Closest Approach and Impact Parameter

Transcript

0:00Introduction

Hello everyone! Here is a strange fact. Fire an alpha particle of a few M e V straight at a gold nucleus, and it never touches the nucleus. It slows down, stops for an instant, and flies straight back. So where exactly does it stop? And if you aim a little to one side, at what angle does it fly off? Two short formulas answer both, and they come up often in JEE questions. So, let's begin.

0:28Head-on: kinetic energy becomes potential energy

So, look at the bottom path in the figure: an alpha particle aimed straight at the nucleus. Alpha is the Greek letter that looks like a rounded a, and the alpha particle has charge two e, where e is the charge of a proton. The nucleus has charge Z e, with Z its atomic number, so the two repel. Okay? As it comes closer, the repulsion slows it down. So at the closest point, what is its speed? Yes, exactly, it is zero! For one instant, the particle stops. So all its kinetic energy, capital K, has become potential energy: K equals one over four pi epsilon naught, times two e times Z e, each in brackets, over r naught. Pi is our familiar three point one four, and epsilon naught, the curly e with a small zero below it, is the permittivity of free space. And r naught, the r with a small zero below it, is the distance of closest approach. With me? Rearrange, and r naught equals one over four pi epsilon naught, times two Z e squared, over capital K; the small two up top means squared. Here the nucleus is taken as fixed. Clear?

1:37The shortcut, and what r₀ depends on

Now, here is a topper's trick. e squared over four pi epsilon naught equals one point four four M e V femtometre. M e V is the mega electron volt, and f m, the femtometre, is ten to the power minus fifteen of a metre. Okay? So for an alpha particle, two times one point four four gives two point eight eight. r naught equals two point eight eight times Z, over K in M e V, directly in femtometres. No joules, no powers of ten. Nice, right? And for any other projectile, of charge small z times e, just replace the two by small z. For a proton, small z is one. Got it? Now, r naught goes as one over K, the kinetic energy. So if I double the speed of the alpha particle, r naught becomes half, right? No, no, careful, that is the trap here! Kinetic energy is half m v squared, with m the mass and v the speed, so double the speed and K becomes four times. So r naught becomes one quarter, as the screen shows: r naught is proportional to one over m v squared; the sign like an infinity sign with one end open means proportional to. Okay? And since the particle turns back without touching, the radius of the nucleus cannot be bigger than r naught. That gives an upper limit for nuclear size. Clear?

3:00Aim off-centre: the impact parameter b

Now, what if we do not aim straight at the nucleus? Look at the double-headed arrow marked b in the figure. b is the impact parameter, the perpendicular distance of the initial line of motion from the centre of the nucleus. With me? The particle is turned through an angle theta, the Greek letter like an o with a line across it, the scattering angle. And b equals one over four pi epsilon naught, times Z e squared over K, times cot of theta over two; cot is short for cotangent. Okay? Now compare the part before the cot with r naught. How are they related? Yes, it is exactly half of r naught! r naught had two Z e squared, and here we have only Z e squared. So b equals r naught over two, times cot of theta over two. Right? Now let me ask you this: if b is zero, a perfect head-on aim, what is theta? Theta is one hundred and eighty degrees, that is correct: the particle comes straight back, and the small circle means degrees. And for a large b, theta tends to zero, which is what the arrow means: the particle hardly bends. So, smaller b, larger theta. Got it? And one special case: when b equals r naught over two, cot of theta over two is one, so theta is ninety degrees. Keep this for our second example. Clear?

4:26What is conserved, and how many scatter

Now a quick question: during the scattering, is the linear momentum of the alpha particle conserved? No, it is not, because the nucleus keeps pushing the particle and turns its direction. Right? But that push always acts along the line to the nucleus, so it is a central force. So the angular momentum about the nucleus and the total energy are conserved, and the path is a hyperbola. Okay? And N of theta, the number scattered at angle theta, is proportional to Z squared times t, over K squared times sine to the power four of theta over two, where t is the thickness of the foil. With me?

5:03Example 1: closest approach of 5.5 MeV alpha to gold nucleus

Now our first example. Find the distance of closest approach of a five point five M e V alpha particle to a gold nucleus, with Z equal to seventy nine, in a head-on collision. Pause the video and try it first. Done? Okay, let's check. The energy is in M e V, so first we convert it to joules, right? No need at all, that is the beauty of our shortcut! r naught is two point eight eight times seventy nine, over five point five, in femtometres. Okay? Two point eight eight times seventy nine is two hundred and twenty seven point five two. Divide that by five point five, and we get about forty one point four femtometres; the wavy equals sign means approximately. Right? And remember our puzzle from the start? The particle turned back about forty one femtometres away, so the radius of the gold nucleus is at most forty one point four femtometres; that is the less than or equal to sign.

6:03Example 2: impact parameter of 5 MeV alpha at 90 degrees

Now a slightly harder one. Find the impact parameter of a five point zero M e V alpha particle scattered through ninety degrees by a gold nucleus. Pause the video and try it first. Done? Okay, let's check. Theta is ninety degrees, so we put cot of ninety degrees in the formula, right? No, no, wait, that is the classic slip! Cot of ninety degrees is zero, which would make b zero, a head-on hit. The formula has theta over two, so we need forty five degrees; the double arrow means which gives. Got it? Now substitute, and notice it is one point four four here, not two point eight eight, because b has only Z e squared. So b is one point four four times seventy nine, over five point zero, times cot of forty five degrees, which is exactly one. One point four four times seventy nine is one hundred and thirteen point seven six. With me? Dividing that by five point zero gives twenty two point seven five, so b is about twenty two point eight femtometres. Okay? And a quick check with our special case: at ninety degrees, b is r naught over two. Here r naught is two point eight eight times seventy nine, over five point zero, about forty five point five femtometres. Half of that is about twenty two point eight, a perfect match. Clear?

7:27Quick revision: r0 shortcut, double-speed trap, b vs theta

So, quick revision. Head-on, r naught equals two point eight eight times Z over K femtometres, with K in M e V, okay? Then, r naught goes as one over m v squared, so double the speed and r naught becomes a quarter. And b equals r naught over two, times cot of theta over two: smaller b, larger theta. Now go and try the practice problems on this page. All the best!