Wave Nature of Radiation
Open the lesson
Are X-rays faster than radio waves? Learn how wavelength, frequency and wavenumber are linked by c = nu lambda and nu bar = 1/lambda, the order of the electromagnetic spectrum from gamma rays to radio waves, the visible range 400 to 750 nm, the cm-1 and angstrom unit traps, and two JEE-style solved problems.
Chapters
- 0:00Introduction
- 0:37Wavelength, frequency and wavenumber
- 2:00The electromagnetic spectrum
- 3:37JEE trap: units of wavenumber and wavelength
- 4:51Example 1: frequency and wavenumber of 580 nm light
- 5:59Example 2: wavenumber to wavelength, frequency and region
- 7:35Quick revision: wavenumber, EM spectrum and unit traps
Transcript
0:00Introduction
Hello everyone! Okay, one small puzzle before we start. X-rays, and the radio waves from an FM station, are both electromagnetic radiation. And X-rays have a much higher photon energy than radio waves. So, more energy, more speed, yes? X-rays must travel faster! Is that right? Hmm, hold that thought. By the end of this class, you will know the answer, and exactly why. And look, in JEE, most numericals on light and photons start with today's simple formulas. So, let's begin.
0:37Wavelength, frequency and wavenumber
So, every electromagnetic wave has two key numbers. Its wavelength, lambda, the Greek letter lambda, and its frequency, nu. Careful, this Greek letter nu looks like a v, but it is nu, okay? They are tied together by one formula: c equals nu times lambda. Each wave is lambda long, and nu waves pass every second, so in one second the wave moves nu times lambda. That is its speed, c. Clear? And in vacuum, this speed c is three point zero times ten to the power eight metre per second, written m s to the minus one. Write this number down; you will use it in every numerical. Now think. In vacuum, c is fixed. So if the wavelength gets longer, what happens to the frequency? Yes, exactly! It gets smaller. Nu times lambda must stay equal to c, so a longer lambda means a smaller nu. The double arrow means so. Remember this; the whole spectrum is built on it. One more quantity: the wavenumber. Nu with a bar on top, we say nu bar, equals one over lambda. It is the number of waves in one unit of length, okay? Same Greek letter, but with the bar it is a completely different quantity. Don't mix them up, got it?
2:00The electromagnetic spectrum
Alright, now let's line up all the radiations, in order of increasing wavelength. The little less than signs show the order. Shortest first: gamma rays, with the Greek letter gamma, then X-rays, then ultraviolet, then visible light. And then, longer and longer: infrared, microwaves, and finally radio waves, the longest of all. Gamma to radio, okay? The visible part is a small window: four hundred to seven hundred and fifty nanometres, from violet to red. So violet is at the short end, and red at the long end. Now use what we just learnt. From gamma rays towards radio waves, the wavelength increases. So does the frequency go up, or down? Down, exactly! Longer lambda, lower nu, and lower photon energy too. So gamma rays have the highest frequency and photon energy, and radio waves the lowest. Now, back to our puzzle. X-rays have more photon energy than radio waves. So, are they faster? No! In vacuum, all of them travel at the same speed, c. X-rays just have a shorter wavelength and a higher frequency; nu times lambda is the same c for both. See, puzzle solved! More photon energy means a higher frequency, not a higher speed. And one more property: they are not deflected by electric or magnetic fields. Remember the cathode rays of chapter one? They bent towards the positive plate. These radiations do not bend at all.
3:37JEE trap: units of wavenumber and wavelength
Now, a classic JEE trap. Wavenumber is often given per centimetre, written c m to the minus one. One centimetre is one hundredth of a metre. So one per centimetre is one hundredth per metre, right? No, no, careful! One per centimetre means one wave in every centimetre. And one metre has a hundred centimetres, doesn't it? So that is a hundred waves per metre: one c m to the minus one equals one hundred m to the minus one. Clear? Wavelengths are also given in angstrom: the capital A, with a small circle on top, okay? One angstrom is ten to the power minus ten metre, which is zero point one nanometre. Now a topper's trick. Nu equals c over lambda, and one over lambda is nu bar. So nu equals c times nu bar, directly! No need to find lambda first. Really fast. But the units must match. If nu bar is in per centimetre, take c in centimetres per second: three times ten to the power ten, c m s to the minus one. Same speed, just written in centimetres. Then the centimetres cancel cleanly. With me?
4:51Example 1: frequency and wavenumber of 580 nm light
Now, example one. Find the frequency and wavenumber of yellow light of wavelength five hundred and eighty nanometres. Pause the video, and try it first. Done? Okay, let's check. Many students put five hundred and eighty straight into c equals nu lambda. Will that work? No! Our c is in metres per second, so lambda must be in metres. Nano means ten to the power minus nine, so lambda is five hundred and eighty times ten to the power minus nine metre. Okay? Now, nu equals c over lambda: three times ten to the power eight, over five hundred and eighty times ten to the power minus nine. The wavy equals sign means approximately: about five point one seven times ten to the power fourteen hertz. And the wavenumber, nu bar, is one over lambda: one over five hundred and eighty times ten to the power minus nine. About one point seven two times ten to the power six, per metre. That's it, two quick divisions, and done!
5:59Example 2: wavenumber to wavelength, frequency and region
Now a slightly harder one. Here, a radiation has wavenumber two times ten to the power four, per centimetre. Find its wavelength and frequency. And in which region does it lie? Pause and try. Hint: watch the units. Done? Okay, let's check. Lambda is one over nu bar: one over two times ten to the power four, in centimetres. That is five times ten to the power minus five centimetre. Now divide by a hundred to get metres: five times ten to the power minus seven metre. And that is five hundred nanometres. Keep this number; we need it at the end. Now the frequency, with our trick: nu equals c times nu bar. Take c as three times ten to the power eight, and multiply. Correct? Oh no, careful! That is the trap. Nu bar is per centimetre, so c must be in centimetres per second: three times ten to the power ten. The two brackets multiply, and the centimetres cancel. Three times two is six, and ten to the power ten times ten to the power four is ten to the power fourteen. So nu is six times ten to the power fourteen hertz. Last part. Five hundred nanometres: which region is this? Yes, visible light! The less than signs say it: five hundred lies between four hundred and seven hundred and fifty nanometres. See? Lambda first, and the region is just one comparison.
7:35Quick revision: wavenumber, EM spectrum and unit traps
So, quick revision. The speed c equals nu lambda, nu bar equals one over lambda, and nu equals c times nu bar, with c three times ten to the power eight metre per second. From gamma rays to radio waves, the wavelength goes up, the frequency and photon energy go down, and all of them travel at c in vacuum. And watch the units: one per centimetre is a hundred per metre, and one angstrom is zero point one nanometre. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!
More videos in Unit 2 · Atomic Structure
- Discovery of Subatomic Particles
- Atomic Models
- Atomic Number, Mass Number and Related Terms
- Counting Particles and Average Atomic Mass
- Planck's Quantum Theory
- Counting Photons and Energy Balance
- The Photoelectric Effect
- Bohr's Postulates and the Orbits
- Transitions, Ionisation and Excitation
- Other Orbital Quantities and Limitations
- Hydrogen-like Ions: A Drill
- Line Spectra and the Rydberg Formula
- First Lines, Series Limits and Ratios
- How Many Lines?
- de Broglie Wavelength
- de Broglie Waves in Bohr Orbits
- Heisenberg's Uncertainty Principle
- Wave Functions and Orbitals
- The Four Quantum Numbers
- Shapes of Orbitals and Nodes
- Energies of Orbitals
- Rules for Filling Orbitals
- Exceptions and Exchange Energy
- Configurations of Ions
- Unpaired Electrons and Magnetic Moment
- Counting Electrons by Quantum Numbers
- Overview
- Mixed Problems: Set 1 (Particles, Photons and the Photoelectric Effect)
- Mixed Problems: Set 2 (Bohr Model, Spectra and de Broglie)
- Mixed Problems: Set 3 (Quantum Numbers, Nodes and Configurations)
- Overview
- Part A: JEE Main Level
- Part C: Concept Checks