Set 1

Transcript

0:00Introduction

Hello everyone! The Balmer series of hydrogen gives the famous visible lines: red, blue-green and violet. So every line of the Balmer series is visible, right? Hold that thought; problem seven of today's set answers it. This is mixed set one on atoms and nuclei, fifteen problems in JEE Main style. So, let's pick up the tools, and spot the traps.

0:25Tool 1: Bohr's results, one electron only

Tool number one is the Bohr model, for hydrogen-like atoms only: one electron round a nucleus of charge Z e, where Z is the atomic number and e the charge of a proton. Okay? The radius of orbit n, written r n with a small n below, is zero point five two nine times n squared over Z, in angstrom, the capital A with a small circle on top; n is the principal quantum number. The speed, v n, is two point one nine times ten to the power six, times Z over n, metres per second. Right? And the energy, E n, is minus thirteen point six times Z squared over n squared, in electron volts, the e V on screen. Clear? Now try problem three. Helium plus is H e with a small plus up high, and lithium two plus is L i with a small two plus. Is the electron faster in orbit two of helium plus, or in orbit three of lithium two plus? It is neither; they are exactly equal, because the speed goes as Z over n: two over two for helium, three over three for lithium. So the ratio is one to one, and the colon on screen is read as, is to. With me?

1:40Tool 2: the spectrum of hydrogen

Tool number two is the spectrum; look at the figure. Each horizontal line is a level n, and each arrow, a fall to a lower level, gives one photon of wavelength lambda, the Greek letter. The Rydberg formula says one over lambda equals capital R, the Rydberg constant, times Z squared, times the bracket, one over n one squared, minus one over n two squared; n one, with a small one below, is the lower level, and n two the upper. And one over R is ninety one point two nanometres. Okay? Each series is named by the level it lands on: Lyman on one, in the ultraviolet, Balmer on two, and Paschen on three, in the infrared. Right? Now, back to our opening question, problem seven. Statement one: all Balmer lines are visible, as even the figure labels them. Statement two: all Paschen lines are infrared. Which of the two is true? Now look at the third line, and here is the trap. The visible Balmer lines end at H delta, the violet line, written H with a small Greek delta below it. Beyond it, up to the series limit at three hundred and sixty four point six nanometres, the lines are ultraviolet. So not every Balmer line is visible. Clear? But every Paschen line does lie in the infrared. So Statement one, the Roman one on screen, is false, and Statement two is true. With me?

3:11Tool 3: binding energy and energy released

Tool number three is the nucleus. Any nucleus of two or more nucleons is lighter than those nucleons taken separately, and the missing mass is the mass defect, delta m, the little triangle being the Greek capital delta. Delta m equals Z times m H, plus bracket A minus Z, times m n, minus M atom; capital A is the mass number, m H and m n are the masses of a hydrogen atom and a neutron, and M atom is the atomic mass. Okay? That mass becomes the binding energy, E b, with a small b below: delta m in atomic mass units, u, times nine hundred and thirty one point five M e V, mega electron volts. Divide by A to get E b n, the binding energy per nucleon. Right? The figure plots E b n against the mass number, and the two arrows, fission and fusion, both climb to larger E b n. That releases energy: Q, the Q value, is the total binding energy after, minus the total before. Clear? Now try problem eleven. Take a nucleus of mass number one hundred, at eight point two M e V per nucleon, splitting into sixty and forty, at eight point six and eight point four. How much energy is released? Now be careful here: never subtract the per nucleon values directly; first multiply each one by its mass number. Sixty times eight point six, plus forty times eight point four, minus one hundred times eight point two. Got it? That is five hundred and sixteen, plus three hundred and thirty six, minus eight hundred and twenty, which gives thirty two M e V released. Yes, the products are more tightly bound, so energy comes out.

5:05Example 1: closest approach of 8 MeV alpha to uranium

Okay? Now, example one, problem one. Find the distance of closest approach of an eight point zero M e V alpha particle, alpha being the Greek letter like a rounded a, to a uranium nucleus, Z equal to ninety two, head-on. Pause the video and try it first. Done? Okay, let's check. Head-on, the alpha slows down, stops for an instant, and turns back. At that point, where has its kinetic energy gone? Yes, exactly, all of it is now electric potential energy, with the nucleus taken as fixed. The alpha has charge two e and the nucleus Z e, so the kinetic energy K equals one over four pi epsilon naught, times two e times Z e, each in brackets, over r naught, the distance of closest approach. Epsilon naught, the curly e with a small zero, is the permittivity of free space. Okay? Now use the shortcut: e squared over four pi epsilon naught is one point four four M e V femtometre, f m. So r naught equals two point eight eight Z, over K in M e V, in femtometres. Careful with that two point eight eight: it is two times one point four four, because the alpha carries charge two e. For a proton you would use one point four four. Right? Two point eight eight times ninety two gives two hundred and sixty four point nine six, and we divide that by eight point zero. Clear? So r naught is about thirty three point one femtometres; the wavy equals sign means approximately. The alpha turns back before it reaches the nucleus, so the nucleus can be no bigger than this.

6:52Example 2: binding energy per nucleon of lithium-7

Got it? Now a harder one, example two, problem ten. Find the binding energy per nucleon of lithium seven; seven up top is the mass number, three below the atomic number. Its atomic mass, and m H and m n, are on screen. Pause the video and try it first. Done? Okay, let's check. These are atomic masses, electrons included. So do we use m p, the proton's mass, or m H? We use m H, the mass of the whole hydrogen atom, so that the electrons cancel. Z is three, so three protons, and A minus Z gives four neutrons. With me? So delta m is three m H, plus four m n, minus capital M, the atomic mass. That is seven point zero five eight one three five, minus seven point zero one six zero zero four. Okay? That leaves zero point zero four two one three one atomic mass units. Right? Multiply by nine hundred and thirty one point five M e V for each atomic mass unit, and we get about thirty nine point two M e V in total. Clear? Now per nucleon: divide by seven, all the nucleons, not by three, the protons. Thirty nine point two over seven gives five point six M e V per nucleon, and that is the answer. Very good work if you got it!

8:22Quick revision: Bohr radius, spectra and binding energy

So, quick revision. For one electron, r n is zero point five two nine n squared over Z angstrom, and E n is minus thirteen point six Z squared over n squared electron volts. Each spectral series is named by the level it lands on, and the Balmer limit is ultraviolet. For nuclei, r naught is two point eight eight Z over K femtometres for an alpha particle head-on, E b is delta m times nine hundred and thirty one point five M e V, and Q is binding energy after minus before. Now try all fifteen problems on this page yourself, in twenty five minutes, and then check the solutions. You can do this, all the best!