Counting Photons and Energy Balance

Transcript

0:00Introduction

Hello everyone! Okay, one small question before we start. Think of a hundred watt bulb glowing in your room. Its light comes out as tiny packets of energy, called photons. So tell me, how many photons of visible light does it throw out every second? Take a guess. Hold on to it, okay? Today we will count them. And look, in JEE, counting photons and balancing their energy gives quick, sure marks, if you know two simple ideas. So, let's begin.

0:33Counting photons from a source

So, first, a quick recall from the last class. One photon carries energy E equals h c over lambda. Here h is Planck's constant, c is the speed of light, and lambda, the Greek letter, is the wavelength. Clear? Now take a source, like our bulb, with power P, okay? Power is the energy given out every second. One watt, the capital W, is one joule per second, written J s to the minus one. So every second, the source gives out P joules. And each photon carries h c over lambda. Now tell me, how many photons come out every second? Yes, exactly! Just divide. Total energy per second, divided by the energy of one photon. So capital N, the number of photons per second, equals P over h c by lambda, which is P lambda over h c. Write this down. This one line is the heart of today's lesson. Now think. Two sources with the same power, but one gives out light of a longer wavelength. Which one gives out more photons per second? The longer wavelength, absolutely! Look, lambda sits on top in the formula. Longer wavelength means each photon carries less energy, so you need more photons for the same power. The double arrow means so, okay?

1:57Topper's trick: the 1240 shortcut

Now a topper's trick that saves a lot of time. Remember this from the last class: the photon energy in electron volt, written e V, equals twelve forty over the wavelength in nanometres. The brackets just tell you the units. No powers of ten at all! And to change electron volt into joules: one e V is one point six times ten to the power minus nineteen joule. Keep this number ready, it comes in almost every photon question. Got it? Then simply, N equals P over E, with P in watts and E in joules. Same formula as before, just in two easy steps. Quick and simple.

2:38Energy balance: one photon in, two out

Alright, now the second idea: energy balance. Suppose a molecule absorbs one photon of wavelength lambda, and then re-emits two photons, of wavelengths lambda one and lambda two. The small one and two written below are just labels, okay? Now, one photon goes in, and two come out. So the wavelengths simply add, right? Lambda equals lambda one plus lambda two? No, no, careful! This is the JEE trap. What is conserved is energy, not wavelength. So the energy that goes in, h c over lambda, equals the energy that comes out: h c over lambda one, plus h c over lambda two. Now h c is in every term, so cancel it. One over lambda equals one over lambda one, plus one over lambda two. See, the reciprocals add, not the wavelengths. With me? One more case. Suppose a photon breaks a chemical bond, and it has more energy than the bond needs. Where does the extra energy go? It is not lost at all! The energy beyond the bond energy appears as kinetic energy of the fragments. On screen, K E means kinetic energy, okay? So the photon energy, h c over lambda, equals the bond energy plus the K E of the fragments. Energy balance again.

3:58Example 1: one photon absorbed, two photons re-emitted

Right, example one. Here, a molecule absorbs a photon of two hundred and fifty nanometres, and re-emits two photons. One of them is five hundred nanometres. Find the wavelength of the other. Pause the video, and try it first. Done? Okay, let's check. Energy balance: one over two fifty equals one over five hundred, plus one over lambda. Every wavelength here is in nanometres, as the bracket says, so no conversion is needed. Nice and easy. So one over lambda is one over two fifty, minus one over five hundred. Make the bottoms equal: one over two fifty is two over five hundred. Two minus one is one, so one over lambda is one over five hundred. So lambda is five hundred nanometres! Interesting, right? Both the new photons have the same wavelength. Now, a quick sense check. Each new photon carries only part of the energy that went in. So should its wavelength be longer, or shorter, than two hundred and fifty nanometres? Longer, yes! Less energy means a longer wavelength. And five hundred is more than two fifty, written with this greater than sign. So our answer makes sense. Always do this quick check, okay?

5:20Example 2: photons per second from a 100 W bulb

Now a harder one, JEE level, and it is our bulb from the start! Here, a hundred watt bulb emits five percent of its power as light, of wavelength six hundred nanometres. How many photons does it emit per second? Pause and try. Hint: read every word of the question. Done? Okay, let's check. The bulb is a hundred watt, so P equals a hundred, right? Straight into the formula? Wait, no! That is the trap. Only five percent of the power comes out as this light, and only that part makes these photons. So the light power is five percent of a hundred watt: just five watt. Clear? Now one photon, with our topper's trick. Twelve forty over six hundred: about two point zero seven electron volt. The wavy equals sign means approximately, okay? Into joules: multiply by one point six times ten to the power minus nineteen. About three point three one times ten to the power minus nineteen joule, for one photon. Now we just divide. N equals P over E: five, over three point three one times ten to the power minus nineteen. About one point five times ten to the power nineteen photons, every second! See, that is the answer to our puzzle. Such a huge number, from just five watt of light. Got it?

6:48Quick revision: photons per second and energy balance

So, quick revision. Photons per second: N equals P lambda over h c, or simply P over E, and use only the power given out as light. One photon in, two out: one over lambda equals one over lambda one, plus one over lambda two. The reciprocals add, not the wavelengths. And when a photon breaks a bond, the extra energy becomes the kinetic energy of the fragments. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!