Overview

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Fire an alpha particle straight at a gold nucleus. It slows down, stops for a moment, and comes straight back. So, how close did it get? In S I units, you would juggle the electron's charge, epsilon naught and joules, with powers of ten everywhere, right? Today we collect a few numbers that turn such sums into one line, and this whole unit runs on them. Questions like this do come up in JEE, so let's begin.

0:33What this chapter covers

First, a quick map of this chapter. We start with Thomson's model, and the experiment that broke it: alpha particles fired at a thin gold foil. Alpha is the Greek letter on screen. Okay? Next, two lengths. The r with a small zero below it, called r naught, is the distance of closest approach, and small b is the impact parameter. Our first example comes from here. Right? And last, Rutherford's planetary atom, with electrons going round the nucleus like planets, and its two difficulties. Our second example comes from there. Clear?

1:10The key constant: e²/4πε₀

Now, the most useful constant of all. Small e is the charge of a proton, and epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space. So, e squared over four pi epsilon naught equals one point four four mega electron volt femtometre, written M e V f m. With me? The same constant is also written as one point four four electron volt nanometre. Same number, different units. How can that be? Look at the prefixes. One M e V is ten to the power six electron volts; the small raised number is the power of ten. And one femtometre is ten to the power minus fifteen metre. Okay? Now multiply. Ten to the power six, times ten to the power minus fifteen, gives ten to the power minus nine. So one M e V femtometre is ten to the power minus nine electron volt metre, which is exactly one electron volt nanometre! So it is one point four four either way. Got it? Here is the topper's trick: pick the form that matches the problem. Nuclei are a few femtometres across, so for nuclei use M e V and femtometres. Atoms are about a tenth of a nanometre, so for atoms use electron volts and nanometres. Clear?

2:30The rest of the toolkit

Alright, the rest of the toolkit. Here h is the Planck constant and c is the speed of light, and h c equals one thousand two hundred and forty electron volt nanometre. You met it with photons. Okay? Next, capital R, the Rydberg constant, is one point zero nine seven times ten to the power seven, per metre. The cross means times, and the m with a raised minus one means per metre, right? You will use it for spectral lines. Then the letter a with a small zero, called a naught, is the Bohr radius, point five two nine times ten to the power minus ten metre. You will meet it in the next chapter, the Bohr model. With me? And the ground state energy of hydrogen is minus thirteen point six electron volts. Let me ask you this: why is there a minus sign? Because the zero of energy is taken with the electron at rest, infinitely far away. The bound electron has less energy than that, so its energy is negative, as the bracket on screen says: it is bound, and energy must be supplied to free it. Got it? And last, the small u on screen is the atomic mass unit. One atomic mass unit equals nine hundred and thirty one point five M e V over c squared. You will need it for binding energy, in the chapter on the nucleus. Clear?

3:54Example 1: closest approach of a 5.5 MeV alpha to gold

Now example one, from this chapter's part on closest approach. Find the distance of closest approach of a five point five M e V alpha particle to a gold nucleus, in a head-on collision. Capital Z, the atomic number, is seventy nine for gold. Pause the video and try it first. Done? Okay, let's check. In the figure, head-on, the alpha particle comes straight at the nucleus, of charge plus Z e, stops, and turns straight back, so theta, the scattering angle, is one hundred and eighty degrees. The curved paths and the formula for b come later. Now, at the turning point, where has all the kinetic energy gone? Into electric potential energy, exactly! The alpha particle has charge two e, and the nucleus, taken as fixed, has charge Z e, as the brackets on screen show. So capital K, the kinetic energy, equals one over four pi epsilon naught, times two e times Z e, over r naught. With me? Now use our constant. So r naught is one point four four, times seventy nine, over five point five femtometres, right? No, no, wait, that is the classic trap! The alpha particle carries charge two e, not e. So r naught equals two times Z, times one point four four, over K, which is two point eight eight times Z, over K, in femtometres, with K in M e V, as the brackets say. For a proton, the two becomes a one. Okay? Now put in Z equals seventy nine and K equals five point five mega electron volts. Two point eight eight times seventy nine gives two hundred and twenty seven point five two. Right? Divide that by five point five, and r naught is about forty one point four femtometres. The wavy equals sign means approximately. That is our puzzle from the start, solved in one line! Got it?

5:54Example 2: hydrogen orbit radius in Rutherford model

Now example two, a little harder, from this chapter's part on Rutherford's model. The ionisation energy of hydrogen is thirteen point six electron volts. Using Rutherford's model, find the radius of the electron's orbit. Pause the video and try it first. Done? Okay, let's check. In Rutherford's model, the electron goes in a circle of radius r around the proton. Its total energy, capital E, equals minus e squared over eight pi epsilon naught r. Note the eight: the total energy is half the potential energy. Clear? Now, ionisation energy is the energy needed to free the electron. So what is the total energy of the orbiting electron? Minus thirteen point six electron volts, yes, exactly! Supply thirteen point six electron volts, and the electron just reaches zero, free. The double arrow on screen means implies. Okay? Now, this is an atom, so we use our constant in electron volt nanometre. So r is one point four four over thirteen point six, about point one zero six nanometre, right? Oh no, careful, that forgets the eight! Eight pi in place of four pi means half our constant. So r equals one point four four electron volt nanometre, over two times thirteen point six, which is twenty seven point two electron volts. With me? One point four four over twenty seven point two gives about point zero five three nanometre. Now look back at our toolkit. The Bohr radius, a naught, is point five two nine times ten to the power minus ten metre, which is also about point zero five three nanometre! Nice, isn't it?

7:44Quick revision: 1.44 MeV fm, hc = 1240 eV nm, the two traps

So, quick revision. First, e squared over four pi epsilon naught is one point four four, in M e V femtometre for nuclei, and in electron volt nanometre for atoms. Then h c is one thousand two hundred and forty electron volt nanometre, one atomic mass unit is nine hundred and thirty one point five M e V over c squared, and hydrogen's ground state is minus thirteen point six electron volts. And the two traps: the alpha particle has charge two e, and the orbit's total energy has eight pi epsilon naught. Keep these numbers handy. Next comes the scattering experiment itself. All the best!