 Let's solve some numericals on quantum nature of light on photons. Here's the first one We have a bulb of 100 watt 100 watt light bulb. That's giving out a light of frequency 4 times 10 to the 14 hertz Now according to the quantum nature the bulb is releasing photons It's emitting photons and so the question is how many photons are being emitted per second by this light bulb? How do we solve this? The first thing that comes to my mind is I know how to calculate energy of a single photon That is from Planck's equation. It's going to be the Planck's constant H multiplied by the frequency of light So that's energy of a single photon and I know Planck's constant I know the frequency so I can calculate the energy of each photon. Okay now from that How do I calculate the number of photons emitted per second? How do I do that? Well, I'm given The power output meaning I know that this bulb is emitting 100 joules of energy per second. That's the meaning of what? Let me write that down. So I know a power is 100 Joules per second this much energy is given out per second Now using this and the energy and I know the energy of the photon How do I figure out the number of photons emitted per second? Can you pause and think about it? Alright, so to relate these two if I can relate these two I'm done This is sort of like total energy and this is like energy of one photon So I can just relate them so the way I like to relate them is let's say this number number of photons emitted per second Let's say this is n if there are n photons coming out per second and if each photon has this much joules of energy Then the total energy coming out per second must be n times e Right think about it if this was two joules and there are a hundred photons coming out per second That means total energy would be 100 times 2 200 So I know that total energy coming out per second has to be n times E and I know that has to be hundred joules per second So now I can just equate this and I can calculate what n is going to be So why don't you pause the video and see if you can and continue this and solve for for n All right, let's do this. So let me separately first calculate what the energy of the photon is going to be Let me just substitute over here Okay, this is going to be What's the Planck's constant 6.63? I'm just gonna 6.6 times 10 to the power minus 34 joules second and the frequency is 4 times 10 to the power 14 Hertz Hertz is one more second It's always nice to whenever possible use units and make sure that we're getting the right unit So this cancels out you get joules that we want and that gives us how much 6 4s are 24 4 here to gets carried 6 4s are 24 again 26 0.4 times 10 to the power minus 34 plus 14 is minus 20 Joules So now I know this number so I can calculate and n is going to be 100 joules per second the power divided by the number of energy of each photon that's given here. So that is 26 0.4 times 10 to the power minus 20 Joules and that gives me I think I now need my calculator So let me begin my calculator. I have hundred divided by 26.4 That gives me 3.8. So let me just get this. All right, so I get three point. Let me use blue Oops, okay three point. What is that 8 right 3.8 times? So this is 3.8 10 to the power 20 Joules cancels and so I get per second so 3.8 times 10 to the power 24 tons per second and There you go. This is how many photons are emitted per second It's an extraordinary large number and these are very realistic values So if you have a tube light or a bulb right in front of you switch it on and just imagine These many photons are coming out per second. Okay, let's try another one This time we have a laser light of the same frequency as before I'm taking the same frequency so that I don't have to recalculate. We can use the same energy Calculation we did before for each photon. So the photons have the same energy because the frequency is the same So this time we have a laser light of this frequency which is incident on a wall and this time Intensity is given to us and the question is again. What is the number of photons this time hitting the wall? per second So how do we solve this the first step is the same as before earlier? We've done that. I know the energy of every single photon. That is something I know But I don't know the power because if I did then I can do the same thing as I did before this time Intensity is given. What's the meaning of intensity? What's the difference between intensity and power? But as you can see from the units itself intensity is power per area So it's telling me that five joules of energy per second is incident per meter square If I had one meter square area, then this much energy would be incident per second But that energy is incident only on two centimeter square Hmm. So the first step would be to calculate how much is the power in Two centimeters square. I know the power in one meter square is this much What is the power in two centimeters square? Once I figure that out then I can solve this just like before in like just like in the previous numerical So why don't you pause the video and see if you can try doing this on your own now? Alright, one of the things that you might be seeing is I'm not I'm not using some formula for intensity and substituting because I like To do it logically and whenever you try whenever you do things logically it makes a lot more sense And of course it becomes more fun also to do this All right, so I know intensity from that. How do I calculate power? So the first step I know is I want to calculate is how much is the power incident in that two centimeters square So power would be intensity Intensity this is power per area. So to calculate power. I have to do intensity into area. Does that make sense? Intensity is power per area Okay So what is that? I know the intensity is 5 watt per meter squared 5 watt over meter square multiplied by area Area is 2 centimeters squared Now I have a centimeter squared and have meter squared. So let's convert that and again the way I like to do this keeping it simple I know one centimeter is ten to the power minus two meters Right one meter is a hundred centimeters. So one centimeter is ten power minus two meters But I need a centimeter squared. So I need to square this So one centimeter squared would be ten to the power minus four meter squared So that means I can just replace this just delete this and I can say this is two times ten to the power minus four meter squared So that will give me meter square cancels. I get five times two is ten Ten times ten to the power minus four is ten power minus three make sense So that many watts or joules per second That is how much energy that is dumped on this wall every Second Okay, now from that how do I calculate how many photons are dumped on this wall per second? Well, just like before I'm gonna say Let's let's say there are n number of photons hitting the wall per second Whenever you don't know something in physics or maths or any problem call it n or call it x Now if there are n photons hitting per second just like before and if the energy of photon is e Then the total energy hitting per second must be n times e So this should equal n times e and therefore n Number of photons hitting the wall per second should be ten to the power minus three Joules per second divided by e Which is we've calculated before that's 26.4 times ten to the power minus 20 How much will this be? Well again? I think I need my calculator Bring that in So I'm gonna do one divided by 26 point one divided by twenty six point four That gives me point zero three eight or same thing as before I didn't need okay point zero three eight point zero three eight point zero three eight point zero three eight okay point Zero three eight times ten to the power. I have a minus three I have a minus twenty so plus twenty you get minus seventeen No, sorry plus seventeen plus seventeen Plus twenty minus three is plus seventeen and this is Joules so Joules cancels out per second. So these many photons per second And of course we can convert this into we can call this as thirty eight So one two three, so I'm borrowing three tens that means ten to the power seventeen minus three is four Fourteen so ten to the power fourteen Photons per second so that many photons are hitting the wall per second Finally, I have a couple of bonus questions for you, which I want you to think about without you looking at any of this What if I were to increase the frequency of light? But I kept the intensity exactly the same What would you expect to happen to this number would now there be more photons hitting this wall per second? Or do you expect less number of photons to hit the wall per second? Can you pause and think about that? All right, here's how I'm thinking if I increase the frequency then the energy of each photon would increase That means that each photon now is carrying more energy than before and so I would need less number of photons to carry this This much energy that means the number of photons should reduce does that make sense? And if even if you think look at over here that makes sense Energy of each photon has increased, but the intensity has stayed the same so this number stays the same So now the number of photons would reduce Okay, another question again don't look at this but just think logically. What if I keep the frequency the same but I? increase the intensity Now what will happen to the number of photons hitting this wall? What do you think will happen? All right now because the frequency is kept the same the energy of the photon would stay the same Each photon would carry the same energy, but I'm increasing the intensity So I've increased the total energy input and therefore that means it there'll be more photons hitting the wall So you can see if I keep the frequency the same and when I increase the intensity I'm increasing the number of photons that are hitting the wall