Planck's Quantum Theory

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. At a sweet shop, you can buy one samosa, two samosas, three samosas. But can you buy two and a half samosas? No chance, right? The shopkeeper sells only whole pieces. Now the surprise. Light energy behaves just like these samosas: it comes only in whole packets! Today we will see why Planck proposed this, and how much energy one packet carries. And look, almost every photon numerical in this chapter starts from today's formula. So, let's begin.

0:38Why packets? The hot body puzzle

So, let's see where this strange idea came from. Start with a hot body, or, as the brackets say, a black body. It gives out radiation as a continuous spectrum: no gaps, one smooth spread. Clear? Now, the shape of this spectrum, how much radiation comes out at each wavelength, was a big puzzle. The classical wave theory could not explain this shape, okay? So Planck came up with a bold new idea. Remember our samosa shop? Can you guess what he proposed? Yes, exactly! Packets! Planck proposed that radiation is emitted or absorbed, given out or taken in, only in small separate packets. Never in any amount you like, just like our samosas. Each packet is called a quantum; many packets are quanta. Later, these packets of light were called photons. So whenever you read photon, think: one packet of light energy. Got it?

1:35Energy of one packet

Alright, now the energy of one packet. Planck's answer is one short formula: E equals h nu. E is the energy of one quantum. Nu, this Greek letter that looks like a v, is the frequency. And h is a constant. That's it, energy equals h times frequency. This h is Planck's constant: six point six two six times ten to the power minus thirty four, in joule second, written J s on screen. Write it down. Okay? But questions often give the wavelength, lambda, this Greek letter on screen. Remember c equals nu lambda from the last class, with c the speed of light? So, can you write E using lambda? Simple! Nu is c over lambda; put that into h nu. The double arrow means this gives: E equals h c over lambda. Same energy, written with the wavelength. Very useful. Now tell me. If the wavelength gets shorter, does each photon carry more energy, or less? More, absolutely! Lambda sits in the denominator. This curly sign means proportional to: E is proportional to nu, and E is proportional to one over lambda. Shorter wavelength, bigger packet. We will need this in example two. Next, a beam with n photons, each carrying h nu, has total energy E equals n h nu, okay? Now, back to our samosa shop. Can light of frequency nu carry two and a half h nu of energy? No! Just like samosas. n counts photons, so n is a whole number: h nu, two h nu, three h nu, and the three dots mean and so on. Never two and a half h nu. See, puzzle solved: light energy really does come in whole packets!

3:31Topper's shortcuts: eV and one mole

Now, two time-saving shortcuts that toppers use. Joules give tiny numbers for one photon, so we often use the electron volt, written e V on screen. Shortcut one: E in electron volts equals twelve hundred and forty over lambda in nanometres. The brackets just tell you the units, okay? But careful. Suppose a question gives lambda in metres. Can you put that straight into this shortcut? No, no! The twelve hundred and forty works only with lambda in nanometres, and the answer comes out in electron volts, not joules. So first change lambda to nanometres. With me? Shortcut two: one mole of photons, each of E electron volts, carries ninety six point five times E kilojoules. Keep both shortcuts ready; we will use them in the examples.

4:21Example 1: energy of a 400 nm photon in J and eV

Now, example one. Find the energy of a photon of wavelength four hundred nanometres, in joules and in electron volts. Pause the video, and try it first. Done? Okay, let's check. For the answer in joules, lambda must be in metres. Nano means ten to the power minus nine, so four hundred nanometres is four times ten to the power minus seven metre. Now E equals h c over lambda. Six point six two six times ten to the power minus thirty four, times three times ten to the power eight, that is the speed c, all over four times ten to the power minus seven. First, the power of ten. Minus thirty four, plus eight, and then minus minus seven. What do you get? Minus nineteen, yes! And the front numbers give about four point nine seven. So E is about four point nine seven times ten to the power minus nineteen joule; the wavy equals sign means approximately. So tiny, right? Now the same energy in electron volts. Which shortcut gives it in one line? Twelve hundred and forty over lambda in nanometres, exactly! Twelve hundred and forty over four hundred: three point one electron volts. That's it, one line!

5:44Example 2: longest wavelength to break the O2 bond

Now a harder one, JEE level. The O double bond O energy, see the two lines between the O's, is four hundred and ninety eight kilojoules per mole, written k J mol to the minus one. Find the longest wavelength that can break one O two molecule; the small two below means two oxygen atoms. Pause and try. Hint: one photon breaks one molecule. Done? Okay, let's check. Many students take lambda equals h c over E, and put E as four hundred and ninety eight kilojoules straight away. Will that work? No, careful! That is the trap. Four hundred and ninety eight kilojoules is for one whole mole, but one photon breaks just one molecule. So, times ten to the power three for joules, and divide by Avogadro's number, six point zero two two times ten to the power twenty three. Per molecule, E is about eight point two seven times ten to the power minus nineteen joule. Now lambda equals h c over E, okay? h times c is about one point nine nine times ten to the power minus twenty five; divide by eight point two seven times ten to the power minus nineteen, in metres. That gives about two point four times ten to the power minus seven metre. In nanometres, that is two hundred and forty. Clear? But the question asked for the longest wavelength. Why longest, and not shortest? Longer wavelength, smaller packet; the double arrow on screen says exactly that. At two hundred and forty nanometres, one packet has just enough energy to break the bond. Any longer, and the packet is too small. So this is the longest. Now the region. Two hundred and forty is less than four hundred nanometres, where visible light starts; see the less than sign. So this wavelength lies in the ultraviolet. And now, a topper's check, with shortcut two used backwards. Four hundred and ninety eight, divided by ninety six point five: about five point one six electron volts per photon. Then twelve hundred and forty over five point one six: about two hundred and forty nanometres again! Same answer, no powers of ten. Really fast, right?

8:11Quick revision: E = hv, 1240 shortcut, mole of photons

So, quick revision. Radiation is emitted or absorbed only in packets: quanta, or photons. One packet carries E equals h nu, or h c over lambda, with h six point six two six times ten to the power minus thirty four joule second; and n photons carry n h nu. Shortcuts: twelve hundred and forty over lambda in nanometres gives electron volts, and one mole of photons of E electron volts carries ninety six point five times E kilojoules. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!