Calculate the no. of photons per second

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 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 all right 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 2 joules and there are 100 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 100 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 going to 6.6 times 10 to the power minus 34 joules second and the frequency is 4 times 10 to the power 14 hertz herds is one over second it's always nice to whenever possible use units and make sure that we're getting the right units so this cancels out we get joules that we want and that gives us how much six fours are 24. so we get a four here two gets carried six fours at 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 n 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 26.4 times 10 to the power minus 20 joules and that gives me i think i now need my calculator so let me bring in my calculator i have 100 divided by 26.4 and 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 eight right three point eight times so this is three point eight ten to the power twenty 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 well 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 centimeters square so the first step would be to calculate how much is the power in two centimeters square i know the power in 1 meter square is this much what is the power in 2 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 all right 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 centimeter 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 square and i have a meter square so let's convert that and again the way i like to do this keeping it simple i know 1 centimeter is 10 to the power minus 2 meters right one meter is 100 centimeters so 100 power minus 2 meters but i need a centimeter squared so i need to square this so 1 centimeter squared would be 10 to the power minus 4 meter squared so that means i can just replace this i can just delete this and i can say this is 2 times 10 to the power minus 4 meter squared so that will give me meter square cancels i get 5 times 2 is 10 10 times 10 to the power minus 4 is 10 power minus 3. 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 going to 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 heating 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 10 to the power -3 joules per second divided by e which is we've calculated before that's 26.4 times 10 to the power minus 20. how much will this be well again i think i need my calculator so they bring that in so i'm going to do 1 divided by 26 point 1 divided by 26.4 that gives me 0.038 or same thing as before i didn't need okay 0.038.038.03 8.038 okay 0.038 times 10 to the power i have a minus 3 i have a minus 20 so plus 20 you get minus 17. no sorry plus 17 plus 17 plus 20 minus 3 is plus 17. and this is joules so use cancels out per second so these many photons per second and of course we can convert this into we can call this as 38 so one two three so i'm borrowing three tenths that means 10 to the power 17 minus 3 is 4 sorry 14. so 10 to the power 14 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 it 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 will 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