Atomic Models

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Fire alpha particles at a very thin gold foil. Most go straight through, as if nothing is there. But about one in twenty thousand comes straight back! How can a foil be almost empty, and still throw a particle back? Strange, right? Today we solve this, with Thomson's model and then Rutherford's. And look, for JEE, this is the base of the whole atomic structure unit. So, let's begin.

0:34Thomson's model: the plum pudding atom

So, the first model, from J J Thomson. The atom is a uniform sphere of positive charge, okay? Its radius: the small wavy sign means of the order of, so about ten to the power minus ten metre, m for metre. Clear? And the electrons? They are embedded in this sphere, like plums in a pudding: the plum pudding model. The negative electrons balance the positive sphere, so the whole atom is neutral. Simple enough, right? Now, fire our alpha particles at this atom. Alpha is the Greek letter on the badge; they are helium nuclei. What do you expect? See, the positive charge is spread out over the whole sphere. The double arrow means implies: so all alpha particles pass through, with only small deflections. So Thomson explains why the atom is neutral, but not the alpha scattering results. So, we need a new model.

1:26Rutherford's gold foil experiment

Now, Rutherford's experiment, done by his students Geiger and Marsden around nineteen oh nine. They fired a beam of alpha particles, helium nuclei, at a thin gold foil. In the picture, the beam comes from the left, and the dashed circle is one atom of the foil. Most of them, the blue lines marked undeflected, go straight through. Not even a small turn. What does that tell you about the atom? Yes, exactly! Most of the atom is empty space. Nothing in the way, so nothing to turn them, right? Next, a few particles, the green paths marked deflected, turn by small angles. Something positive pushes them, so the positive charge must be concentrated, not spread out like Thomson's pudding. The little plus sign means positive charge, okay? And now, the shocker. About one in twenty thousand, the orange path marked bounced back, comes straight back! Our puzzle. So tell me, what can throw an alpha particle back? Only a tiny, dense, positive nucleus, holding almost all the mass: the orange dot in the picture. This is Rutherford's nuclear model, from nineteen eleven. Puzzle solved: the atom is almost empty, but its centre is tiny and very dense. All clear?

2:45How small is the nucleus?

Alright, so how tiny is this nucleus? Its radius is of the order of ten to the power minus fifteen metre. In the bracket, the same in centimetres: ten to the power minus thirteen. Careful: minus fifteen goes with metres, minus thirteen with centimetres. And the atom, as in Thomson's model: radius of the order of ten to the power minus ten metre. So, what fraction of the atom's size is the nucleus? Divide: ten to the power minus fifteen, over ten to the power minus ten, is ten to the power minus five. So the nucleus is about ten to the power minus five of the atom's size. Tiny! Example one needs this.

3:27Scattering angle and closest approach

Now two results that JEE problems use directly. First, the number of alpha particles scattered through an angle theta, the Greek letter theta. The sign like a stretched alpha means proportional to: one over sine to the power four of theta by two, half the angle. So, as theta gets bigger, do more particles scatter, or fewer? Fewer, yes! Bigger theta, bigger sine of theta by two, so one over sine to the power four gets smaller. That is why very few bounce back. Got it? Second: an alpha particle aimed head on at the nucleus, with kinetic energy capital K, turns back at its closest distance, r nought, r with a small zero. r nought equals one over four pi epsilon nought, times two Z e squared, over K, okay? Epsilon nought is the permittivity of free space. Two e is the alpha's charge, and Z e is the nucleus's charge, Z being its atomic number. Topper's trick for ratio questions: for the same K, r nought is simply proportional to Z, nothing else. More charge on the nucleus, a stronger push, so the alpha turns back farther away. With me?

4:36Drawbacks of Rutherford's model

Now, Rutherford's model has a problem: its electrons move in circles around the nucleus. Suppose the speed stays steady. No change in speed, so no acceleration, right? No, no, careful! In a circle, the direction keeps changing, so the velocity keeps changing. An electron moving in a circle is accelerating. Absolutely. And classical electromagnetism says it must radiate energy. So the electron keeps losing energy, and spirals into the nucleus. How long would that take? About ten to the power minus eight second, s for second. Almost instant! You, me, this table, all made of atoms. Are we collapsing? No! Atoms are stable. So Rutherford's model cannot explain the stability of atoms, nor the line spectra of elements. Bohr's model, in chapter three, addresses both.

5:33Example 1: nucleus the size of a marble, how big is the atom

Now, example one. If a nucleus were magnified to the size of a one centimetre marble, how large would the atom be? Pause the video, and try it first. Done? Okay, let's check. From the bracket: ten to the power minus ten metre, versus ten to the power minus fifteen metre. So the atom is about ten to the power five times the nucleus; the wavy equals sign means approximately. Magnify both by the same factor. The marble is one centimetre, so the atom becomes ten to the power five centimetres. Now, how many kilometres is that? Ten to the power five centimetres is ten to the power three metres: one kilometre across! Look, one small marble in the middle of a ball one kilometre wide. No wonder most alpha particles went straight through, right?

6:25Example 2: closest approach of an alpha to a gold nucleus

Now a harder one, JEE level. An alpha particle of kinetic energy five M e V, mega electron volts, approaches a gold nucleus head on; for gold, Z equals seventy nine. Find the distance of closest approach. Take one over four pi epsilon nought as nine times ten to the power nine, S I units. Pause the video, and try it. Done? Okay, let's check. The formula needs capital K, the kinetic energy. The question says five, so we just put K equal to five, right? No, no, careful! Five is in M e V, but everything else is in S I units, so K must be in joules. Five M e V is five times one point six times ten to the power minus thirteen joule, capital J, okay? On top: nine times ten to the power nine, times two, times seventy nine, times one point six times ten to the power minus nineteen, whole squared. Below: five times one point six times ten to the power minus thirteen. That two comes from the alpha's charge, two e. Don't forget it! Now simplify smartly. One factor of one point six cancels, top and bottom, and the five stays below. Powers of ten: plus nine, minus thirty eight from the square, plus thirteen from below. So what power of ten do we get? Quickly, tell me. Minus sixteen, correct! So we get four hundred fifty five point zero four, times ten to the power minus sixteen, approximately four point six times ten to the power minus fourteen metre. The alpha comes this close, then the tiny nucleus throws it back. Our puzzle, in numbers!

8:09Quick revision: Thomson, Rutherford and the drawback

So, quick revision. Thomson: positive charge spread out, so it cannot explain alpha scattering. Rutherford: mostly empty space, with a tiny, dense, positive nucleus, of the order of ten to the power minus fifteen metre. Drawback: an electron moving in a circle should radiate and spiral in. Now go and try the practice problems on this page. All the best!