Set 2

Transcript

0:00Introduction

Hello everyone! Welcome to Set two of our mixed JEE Main problems on dual nature: fifteen problems, and twenty five minutes to solve them. Here is a puzzle from this set. Take a photon and an electron, each with a wavelength of half a nanometre. Same wavelength, so the same energy, right? Hmm, hold that thought; we will settle it in the second example. So, the tools first, then two full problems. Let's go.

0:29Photons and Einstein's equation

So, light comes as photons. Capital E, the energy of one photon, equals h nu, which equals h c over lambda. Here h is Planck's constant, nu, the Greek letter that looks like a v, is the frequency, lambda is the wavelength, and c is the speed of light. Okay? Keep this number ready: h times c equals twelve hundred and forty electron volt nanometre. So the photon energy in electron volt is twelve hundred and forty over lambda in nanometres; that is what e V and n m in brackets mean. Right? Now, in the figure, the wavy orange line is the photon, bringing h nu, and it gives all of it to one electron, the dark blue dot. To get out, the electron spends at least phi naught, the work function, the gap below the surface. Phi is the Greek letter drawn as a circle with a line through it, and the small zero below it is read naught. So how much energy is left for its motion? Yes, exactly, h nu minus phi naught! That is K max; the small max below the K means maximum, for the most loosely bound electrons at the dashed line. The figure writes it as h nu equals phi naught plus K max. And the stopping potential, V naught, measures it: K max equals e V naught, with e the size of the electron's charge. Clear? And never forget the condition. If h nu is less than phi naught, the sign on screen, no electron comes out at all, however intense the light. So every metal has a threshold frequency, nu naught, used in the last form in the figure, h times, nu minus nu naught. Got it?

2:16Intensity or frequency?

Now, a classic question. Keep the frequency fixed, and make the light brighter. Do the electrons come out faster? No, they do not, and this is a common trap. The left graph is photocurrent against collector potential at one frequency, for three intensities, I one, I two and I three; the greater than signs below say I three is the largest. But all three dark blue curves start at the same point, minus V naught. So the stopping potential, and K max, do not depend on the intensity. With me? So what does the brighter light change? More intense light means more photons, so more electrons per second, and a higher saturation current, the flat top of each curve, proportional to the intensity; that sign, like a stretched alpha, means proportional to. On the right, the orange curves are at one intensity, for three frequencies; the caption below says nu three is the largest. They share the saturation current, but the larger frequency has the larger stopping potential, from minus V naught one out to minus V naught three. Okay? So, the shortcut for this whole set: work in electron volt. Since K max equals e V naught, the stopping potential in volt is the same number as K max in electron volt, with no joules needed. Clear?

3:40Matter waves, and the photon

Now, matter waves. Every moving particle has a de Broglie wavelength, lambda, equal to h over p, where p is the momentum. With m the mass, it is also h over the square root of two m K, where capital K is the kinetic energy. And for a charge q accelerated from rest through a potential difference, capital V, it is h over the square root of two m q V, for speeds well below that of light, which the brackets call non-relativistic. Okay? For an electron, the constants are worked out: lambda equals one point two two seven over root V, in nanometres, with V in volt. Use this shortcut for electrons only. Right? Now, a photon also has a wavelength. So can we use h over the square root of two m K for a photon too? We cannot, because a photon has zero rest mass. For a photon, lambda equals h c over capital E, straight from E equals h c over lambda. Clear? So remember this trap: never use one formula for the other, h c over E for a photon, and h over root two m K for a particle. Our puzzle depends on exactly this.

4:54Example 1: stopping potential for 250 nm light, 3.0 eV metal

Got it? Now, Example one, an easy start. Light of wavelength two hundred and fifty nanometres falls on a metal of work function three point zero electron volt. Find the stopping potential. Pause the video and try it first. Done? Okay, let's check. Photon energy first, with our shortcut: twelve hundred and forty over two hundred and fifty gives four point nine six electron volt. That is more than three point zero, so electrons do come out. Okay? Now the stopping potential. The photon brings four point nine six electron volt, so the stopping potential is four point nine six volt, right? Oh no, careful, the work function is paid first! K max equals four point nine six minus three point zero, which gives one point nine six electron volt. With me? And now the shortcut pays off: K max in electron volt is the stopping potential in volt. So V naught equals one point nine six volt; the capital V after the number is the unit, volt. That's it, and no joules anywhere!

6:05Example 2: photon vs electron energy at the same wavelength

And now Example two, a hard one, and our puzzle from the start. Here, a photon and an electron each have a wavelength of zero point five nanometres. Find the ratio of the photon's energy to the electron's kinetic energy. Pause the video and try it first. Done? Okay, let's check. Same wavelength, so we use h c over lambda for both, and the ratio is just one, isn't it? No, no, wait, that is exactly the trap! The formula h c over lambda is only for the photon. So the photon's energy is twelve hundred and forty over zero point five, which gives two thousand four hundred and eighty electron volt. Okay? Now the electron, with its own formula: lambda equals one point two two seven over root V, in nanometres. An electron accelerated from rest through V volt has kinetic energy capital K equals e times V, so K in electron volt is the same number as V in volt. With me? So turn the formula around: root V equals one point two two seven over zero point five, and K is that, the whole squared, in electron volt. Work it out, and K comes to six point zero two electron volt. Right? Finally the ratio, capital E over capital K, is two thousand four hundred and eighty over six point zero two, which comes to about four hundred and twelve; the wavy equals sign means approximately. So our puzzle is settled: same wavelength, but the photon carries roughly four hundred and ten times the electron's energy. Got it?

7:46Quick revision: E = 1240/lambda, Kmax = eV0, lambda = h/p

So, quick revision. First, the photon energy in electron volt is twelve hundred and forty over lambda in nanometres, and K max equals h nu minus phi naught, which equals e V naught. Second, intensity sets the number of electrons, and frequency sets K max. And third, a photon's wavelength is h c over E, a particle's is h over the square root of two m K, and we never mix the two. Now set a timer for twenty five minutes, and try all fifteen problems of this set on this page. All the best!