CSEC PHYSICS PAST PAPER PRACTICE [ PART 2 ] JANUARY 2025 PAPER 2

Añadido: 2026-05-21

[00:01:59]Oh.

[00:03:54]>> Mhm.

[00:04:41]>> [music] >> Mhm.

[00:06:03]>> Good evening. Good evening, everyone.

[00:06:06]If you can hear me, just put a one in the chat or an emoji so I can know that you're hearing me and then I can start.

[00:06:18]Also, please share the link and also subscribe if you have not subscribed.

[00:06:29]Good evening, Joel. Nice seeing you in the chat.

[00:06:32]If you can hear me, just send an emoji or a one.

[00:07:27]All right, so we are will be using Zoom on Thursday.

[00:07:36]So, don't worry about the Zoom. We'll be using Zoom on Thursday.

[00:07:40]This is just a practice paper.

[00:07:43]So, any comments you can share it in the chat.

[00:07:47]All right, so let's see if we can begin.

[00:07:50]So, we would have done part one.

[00:07:55]We're doing part one of this paper, the January 2025 paper, and we'd have done question one to three.

[00:08:06]And tonight, this evening, we'll be doing question four.

[00:08:13]Four to six.

[00:08:15]All right, so let's get right into it.

[00:08:21]So, section B, question four.

[00:08:25]First question is coming from our waves topic, and specifically, it talks about the laws of reflection.

[00:08:36]So, if you remember the diagram for reflection, you know that we have a incident ray, we have a reflected ray, we have a normal, and we know that we have a a boundary.

[00:08:58]And we also have a point of incidence.

[00:09:10]So, the first law speaks to all these areas of the diagram. So, if we're to write the first law, we just simply need to write that the incident ray, incident ray, which is I, the refract reflected ray the reflected ray which is R and the normal and the normal at the point of incidence are all on the same plane.

[00:10:29]So, that's the first law. It speaks to all these components on the diagram for reflection.

[00:10:36]And a lot of times persons tend to substitute the reflection for refraction, but you have to pay attention to which law are that they are speaking about. All right. So, once it's reflection, you know that it has to do with the incident ray, the reflected ray, the normal, and the point of incidence. So, all these elements are on the same plane.

[00:11:00]So, we move into law number two.

[00:11:03]Now, law number two has to do with the angles.

[00:11:08]So, the angle that is between the normal and the incident ray and also the angle that is between the normal and the reflected ray.

[00:11:17]So, law number two speaks to those angles. And we look at law number two um we should write that the angle of incidence So, the angle of incidence is equal to the angle of reflection.

[00:11:55]So, not refraction, but reflection.

[00:11:58]Right? And that's the simple answer for law number two. It has to do with the angles, and we know that since it's being reflected, those two angles must be the same. And so, that's what we would write for the second law.

[00:12:11]All right, moving on to the next part of the question.

[00:12:14]So, the next part of the question says that one property of an image formed by a plane mirror is that it is virtual.

[00:12:22]So, they have given us one property already.

[00:12:24]And now, they're asking us to state one other property.

[00:12:30]So, one other property that is not the one that they just gave us.

[00:12:37]So, when we look at images formed in a plane mirror, there are different properties.

[00:12:44]And so, we know that images formed in a plane mirror, they can be inverted.

[00:12:52]So, they can be laterally inverted.

[00:12:54]And also, we know that the image that is formed is the same size as the object.

[00:12:59]So, I'm going to use that one.

[00:13:01]So, one other property of the image is that the image size is the same as the object size.

[00:13:25]All right.

[00:13:27]So, that's one other property. And as I said, you could have used the other one which says that it is laterally inverted.

[00:13:35]And you also have the one that speaks about the distance.

[00:13:39]So, they are the same distance apart.

[00:13:43]So, that is the object in the object is the same distance from the mirror that the image is from the mirror.

[00:13:49]All right. So, any of those could be stated and you would have gotten the one mark.

[00:13:56]All right, moving on to the next part of the question. So, this part of the question here says we should state Snell's law.

[00:14:06]Uh would anyone want to try that?

[00:14:16]So, what does Snell's law speak to?

[00:14:22]So, Snell's law falls under the category of the laws of refraction.

[00:14:29]All right.

[00:14:31]So, when we think about Snell's law, we think about the laws of refraction.

[00:14:36]So, Snell's law talks about the sign of the angle. So, if I was supposed to draw the diagram to represent refraction, the diagram would look something like this.

[00:14:50]So, we'd have our incident ray.

[00:14:53]All right.

[00:14:54]We'll have our normal. So, I'm going to put the normal in a different different color.

[00:15:04]And the normal is always perpendicular to the boundary.

[00:15:07]So, when the incident ray passes from one medium and goes into the next medium, we notice that the angle will change.

[00:15:21]All right. So, the angle will change.

[00:15:29]So, Snell's law focuses on these angles, the angle of incidence and the angle of refraction.

[00:15:37]So, what Snell's law says is that the for light traveling from one medium to another So, for light traveling from one medium to another What does the rest of it say?

[00:16:14]It compares both angles. So, when you compare both angles, we are talking about the ratio. All right? So, we can say that the ratio The ratio of the sign of the angle So, the ratio of the sign of the angle of incidence So, that is the first angle here.

[00:16:53]It's the ratio of this sign of the angle of incidence and the sign of the angle of refraction So, that is the second angle of refraction.

[00:17:18]Right?

[00:17:20]So, the ratios um of the signs of the angle of incident and the angle of refraction is actually equal to a constant. So, it is a constant.

[00:17:36]Right? And you should be familiar with this. So, sin I divided by sin R is equal to the constant, which we refer to as the refractive index. So, this is the constant.

[00:17:55]So, as long as you're able to tell that for light traveling from one medium to another, and you compare the two angles, right? And you mention that they are equal to a constant, then you will receive the three marks.

[00:18:09]All right? So, pretty simple theoretical part of um the topic reflection and refraction.

[00:18:16]So, ensure you study the laws of reflection and the laws of refraction.

[00:18:22]All right, moving on.

[00:18:24]We are going to look at part C. So, here in part C, we have blue and red light are incident on a Perspex block as shown in the figure.

[00:18:33]So, one ray is blue light, one ray is red light, and they've given us some angles. Now, it says here that we are to calculate the angle of incidence for each color.

[00:18:46]Now, anytime you get a diagram like this that has reflection, sorry, that has refraction, or light passing from one medium to the next, you need to ensure that you draw in your your um normal.

[00:19:01]So, you want to put in our normal because all our angles should be taken from the normal.

[00:19:09]Right? So, once we can figure out what the angles are from the normal, then we can work it out. So for blue light, the angle has to be between the normal and blue light. So that's going to be our first angle.

[00:19:25]And then for red light, it has to be the angle between the normal and red light.

[00:19:33]So what we're trying to find now is what's the angle between the normal and blue light.

[00:19:41]So the question here is what is that angle?

[00:19:44]Now since this forms a 90°, so since the normal is perpendicular to the the boundary and it forms a 90° angle, if we subtract the 53 from the 90, then we would find what the angle here is for the blue light. So what we're going to do here is that we're going to subtract 53° from 90 and we should end up with 37° All right. So this angle here for blue light would be 37°.

[00:20:23]Here is a simple Now it asks us to find the angle of incidence for red light. So red light is has a larger angle here. So between the normal and red light, we need to find out what this angle is, right?

[00:20:42]So looking at the diagram, what angle would that be?

[00:20:45]So now we know that this angle is 37 and they have already given us this angle which is 17. So if we should add these two angles together, we would get the angle of incidence for red light.

[00:20:58]So we would take 17° and we would add it to 37° right? And we would get our angle for blue light, which would be 54 degrees.

[00:21:14]All right?

[00:21:15]It's quite simple.

[00:21:17]So, the key thing to remember here is to establish your normal at the point of incidence, and then you can work your way from there because once you have a line perpendicular to the boundary, you know that this will form a 90° and then from there you can work out the question.

[00:21:38]All right? Moving on.

[00:21:41]So, this part of the question asks us to calculate.

[00:21:46]So, we're going to do some calculations now.

[00:21:49]So, we have to calculate the angle of refraction of blue light given that the refractive index of blue light is 1.

[00:22:01]53.

[00:22:03]So, they have given us the refractive index of blue light.

[00:22:08]And based on what we've calculated here, right? We know what the sign of the angle of incidence is for blue light.

[00:22:18]So, the angle of incidence for blue light is 37, right? So, angle of incidence for blue light is 37°, so that's one thing we know. And the question would have given us the refractive index of blue light, so that means the refractive index for the light passing from the air to Perspex is going to be 1.53.

[00:22:43]Right?

[00:22:44]And we also know that the refractive index for light in air is actually 1.0.

[00:22:55]So, with all this information, we can calculate the angle of refraction. So, what we're trying to find is the angle of refraction.

[00:23:02]Right? So, we're trying to find the angle of refraction.

[00:23:05]Now, once we see these um variables or these terms, we know that we should remember our formula that describes what happens with the sign of I over the sign of R, which is equal So, these ratios are equal to these ratios, which is the ratio of the refractive index.

[00:23:35]All right? And also, these can also be equal to the speed of light. [snorts] But, you can only use two pairs at a time. So, in this case, because we're given the angle of incidence of blue light, and we know the refractive index for both of the materials, then we only have one unknown.

[00:23:55]And so, now all we have to do is substitute in the information and solve.

[00:24:01]So, the sign of I, which is 37, so 37°, over the sign of R is equal to the refractive index of the second medium, which is 1.53 over the refractive index of the first medium, which is here, which is 1.0.

[00:24:26]So, all we have to do now is to cross multiply here. So, we're going to multiply this by that, and we're going to get 1.0 sign 37°, all right?

[00:24:43]Equal, and then we are going to cross multiply here.

[00:24:47]All right? So, when we cross multiply there, we're going to get 1.5 3 sign R. And now, it's just a matter of rearranging the equation. So, we want to get rid of the 1.53 from over here. So, we're going to divide here by 1.53.

[00:25:05]And we're going to do the same thing on both sides. So, 1.53.

[00:25:11]So, this will cancel out that. That will cancel out that.

[00:25:15]Right? So, now we are left with So, flipping this around, so sine R is going to be equal to 1.0 sine 37° all divided by 1.53.

[00:25:31]Three. So, all we need to do is just put this part in the calculator and let's see Let's see what we get.

[00:25:37]So, it's 1.0 sine 30 seven. So, we put in one.0 sine 37 and that's the answer.

[00:25:54]And then we're going to take that and we're going to divide it by 1.53.

[00:26:02]So, when we divide that by one.53 we get an answer of 0.393. So, rounding off to three decimal places to maintain accuracy. So, sine R is going to be equal to 0.

[00:26:21]393.

[00:26:23]393.

[00:26:25]Right? And now we need to find what R is. So, R will be equal to the sine inverse of 0.9 0.393.

[00:26:42]And so, our answer will be So, when you find the sine inverse of the answer So, shift on the calculator, sign, so Let's do it like that.

[00:26:56]So you press shift on the calculator, sign inverse of 0.393.

[00:27:04]All right? And so your answer will be 23°.

[00:27:09]So answer will be 23°.

[00:27:13]And that's how you get the four marks.

[00:27:18]All right?

[00:27:23]So I hope you're all following following.

[00:27:27]All right, moving on to the next part of the question.

[00:27:30]Um we're going to calculate the critical angle.

[00:27:34]We're going to calculate the critical angle for blue light. So we're still focusing on blue light.

[00:27:40]So in order to calculate critical angle, you must recall the formula sin C, where C represents the critical angle, is equal to 1 over the refractive index of the second medium.

[00:27:55]All right? So the light is traveling from one medium to the next.

[00:27:59]So one is representing medium that is air.

[00:28:04]And the other medium would be the refractive index of the second medium.

[00:28:09]So we are given the refractive index of blue light.

[00:28:12]And that's 1.53.

[00:28:15]So sin C is equal to 1, which is the refractive index of the light in air divided by the refractive index of blue light in the Perspex.

[00:28:28]So that means if we are now to find C, we're just going to find the sin inverse of 1 over 1.53.

[00:28:43]And you just enter that in your calculator.

[00:28:47]So, let's clear that.

[00:28:50]So, we're going to find the sine inverse of quick shift open bracket 1 divide by 1.53.

[00:29:01]All right? And so, the answer is going to be 4.40.8.

[00:29:06]All right?

[00:29:07]So, 40.8 degrees as the critical angle.

[00:29:15]All right? Or you could have rounded it off to the whole the whole number, which is 41 degrees.

[00:29:23]And that's how you would get your three marks for that question. So, in total for this question with waves uh refraction and reflection, you would have amassed a total of 15 marks if you would have mastered this topic.

[00:29:39]So, it's good for you to ensure that you master the topics so that any topic comes, you are able to do it.

[00:29:48]All right. Question five now is from the electrical part of the syllabus.

[00:29:55]And this part asks us to write a formula.

[00:30:01]So, we are supposed to write a formula to show how potential difference, um charge, and energy are related.

[00:30:09]And also, we are to state the SI units for each quantity.

[00:30:15]So, first, let's do the formula. So, the formula that tells the relationship is So, potential difference V is equal to the energy divided by the charge.

[00:30:32]All right? So, you should remember that equation from that topic.

[00:30:37]Energy divided by charge. And this is what tells us the relationship between energy and potential difference and charge and potential difference.

[00:30:47]So one has a direct proportional relationship, so the potential difference is directly proportional to the energy and the charge is inversely proportional to the potential difference.

[00:31:00]Now we are going to do the second part, which is to state the SI units for each quantity. So the first quantity is the potential difference. So what What is the SI unit for the potential difference?

[00:31:23]So we need to know the SI unit for the potential difference.

[00:31:28]Um let's see if anyone can get it in the chat.

[00:31:37]What's the SI unit for potential difference?

[00:31:41]And let me write out the others. So we have the charge.

[00:31:48]What's the SI unit for charge and what's the SI unit for energy?

[00:31:56]So the SI unit for potential difference you know is volts.

[00:32:05]Cuz potential difference is referring to the voltage that is coming from the source. So we have volts.

[00:32:13]Um for the charge we know that charge Q is as SI units of coulomb.

[00:32:22]Right? So we are going to write coulomb.

[00:32:28]So it's good to write out the the name and add the symbol beside it.

[00:32:33]Right? So you're aiming for maximum marks.

[00:32:37]Then we have energy. So, we know that energy the SI unit for energy is joules.

[00:32:45]All right?

[00:32:46]And that's the symbol.

[00:32:49]So, once you have done the formula, you've done the SI unit for each, you will have gotten your four marks.

[00:32:57]All right, moving on to the next question.

[00:33:00]Um we have the sir a circuit that can be used to recharge a battery as shown in figure four.

[00:33:07]So, here we have the battery. So, the battery is 12 V. All right? All right, and we have a source. And what we are trying to do is charge a battery here. So, it says on the diagram, we should insert the polarity of the DC supply.

[00:33:23]So, once we're able to figure out the polarity of the battery, then we can figure out the polarity of the DC supply. All right?

[00:33:30]So, for the longer leg of the battery, you know that is positive. And the shorter leg, we know that is negative.

[00:33:37]And we know that current has to flow from either positive to negative, right, if it's conventional current flow, or from negative to positive if it is electron flow.

[00:33:49]So, for the DC source now, we are to label the polarities here.

[00:33:55]So, the polarities have to be positive here and the negative here cuz what we are trying to do is to charge the battery.

[00:34:04]All right? So, the the current must flow in that direction.

[00:34:10]All right?

[00:34:12]So, that is all you will get your mark for that question.

[00:34:16]Moving on to the next part of the question, it says that you should indicate in the space provided um which of the values 6 V, 12 V, or 15 V would be most suitable for the DC supply.

[00:34:29]So, for this DC supply, which of these voltages would actually charge the battery? That's the question.

[00:34:37]Which of these voltages would actually charge the battery?

[00:34:51]So, it couldn't be the 6 V because 6 V is less than 12 V, so the the potential difference here is is not good.

[00:35:01]So, the the Remember, energy likes to flow from high potential to low potential.

[00:35:06]So, this is low to high, so this cannot work. Right?

[00:35:12]Um 12 wouldn't work because then there would be no potential difference there.

[00:35:17]Right? It's all balanced, so so no current will flow.

[00:35:20]Cannot overcome the electromotive force.

[00:35:22]So, the best option to use is 15 V. So, the 15 15 V DC supply creates a high potential and the battery, which is a low potential now, will be charged. So, the current will flow from high to low potential. So, we have 15 V as the answer.

[00:35:44]All right, moving on to the next part of the question.

[00:35:49]All right, so this part of the question is some calculations here.

[00:35:53]So, it says here that a compact fluorescent light bulb consumes um 4.8 W of power when connected to a 120 V AC source.

[00:36:06]All right, our AC supply.

[00:36:07]Now, we're asked to calculate the current flowing through the bulb.

[00:36:12]All right?

[00:36:13]So, what's really happening here is that we have a AC source.

[00:36:21]All right?

[00:36:25]And the AC source is connected to a bulb.

[00:36:32]All right?

[00:36:35]The AC source is connected to a bulb and the voltage is 120 from the source.

[00:36:43]All right? And the bulb consumes 4.8 W of power.

[00:36:52]So, now we're asked to calculate the current flowing through the bulb. So, how are we going to calculate the current flowing through the bulb?

[00:37:01]So, we have what the power we have the power, we have the voltage. So, we need to recall an equation that has all three of these that can help us to find the answer.

[00:37:13]Now, if you remember, power has many equations and it depends on the the variables that are presented to you.

[00:37:22]So, we know that power is equal to I times V.

[00:37:27]So, if you want to find the current, all we need to do is transpose this equation. So, I would be equal to P divided by V.

[00:37:37]So, our current would be equal to 4.8 W divided by 120 V.

[00:37:46]So, the current would end up being Let's do that on the calculator.

[00:37:51]4.8 W divided by 120 gives us answer of 0.04.

[00:38:01]So, it's 0.04 A or ampere, which is the SI unit of current.

[00:38:11]So, that's how we would do that question.

[00:38:17]All right? Moving on to the next part of the question.

[00:38:20]So, it says if the bulb is connected for 5 minutes, calculate the charge passing through it.

[00:38:28]So, now the charge comes into play, which is Q.

[00:38:33]And they've given us the time.

[00:38:36]And we're asked to calculate the charge, right?

[00:38:40]So, we know that the equation that has these two in it is Q that is equal to I times T.

[00:38:51]And since they're asking us to calculate the charge, there's no need to transpose, so Q is equal to the current that we just found, right? Which is 0.04 amps times the time. Now, if you look at the time, time is in minutes. However, the SI unit for time is seconds. So, we need to convert our 5 minutes into seconds in order to use it to calculate our value.

[00:39:17]So, just to go to the side here, um we know that 1 minute is equivalent to 60 seconds. So, the question is how much seconds is 5 minutes?

[00:39:30]So, how much seconds will 5 minutes be?

[00:39:32]So, it's simple, so we just multiply cross multiply, and we get 5 * 60, which gives us 300 seconds.

[00:39:45]All right? So, 0.04 * 300 seconds will give us the charge.

[00:39:53]All right?

[00:39:54]So, if we should do that on the calculator, right? So, we have 0.04 times 300 seconds, which gives us a charge of 12.

[00:40:08]So, it's 12 coulombs for the charge.

[00:40:14]All right?

[00:40:21]All right. So, we're moving on to the next one. So, calculate the energy the bulb transfers in 5 minutes.

[00:40:33]So, how do we calculate the energy the bulb transfers in 5 minutes?

[00:40:42]So, now energy comes into play, time comes into play.

[00:40:47]So, how do we calculate the energy transferred in 5 minutes?

[00:40:53]So, if you remember power is equal to energy over time.

[00:40:59]Right?

[00:41:00]And they ask us to calculate energy. So, therefore, if we transpose this equation it means that energy is going to be equal to P times T.

[00:41:13]So, energy is equal to the power right? Which we found which was given to us, which is 4.8.

[00:41:23]So, the power was 4.8 W times the time, which is the same 5 minutes. And we know that we should convert our time to seconds, and we already converted our 5 minutes, so we know that is 300 seconds.

[00:41:40]So, the energy would then be equal to Let's clear that.

[00:41:48]So, the energy would then be equal to 4.8 * 300 which would be 1,440 J.

[00:42:01]So, it's 1,440 440 J of energy.

[00:42:10]And we would have gotten the 15 marks for this part of the question.

[00:42:19]All right.

[00:42:23]Final question.

[00:42:27]So, now we're at the final question.

[00:42:30]And the final question here is on radioactivity.

[00:42:38]So, let's see who can get three marks here.

[00:42:43]So, radioactivity, um radioactive materials emit nuclear radiation. And the first question is to list three main types of nuclear radiation.

[00:42:55]So, let's see who knows the three types of nuclear radiation.

[00:43:15]What are the three types of nuclear radiation?

[00:43:18]So, I'm seeing looking in the chat to see who will get these three marks.

[00:43:37]No, it's not radium.

[00:43:39]Three types of nuclear radiation.

[00:43:50]Dragon.

[00:44:06]All right. So, let's go into it.

[00:44:11]So, the three types of nuclear radiation is one, we have alpha.

[00:44:16]So, that's we have alpha emissions.

[00:44:21]Number two is beta.

[00:44:25]So, we have beta emissions.

[00:44:28]And then we have gamma.

[00:44:33]Gamma waves.

[00:44:36]So, these are the three types of nuclear radiation.

[00:44:42]All right, moving on.

[00:44:44]Identify the type of nuclear radiation which forms part of the electromagnetic spectrum.

[00:44:50]So, which of these three is a part of the electromagnetic spectrum?

[00:44:57]All right. Great.

[00:45:00]So, gamma waves.

[00:45:08]So, gamma waves are part of our electromagnetic spectrum.

[00:45:17]Next part of the question, determine the type of radiation that is the most ionizing.

[00:45:23]So, which type of radiation is the most ionizing?

[00:45:42]So, no, it's not gamma that is most ionizing.

[00:45:51]Try again.

[00:45:53]All right, so it's alpha.

[00:45:55]So, alpha particles are more ionizing.

[00:46:00]And the reason for that is because they have a charge of plus two.

[00:46:08]All right? So, they have a high charge which is me and they are and they are also larger.

[00:46:15]All right? So, alpha particles they are most ionizing.

[00:46:23]All right. Which type is made up of electrons?

[00:46:29]Which type is made up of electrons?

[00:46:57]All right. Beta.

[00:46:59]Nice. So, beta beta particles are made up of electrons.

[00:47:11]So, alpha particles and remember alpha particles actually is similar to a helium atom and beta particles they are represented by electrons.

[00:47:27]All right? So, alpha particles have a charge of plus two while beta particles have a charge of negative one.

[00:47:38]All right.

[00:47:44]All right.

[00:47:50]All right. So, this part of the question focuses on our Geiger-Müller tube. So, in your studies of radioactivity you would have been exposed to how the radioactive detector works and one of the searches is the Geiger-Muller tube.

[00:48:07]So, it says here that the figure five shows a radioactive source which was placed in front of a radioactive detector. The average background count rate is five counts per second, so that is important.

[00:48:23]So, the background count rate is five counts per second.

[00:48:27]With the source um in place, the rates rise to 69 counts per second. After an hour, so we need to take into account the time.

[00:48:39]After an hour, the count rate drops to 21 counts per second.

[00:48:47]And now we are asked to determine the correct count rates for the activity of the source.

[00:48:58]So, would you answer this question?

[00:49:05]Let's see how many of us have mastered the radioactivity topic.

[00:49:14]How would we determine the counts for the activity of the source?

[00:49:55]>> All right, so let's break it down. So, when we talk about background count rate, we're actually talking about background radiation.

[00:50:03]And background radiation is just um ionizing radiation within our environment. So, these are emissions that are done um naturally. All right, they are naturally um in the background all around us.

[00:50:18]Right? So, the background radiation is recorded there. So, anytime we use a Geiger-Müller tube, there will always be some form of background radiation.

[00:50:28]Now, when the source, so here we have a radioactive source, when the source is placed in front of the tube, right? And that is to ensure that the the um the radiation goes into the tube.

[00:50:44]So, when the source is placed in front of the tube, this source will now emit um its own radiation.

[00:50:53]And it will cause the count rate, that is the disintegration, the decays per per second to go up, right? Because this is now uh radioactive source being placed directly in front of the the tube.

[00:51:07]And it says, "After an hour, the count rate drops to 21 counts per second." So, if we are supposed to find out the count rate for the activity of the source, then we need to find a way to get rid of the background um count rate. So, how would we get rid of the background count rate?

[00:51:52]>> Yes, there's a formula that you have to use.

[00:51:59]But it's also something intuitive.

[00:52:10]How do we get the count rate? The actual count rate.

[00:52:16]So, what we are trying to find here is the actual count rate.

[00:52:27]Right? So, how do we find the actual count rate? So, we have the background count rate and then when we put the source in front of it, we know have a count rate of 69 um counts per second.

[00:52:39]So, this is the total of both the background and the counts from the the source.

[00:52:47]So, in order to get the actual, we would actually have to minus the background count rate from the observed or received count rate.

[00:52:57]So, the received count rate Right?

[00:53:05]Minus the background So, the received count rate minus the background count rate will give us the actual count rate.

[00:53:21]So, the received count rate for radioactive source when it just started is 60 69, sorry.

[00:53:32]Minus the background count rate which is five.

[00:53:37]So, that means the actual count rate here is going to be 60 four counts per second.

[00:53:54]All right? And it said activities, so we are going to look at the activity after the hour.

[00:54:01]And the activity after the hour was 21.

[00:54:05]So, that means we are going to do the same thing.

[00:54:09]So, at the end of 1 hour it would be 21 minus the background count rate, which is five.

[00:54:18]And so, we would get a count rate, actual count rate of 16 counts per second.

[00:54:31]All right?

[00:54:32]So, that's how we would correct the error caused by the the background count rate.

[00:54:39]All right?

[00:54:40]All right. Now, it says we are to calculate the half-life of the source.

[00:54:45]So, now that we have the corrected numbers, we are to calculate the half-life of the source.

[00:54:55]So, if you remember from radioactivity, what's the meaning of half-life?

[00:55:00]If you know what the meaning of half-life is, you can work this out.

[00:55:15]How would we What's the definition for half-life?

[00:55:39]All right. So, it's all long it takes to decay by how much.

[00:55:45]So, you're missing something from the definition there.

[00:55:48]So, how long it takes.

[00:55:57]All right. So, remember half-life is how long it takes for the the sample to be go down by half each time. So, that's the half-life. So, if I have 64, all right, the half-life then will be how long it takes to go from 64 to half of its of this value. So, half of this value would be what?

[00:56:29]All right. So, half of that value would be 32.

[00:56:33]All right? And then we need to go down more because he said the final quantity was 16. So, half of 32 would be 16.

[00:56:43]All right?

[00:56:44]So, how many half-lives did it take to go from 64 to 16?

[00:56:51]How many half-lives did it take to go from 64 to 16?

[00:57:14]So, if we should count the number of half-lives, we would get like one here and the next one here. So, it takes a total of two half-lives. That's correct. So, it takes a total of two half-lives to get from 64 to 16, but we need to actually find the actual time taken.

[00:57:32]So, if it takes an hour, right, to go from this, the starting, to the ending, then what would be the half-life?

[00:57:42]And we want to get this in minutes.

[00:57:47]So, how would we get the half-life?

[00:57:50]If it this entire thing took an hour, which is 60 minutes, right, what is the value of one half-life?

[00:58:16]All right. So, T 1/2 So, one half-life would be equal to 60 minutes, right, divided by two.

[00:58:28]And so, each half-life represents 30 minutes.

[00:58:35]All right. And that's where you get the three marks.

[00:58:41]Nice.

[00:58:46]All right. Final part of the question.

[00:58:50]Almost there.

[00:58:52]So, he says a block of lead is then placed between the source and the detector, right? So, if you remember here, so now they're saying that a block of lead is placed between the source, right, and the detector.

[00:59:13]And the results are shown in the table below. So, it says the count rate with the source is 69, and the count rate with the source and the block is 32. So, what is the count rate with the source and the block removed?

[00:59:32]So, if we should remove this and we should remove this, what would be the count rate?

[00:59:47]What would be the count rate here?

[00:59:50]If we should remove the source and the block.

[01:00:11]All right, somebody added them together.

[01:00:15]And that's not correct.

[01:00:17]So, remember Remember what the question said, right?

[01:00:23]So, the question said that when the source is placed in front, right? So, when the source is placed in front, then we get this.

[01:00:34]Right?

[01:00:35]Then they said that when you add the block of lead in front of the source, right? So, if we add the block of lead in front of the source right here, right? It says the count rate goes down to 32.

[01:00:53]But now, if we remove the source and the block, if you remove the source and the block, meaning that if we take away all of this, what are we left with if we should take away all of that?

[01:01:18]Somebody said zero.

[01:01:20]It's not zero cuz as I said, remember background radiation is natural. It's always there in the back in the environment.

[01:01:29]All right?

[01:01:30]So, no matter if there's a radioactive source or not, the environment actually produces background radiation. And the background radiation recorded when there was nothing in front of the the um the tube was actually five counts per second.

[01:01:48]All right?

[01:01:50]So, we have to remember that.

[01:01:52]So, when there is nothing in front of the tube, it will pick up the background radiation, which is actually five.

[01:02:04]All right.

[01:02:06]Now, we move on to the next question. It says, "The source is known to emit one type of radiation only. Deduce what type of radiation." So, from the previous part of the question, we know that there are three types of emission that we can get. All right? But, based on the scenario, what type of radiation is at the source?

[01:02:29]Is it alpha? Is it beta? Is it gamma?

[01:02:51]So, what type of emission? All right, somebody said gamma.

[01:02:56]Great. So, correct answer is gamma.

[01:03:00]So, you would have gotten So, gamma waves are being emitted, but how do you know? It says, "Give two reasons to support your answer." How do you know that it is gamma waves that are being emitted from the source?

[01:03:14]How do we know that it's gamma wave that are being emitted from the source?

[01:03:31]So, this comes back to um knowing the stopping power of each of them. So, what is what is able to stop each of these radiations?

[01:03:44]Right? So, if you should go back to your notes, you realize that for alpha emissions, right? It takes So, for alpha emissions, it takes a thin sheet of paper to stop alpha particles.

[01:04:05]Right?

[01:04:07]For beta emissions, what what is it that we need to stop beta emissions?

[01:04:17]What is it that we need to stop beta emissions?

[01:04:26]Yes, so one of your reasons there I see you choose they choose lead to put in front of the source and you're saying because of the high penetrating power, so that's why you know that's correct.

[01:04:36]Right?

[01:04:38]But what what is used to stop beta emissions?

[01:04:48]Right, so if we use so several meters of concrete, right? Or we can use aluminum.

[01:04:59]But in order to stop gamma emissions, right?

[01:05:05]In order to stop gamma emissions or attempt to stop gamma emissions, we need something thicker than that.

[01:05:12]All right, we need something thicker than that.

[01:05:23]So, let me fact-check my my thing here.

[01:05:31]All right.

[01:05:32]Right. So, um for beta emission, thanks for the correction there. For beta emission, it's actually foil paper.

[01:05:42]So, foil or aluminum.

[01:05:52]So, 3 to 5 mm of aluminum we can use to stop beta emissions, and then for gamma waves, you can use several meters of concrete.

[01:06:02]Right? Or we can use the lead.

[01:06:09]Right?

[01:06:11]So, the fact that when you put the lead in front of the source, it still had um it still picked up some counts, it means that it had to be a gamma a gamma wave going through.

[01:06:26]Right? Because it was any one of the others, they would it would stop already. They wouldn't record anything.

[01:06:32]So, all we need to do is to write out the reasons in a sentence.

[01:06:38]So, reason number one, lead is required to reduce gamma radiation. So, number one, lead is used to reduce gamma radiation.

[01:07:04]All right, and number two, how would be number two?

[01:07:14]So, the fact that um some of the radiation was able to penetrate the lead.

[01:07:21]So, the lead doesn't completely stop it.

[01:07:24]But, some radiation, so some radiation was able to penetrate the lead.

[01:07:47]So, once you put in those two um reasons, then that's what indicates that it's actually gamma waves.

[01:08:01]All right, and this brings us to the end of our practice session for tonight.

[01:08:15]All right.

[01:08:16]So, thanks for joining.

[01:08:18]And I hope that you will continue studying. Remember that on Thursday, we'll have our live Zoom Zoom um pre preparation session.

[01:08:30]So, ensure you keep following the channel for more updates. So, find the Ministry channel on WhatsApp for the link that will be used for Thursday.

[01:08:41]Thank you for joining.

[01:08:45]Have a good night.