The student connects the battery to an ammeter and motor. There is current of 2.5 amps in the motor. Calculate the energy transferred to the motor in 1.5 minutes. So, the power is equal to the current times the voltage. But power is energy over time, right? So, if we combine these two we find that energy is time times power, so time times current times voltage. So, 2.5 * 3 * 1.5 minutes * 60 seconds in minute. So, 675 joules. Okay, that's the energy transferred. Question 17. student directed light from laser on 20 fraction grating. The student saw series of maxima on the screen. The diagram shows the position of the central maximum at and the first order maximum at Calculate the number of lines per millimeter on the diffraction grating. And they give you the wavelength of light from laser as 6. Sorry, 635 nanometers. Okay, from the geometry, so this is the angle theta, okay? The tangent of theta is 0.78 over 4 right? So, theta is the inverse tangent of 0. 19. So, theta is 11.03 degrees. Okay, that's step one. Now, for constructive interference, we know that the separation, like the distance separating two lines in the diffraction grating, times the sign of this angle we just calculated, theta, is equal to the order of the maximum times lambda. And since we're talking about the first order maximum, that is going to be one, okay? And I'm going to use clean page because don't want everything to be cramped there. So, is therefore 635 * 10 to the -9 which is the wavelength, over the sign of 11.03 degrees. Okay, so this yields 3.32 * 10 to the -6 meters, which is also equal to 3.32 * 10 to the -3 millimeters. Okay, so this is the separation between two lines is in this diffraction grating. If we take 1 over we will get the number of lines per millimeter, okay? So, 1 over 3.32 times 10 to the -3 we will have millimeters to the minus one, and this yields 301.2 lines per millimeter, which you can write as millimeters to the minus one, okay? And wait, we are not done. Excuse me, we have to go back because there's part Explain why there is maximum at So, so the reason is that the path difference the path difference between waves from adjacent from adjacent Wait, excuse me. Let's write it again. Adjacent. Good, okay? Slits is lambda. Okay? so, waves are in phase. So, waves actually arrive at in phase. It's better this way. hence constructive interference occurs. Okay. And this was question 17. Question 18. In 1886, Heinrich Heinrich Hertz directed radio waves towards metal plate. standing wave with five antinodes was produced as shown. Explain how the antinodes were formed. So, waves are reflected on the metal plate. Okay. And meet with the incident waves in phase. Okay. constructive interference constructive interference thus occurs. Okay, part determine the value for the speed of the radio waves in this experiment. They give you the frequency of radio waves as 157 MHz. Okay, we can see from the from the graph, mean from the sketch they give us over here, that we have one wavelength, two wavelengths, and one half of wavelength in those 4.75 right? So, that's one lambda, two lambdas, and lambda over two. So, 5 lambda over two is 4.75. So, lambda is 1.9 Now, we need the speed. So, the speed is equal to the wavelength times the frequency. So, 1.9 meters, right? Let's write it here. times the frequency, which is 157 * 10 ^ 6 because it's MHz. So, seconds goes to the denominator. want you to pause the video for second and think about what the answer should be. This Okay, the the hint is that these are radio waves, okay? Okay, now that you're back, the answer is 2.98 * 10 ^ 8 m/s, which is approximately the speed of light, which is what you expect, right? These are radio waves, and all electromagnetic radiation travels with the speed of light. Okay. this was question Are we done with question 18? No, there's more to it. Excuse me. so, let's do question teacher demonstrated the properties of radio waves. He used an aerial to detect the radio waves. an aerial is another name for antenna, guys, okay? In case you don't know. So, initially the aerial was vertical, and the maximum intensity of radio waves was detected. The teacher rotated the aerial by 90°, so that it was horizontal. He observed that when the aerial was horizontal, the radio waves were not detected. Explain this observation. So, this tells you that radio waves are polarized, pretty much, right? Because it's like when you have one of those filters, the optical filters that you rotate, and then light doesn't pass through. So, it's the same case, but with radio waves. Okay. So, radio waves are polarized. Okay. The aerial detects best when aligned with the oscillations. Oops. aligned when aligned with the oscillations. Okay. And when rotated through 90°, the aerial is perpendicular, and nothing is detected. Okay, so basically if this is the polarization of the radio waves, guess, then when the antenna is placed don't know, the antenna maybe it's like this, it detects them, but if you rotate the antenna and you and you make it look like this, then it doesn't detect anything, okay? So, that that's the story. hope this this helps. Let's move on. Okay, 19. when there is an asterisk, this means that this is long writing problem, which means that it's usually worth six marks. And in these types of problems, you should provide about six points that describe the situation or explain the situation, or provide evidence for or against an argument, and so on. So, question 19. Ultrasound can be used to detect cracks in the wing of an airplane as shown. Okay, we see the transducer and crack. Okay. And the wing. And transducer that emits and receives ultrasound is used to scan the wing. Explain how ultrasound can be used to determine the depth of crack. So, we can write that ultrasound ultrasound pulses are sent to the wing. Okay. The crack reflects some of the ultrasounds back to the transducer. Okay. the time taken for the echo to return is measured. Okay, since the speed of the ultrasound in the material of the wing is known, mean, this is like the condition, right? So, since the speed of the ultrasound in the in the material of the wing is known, okay, the distance traveled by the pulse can be calculated. Okay. So, the pulse travels to the crack and back. Okay. So, the depth of the crack is half the measured distance. actually never heard about this before solving this problem, to be honest. Which is and found it kind of clever, so yeah. Anyway, let's move on. Question 20. device called refractometer, never heard of this before, can be used to determine the concentration of solution of glycerol. The graph shows the refractive index of four different concentrations of glycerol. Okay, so we see that as the concentration increases, the refractive index index, excuse me, is increasing. think glycerol is like syrup, right? Something like that. Let me know in the comments. So, mean, it kind of makes sense that as it it will become thicker, right? So, the refractive index will increase as you increase the concentration, guess. don't know. Let me know in the comments. So, question In refractometer, ray of light is shown through sample of glycerol and refracts at the surface of glass block as shown. The refractive index of the glass block is 1.75. Determine the speed of light in the glass block. So, the index of refraction of the glass, okay, should be equal to the speed of light over the speed of light in glass, which means that of glass is 3 * 10 ^ 8 over 1.75 or 1.71 * 10 ^ 8 m/s. part two, the table shows the concentration of different samples of glycerol. Sample 1, 2, and 3. Deduce which sample is being is being tested in the refractometer. So, okay. First of all, they kind of tried to trick you here. They give you this angle, whereas to apply Snell's law, you need this angle, okay? hope you notice these kind of things in exams, okay? So, you actually need the angle 90 - 55.1 degrees, and similarly here, you need the angle 90 - 64 degrees. Okay. So, Snell's law tells us that the of glycerol times the sign of 90 degrees - 55.1 degrees is equal to 1.75 times the sign of 90 - 64. So, this yields of glycerol, which we don't know, as 1.34. Okay, now we have to go to the graph, okay? So, from graph, okay, 1.34 is over here, pretty much, so that's straightforward. So, the concentration is is about 3, maybe 3.1, something like that. So, from graph, the concentration if is 1.34 is 3.1%, which is sample two, okay? Which is sample two. Okay. Let's move on to part Light rays can be directed through the sample of glycerol at different angles of incidence. One ray of light is shown. Explain why it is not possible for any light from the glycerol to reach the region between and Okay. So, mean, the angle of incidence remember that we have to take the angles with respect to the vertical. So, this angle is almost 90, okay? So, about 90 degrees. Okay. So, the angle of incidence the angle of incidence is about 90 degrees. Okay. So, so 1.34 times the sign of 90 degrees is 1.75 times the sign of whatever this angle is, theta, okay? So, so theta is about 50.35 degrees, and it's the max possible angle. Okay. Meaning that if the angle of incidence is say 88 degrees, then theta is going to be smaller than that value, okay? So, that's the maximum possible angle, therefore nothing can be found between and okay? That was question 20. Okay, question 21. camera detects light using charged coupled device, CCD. One CCD has an array of metal pixels as shown. The pixels release electrons when exposed to UV light, ultraviolet light, okay? The pixels are exposed to UV light of wavelength 380 nm. Show that the energy of photon of this light is about 5.2 * 10 to the negative 19 Okay, so the energy of photon is equal to the Planck's constant times its frequency, and the frequency is the speed of light over lambda, because again, they travel with the speed of light, mean, photons and electromagnetic radiation. So, 6.63 * 10 ^ * is the Planck's constant times 3 * 10 ^ 8 * ^ -1 is the speed of light over lambda. So, seconds cancel. Over lambda, which is 380 * 10 ^ -9 so meters cancel. And if we do this calculation, we find 5.23 * 10 ^ -19 which is about 5.2 * 10 ^ 19 as requested. Part two, the manufacturer states that about 80% of these photons incident on pixel will cause the emission of an electron. Deduce whether the manufacturer's statement is correct, and they give you the intensity of incident light, the diameter of pixel, the number of electrons emitted by each pixel per second, okay? So, it's little bit complicated. I'm trying to think of the easiest way. Okay, so, first of all, let's start with this value over here, okay? So, 1.1 microwatt per square meter is equal to 1.1 times 10 ^ -6, and I'm going to convert watts to joules per second times meters squared. Now, the area of the pixel is pi squared over four. So, 1.13 * 10 ^ -10 square meters for each pixel. Okay, so the energy of each photon is 5.23 times 10 to the negative six joules. Okay, so let's try to combine all these in way that that creates Wait, I'm confused. Okay, so this is all the information we know, okay? So, the intense the intensity of incident light is that. So, if we multiply by the cross section and divide by the energy of each photon, we will be able to find the theoretical value of how many electrons per second can be emitted, okay? So, let's use clean page for that. So, 1.1 * 10 to the negative six joules per second per square meter. So, I'm doing things things little bit different here. So, usually, mean, put the units to the right, but now I'm not going to do that. Okay? So, 1.13, we have to multiply with the cross-sectional area, which is 1.13 * 10 ^ -10 ^ 2, and then divide by the energy of each photon. So, 5.23 * 10 ^ -19 So, you see how everything cancels except the seconds, okay? And we find 237.7 per second, okay? So, this is the number of electrons per second that can be emitted by this pixel. And if we multiply that, 2. 37.7 times 80% we find 190. So, the statement is correct. Okay. hope this made sense. It's little bit of weird problem, but okay, the mean, the number of electrons emitted by each pixel is 195. By the way, let me do one calculation really quick. So, 237.7 times 0.8. it's actually 190.2. But okay, mean, close. So, it's close enough, so the statement is correct, okay? Okay, and now question The threshold frequency of the metal in the pixels is equal to the frequency of green light. The pixels were exposed to white light with the same intensity as the UV light. The UV light source was removed. Explain how this changed the percentage of photons that released electrons. So, let's start by saying that white light contains range of energies. And frequencies, excuse me. Of wavelengths, okay? Maybe it's better to say it this way. Wavelengths. Okay. only portion of the photons only portion of the photons will have frequency higher than the threshold. Higher than the required threshold. Okay. So, only some electrons will have enough energy enough energy to overcome the work function. Okay, so only smaller percentage of photons will release electrons. Okay, so we will have reduction in the number of electrons emitted. Okay, this was long test and this was the most recent unit two from January 2026. Again, there is one more version that is called unit two Maybe at some point will upload that one, too. haven't worked through it yet. thank you for those of you who are watching these videos. Please like and subscribe. 85% of people who watches these videos are not subscribed, so that's kind of sad. And anyway, see you guys in the next one. bye.