Comparing Particles and Photons

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Imagine a photon and an electron, both with a wavelength of exactly one nanometre. Same wavelength, so they must carry the same momentum, and the same energy too, right? Hold on to your guess; in the second example, we will check it with real numbers. Questions that compare two particles come up often in JEE, and one simple idea handles all of them. So, let's begin.

0:29One formula, many comparisons

So, here is our toolkit in one line, from the last lesson. Lambda, the de Broglie wavelength, equals h over p, with h Planck's constant and p the momentum. That is h over m times v, with m the mass and v the speed, or h over the square root of two m K, with capital K the kinetic energy, or, 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, and all of this holds for non-relativistic particles, as the screen says, okay? Now the secret of every comparison: see what is kept the same, and which letters are left. Clear? First, the same momentum, which you know already. Lambda is h over p, so the same p gives the same lambda, whatever the masses. Right? Next, the same speed v for both particles. Now tell me, how does lambda depend on the mass this time? Look at h over m v. With h and v fixed, only the mass is left in the denominator. So lambda is proportional to one over m; this sign, like a stretched alpha, means proportional to. Clear? Now the same kinetic energy. Heavier particle, shorter wave, so lambda is again proportional to one over m, isn't it? Oh no, careful, that is exactly the trap here! For the same kinetic energy, use h over the square root of two m K, okay? Now the mass sits under the square root sign, so lambda is proportional to one over root m. The lighter particle still has the longer wave, but only by the square root of the mass ratio. Got it? And for the same accelerating voltage, lambda equals h over the square root of two m q V, with the voltage fixed. So lambda is proportional to one over the square root of m times q. Here the charge counts too, not just the mass. With me?

2:27Same wavelength, and the photon

Now let us turn it around. If two particles have the same wavelength, then, since lambda equals h over p, they have the same momentum; the double arrow means it follows that. Okay? So let me ask you this. Same momentum, one light particle and one heavy one. Which one has more kinetic energy? The lighter one, yes, exactly! Capital K equals p squared over two m; the small two up top means squared. Same p, and the mass is in the denominator, so the smaller mass gives the larger energy. Right? Now the photon, the particle of light. For a photon, lambda equals h times c over capital E, with c the speed of light and capital E the photon's energy. And a handy number from the photon chapters: h times c equals one thousand two hundred and forty electron volt nanometre; e V is the electron volt, n m the nanometre. Clear? And for a particle with mass, lambda equals h over the square root of two m K, as before. Compare the two: the photon's energy stands alone, but the particle's energy sits under a square root, with its mass. So the same wavelength can hide very different energies. Got it?

3:45Masses and charges to keep ready

Alright, for ratio questions, keep a few numbers ready. Here, m with a small p below is the proton's mass, and m with a small e below is the electron's mass. The proton is about one thousand eight hundred and thirty six times as heavy; the wavy equals sign means approximately. Okay? The deuteron, with the small d below, is about twice as heavy as the proton, and the alpha particle, with alpha the Greek letter, about four times as heavy. Right? And now the charges. Small e is the proton's charge, and q with a small alpha below, the alpha particle's charge, is two e. The proton and the deuteron carry just e each. When the voltage is the same, this factor of two changes the answer. Clear?

4:31Example 1: proton and alpha particle, same voltage

Now, example one. Take a proton and an alpha particle, accelerated through the same potential difference. Find lambda p over lambda alpha; the small p and alpha written below just name the particle. Pause the video and try it first. Done? Okay, let's check. The alpha particle is four times as heavy, and lambda goes as one over root m, so the ratio is root four, which is two, right? No, no, wait, that is the trap we just talked about! One over root m was for the same kinetic energy. Here the voltage is the same, so lambda is proportional to one over the square root of m times q. With me? So write the ratio, and see how it flips. Lambda p over lambda alpha equals the square root of m alpha times q alpha, over m p times q p. Okay? Now put in the numbers. The alpha particle has four times the proton's mass and two times its charge. So we get the square root of four times two, root eight. Right? And root eight is two root two. So the proton's wave is two root two times as long as the alpha particle's. If you wrote two, you forgot the charge. Got it?

5:48Example 2: electron and photon of 1 nm, momenta and energies

Now, back to our puzzle. This is example two. An electron and a photon each have a wavelength of one point zero zero nanometres. Find their momenta, the energy of the photon, and the kinetic energy of the electron. Pause the video and try it first. Done? Okay, let's check. Momentum first. Which formula works for both of them? Lambda equals h over p, so p equals h over lambda, for both. That is six point six three times ten to the power minus thirty four, all over one point zero zero times ten to the power minus nine, since a nanometre is ten to the power minus nine metre. Clear? So p equals six point six three times ten to the power minus twenty five kilogram metre per second, for both; the s to the power minus one means per second. Okay? Now the big question from our puzzle. Same momentum, so the same energy too, right? Let us calculate first. For the photon, E equals h times c over lambda, which is one thousand two hundred and forty over one point zero zero, in electron volts. That gives one thousand two hundred and forty electron volts, or one point two four kilo electron volts. With me? Now the electron. Turn the electron shortcut from the last lesson around: capital K equals one point two two seven over lambda, the whole squared, in electron volts, with lambda in nanometres. With lambda equal to one point zero zero nanometres, that gives about one point five one electron volts. Right? So the answer to our puzzle is no, not at all! One point two four kilo electron volts for the photon, against just one point five one electron volts for the electron. Same wavelength, same momentum, but the photon has far more energy. Got it?

7:43Quick revision: lambda ratio for same p, v, K and V

So, quick revision. Same momentum, same wavelength; same speed, lambda goes as one over m; same kinetic energy, one over root m. Same voltage, one over the square root of m times q, and remember the alpha particle carries two e. And the same wavelength means the same momentum, but the photon has far more energy than the electron. Now go and try the practice problems on this page. All the best!