Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small situation before we start. Light of wavelength three hundred and ten nanometres falls on a metal, and you need the energy of one photon, in electron volts. The long way? h into c, divided by lambda, powers of ten everywhere, and then one more division to change joules into electron volts. And all this while, your clock keeps running, right? But toppers get it in one line. How? Hmm, think about it. We will find out today. And look, atomic structure gives one or two questions in almost every JEE Main shift. So today I will show you how to use this chapter of mixed problems like a real test, and then we will solve two problems together. So, let's begin.

0:47Why this unit matters in JEE Main

So, first, why does this unit matter so much? Atomic structure gives one or two questions in almost every JEE Main shift, okay? Almost every shift! For just one unit, that is a lot of marks. Now tell me, what kind of question is it, usually? Pure theory to remember, or something else? Mostly numerical, that's right! You calculate a number, and that number is your answer. So your method must be clear, and your calculation must be fast. Clear?

1:19What the questions ask

So what exactly do these questions ask? Let's go one by one. First, particle counts. Then photon energies, and the photoelectric effect. Both of today's examples come from here, okay? Next, Bohr-model quantities. Then spectral lines, and de Broglie wavelengths. Alright, a few more to go. And finally, nodes, configurations, and magnetic moments. See, you have already learnt every one of these in Chapters One to Seven. Here they simply come mixed together. Got it?

1:52How this chapter is built

Now, where are the real exam questions? The real JEE Main questions from twenty twenty one to twenty twenty six are at the end of Chapters One to Seven, okay? This chapter is different. It has three sets of fifteen problems, all original, written in the same style as those real questions. So you get fresh practice, on problems you have never seen before. Set one is on particles, photons and the photoelectric effect. Both our examples today come from this set, so you will see its style in a few minutes. Set two is on the Bohr model, spectra and de Broglie. And set three is on quantum numbers, nodes and configurations. So between the three sets, the whole unit is covered, every bit of it. Right?

2:39Treat each set as a real test

Now the most important part: speed. Each problem in this chapter should take one to two minutes. Just one to two minutes, okay? So tell me, fifteen problems, one to two minutes each. How long should one whole set take? Fifteen into one is fifteen, and fifteen into two is thirty. So fifteen to thirty minutes for a set, exactly! Sit down with a clock, and treat every set as a small test. Now, here comes the trap. You are on problem five, you feel a little stuck, and you think, let me just take a quick peek at the solution. You read it, you nod, and it all looks so easy. So that problem is done, right? No, no, careful! That easy feeling is the trap. Reading a solution is not the same as solving the problem. So the rule of this chapter is clear: attempt the whole set before reading any solution. Your own answers, on paper, right or wrong. Then compare, and learn from every mistake.

3:39Topper's trick: the 1240 shortcut

Now, remember our puzzle from the start? The energy of one photon is E equals h c over lambda. Here h is Planck's constant, c is the speed of light, and lambda is the wavelength. Correct, but h and c come with powers of ten, and the answer comes out in joules. So, which shortcut gives the photon energy straight in electron volts? You have already met it in the photon chapter. That's right, twelve forty! E in electron volts equals twelve forty over lambda in nanometres. The brackets just tell you the units, okay? This chapter itself tells you to use it for photon energies. Wavelength in nanometres goes in, energy in electron volts comes out. No powers of ten at all!

4:24Example 1: stopping potential for 310 nm light

Now, let's practise exactly the way this chapter says. This one is from set one, and it is a medium one. Here, a metal of work function two electron volts is lit by light of wavelength three hundred and ten nanometres. Find the stopping potential. Pause the video and try it. Your two minutes start now. Done? Okay, let's check. First, the photon energy, with our shortcut: twelve forty over three hundred and ten. Exactly four electron volts! Look, there is our puzzle from the start, solved in one line, with no powers of ten. Now the stopping potential. The photon brings four electron volts, so the stopping potential is four volts, right? No, no, careful! The metal takes its share first: the work function, phi naught, the Greek letter phi with a small zero. So K max, the maximum kinetic energy, equals h nu minus phi naught. Here h nu, h into the frequency nu, is just the photon energy, our four. Four minus two: two electron volts. Last step: e V naught equals K max. Here e is the charge of the electron, and V naught is the stopping potential. The arrow means, so. Two electron volts of kinetic energy means V naught is two volts. In electron volts, the number stays the same, only the unit changes.

5:52Example 2: stopping potential change, 400 to 300 nm

Got it? Now a hard one, also from set one, okay? By how much does the stopping potential for a metal change, when the wavelength of the light is changed from four hundred nanometres to three hundred nanometres? Pause the video and try it. Done? Okay, let's check. The first reaction of many students: oh no, the work function is not even given! So how can we solve this? Yes, we can! Write e V naught equals h nu minus phi naught. Same metal both times, so the same phi naught. And we take it that both wavelengths do eject electrons, otherwise this method does not work, okay? When we take the change, phi naught simply cancels. We never need it. Clear? Alright, the wavelength changed by one hundred nanometres. So the change is twelve forty over one hundred: twelve point four volts, right? No, no, wait! That is the JEE trap. Twelve forty over lambda works for the wavelength of one photon, and no photon here has a wavelength of one hundred nanometres. So find each photon energy, and subtract. Delta V naught, where the little triangle delta means change, equals twelve forty over three hundred, minus twelve forty over four hundred, in volts. Now, take twelve forty common, outside the bracket: one over three hundred, minus one over four hundred. Common denominator, twelve hundred. Then the top becomes four minus three, right? So it is twelve forty times four minus three, over twelve hundred. With me? Four minus three is one, so delta V naught is twelve forty over twelve hundred: about one point zero three volts. The wavy equals sign means approximately, okay? Shorter wavelength, bigger photon energy, so the stopping potential rises. See: no work function, no powers of ten, and done in about a minute. That is JEE speed!

7:57Quick revision: 1240 shortcut and timed practice sets

So, quick revision. One or two questions a shift, and one to two minutes a problem. For photon energies, twelve forty over lambda in nanometres gives electron volts. And attempt a whole set before reading any solution. Now open set one, start your clock, and begin. You can do this, I promise. All the best!