Set 3

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. In Rutherford's experiment, a detector counts eighty alpha particles per minute at a scattering angle of sixty degrees. Move it to one hundred and twenty degrees, twice the angle. So the count should halve, to forty per minute, right? Hold on to that answer. Today is our third set of mixed problems on atoms, and problems like these come up often in JEE Main. So, let's begin.

0:29Idea 1: the count at each scattering angle

Idea one is our puzzle; note the Greek letter alpha, and the small circle for degrees. In the figure, dark blue paths are alpha particles passing a nucleus of charge plus Z e, the orange dot. Here b, the impact parameter, is how far off centre it is aimed, and theta is the scattering angle; small b gives large theta. Aimed head-on, it stops at a distance r naught from the nucleus, the r with a small zero, and comes straight back, at one hundred and eighty degrees. The two formulas on the right belong to another lesson, so we skip them today. With me? Now, the number scattered at angle theta. N of theta is proportional to Z squared times t, over K squared times sine to the power four of theta over two; the stretched alpha sign means proportional to. Z is the atomic number of the foil, small t is its thickness, and capital K the kinetic energy of the alpha particles. Same foil, same beam, so N is proportional to one over sine to the power four of theta over two. Okay? Now tell me: the angle doubles, so does the count simply halve, to forty per minute? It does not halve at all, and that is exactly the trap. N sub one twenty over N sub sixty, the N with small labels below, equals sine to the power four of thirty degrees, over sine to the power four of sixty degrees. Sine thirty is a half, so its fourth power is one over sixteen; sine sixty is root three over two, giving nine over sixteen. The ratio is one over nine. Clear? So N sub one twenty equals eighty over nine, which gives about eight point nine; the wavy equals sign means approximately. About nine per minute, not forty, and that solves our puzzle. Got it?

2:24Idea 2: total, kinetic and potential energy

Idea two, the Bohr model. Find the potential energy of the electron in the third orbit of lithium two plus, L i with a two plus up top: two electrons removed, just one left. For such hydrogen-like atoms, E sub n, the E with a small n below, equals minus thirteen point six, times Z squared over n squared, in electron volts, the e V on screen. Here n is the principal quantum number, and Z is the atomic number. Okay? And in every orbit, the kinetic energy, capital K, equals minus E, and the potential energy, capital U, equals two times E, the total energy. This pair comes up often. With me? For lithium two plus, Z is three, and n is three. Z squared is nine, and thirteen point six times nine is one hundred and twenty two point four, so E equals minus one hundred and twenty two point four over nine, which gives minus thirteen point six electron volts. Okay? So that is our answer, minus thirteen point six electron volts, right? Minus thirteen point six is the total energy, not the potential energy, so be careful here. U equals two times E, so U is minus twenty seven point two electron volts. The question asked for U, so read it twice. Clear?

3:49Idea 3: the Rydberg formula and series limits

Idea three, which both our examples use. The figure shows hydrogen's energy levels, with their energies on the right, from n equals one at minus thirteen point six electron volts, up to n equals infinity at zero. Each arrow is a jump down that gives out a photon. Dark blue arrows end on level one, the Lyman series, ultraviolet; orange ones on level two, the Balmer series, mostly visible; and black ones on level three, the Paschen series, infrared. Okay? The middle of the figure shows E sub n and this formula. When the electron falls from level n two, the n with a small two below, to a lower level n one, the arrow on screen, one over lambda equals R Z squared, times the bracket, one over n one squared, minus one over n two squared. Lambda is the wavelength, and R is the Rydberg constant. Clear? Now the series limit, the shortest wavelength of a series. Let n two be infinity, the sideways eight on screen; then one over n two squared becomes zero. So lambda limit, with limit written small below, equals n one squared over R Z squared. With me? And the course gives one over R as ninety one point two nanometres, the n m on screen. And for the same jump in a hydrogen-like ion, the wavelength is divided by Z squared. Got it?

5:14Example 1: Balmer-type series limit of He+

Alright, example one. Find the series limit of the Balmer-type series of H e plus, with the plus up top, the helium ion with one electron, for transitions ending on n equals two. Pause the video and try it first. Done? Okay, let's check. Hydrogen's Balmer limit is three hundred and sixty four point six nanometres, and helium has Z equal to two. So divide by two and get one hundred and eighty two point three nanometres, right? That is the trap, because the wavelength is divided by Z squared, not by Z alone. For H e plus, Z is two, and Balmer-type means n one is two, with n two at infinity. Okay? So lambda limit is n one squared over R Z squared: four over four R, just one over R, which is ninety one point two nanometres. Clear? Quick check: hydrogen's Balmer limit, three hundred and sixty four point six, over Z squared, which is four, gives about ninety one point two nanometres. The same answer. Very good!

6:26Example 2: Li2+ jump matching hydrogen 2 to 1 wavelength

Now a harder one. Which transition of lithium two plus gives the same wavelength as the transition two to one of hydrogen? The arrow means a jump from level two down to one. Pause the video and try it first. Done? Okay, let's check. Same jump, same wavelength, so two to one for lithium two plus as well, right? No, that is the trap here, because lithium has a different Z, so the same jump gives a different wavelength. Start from hydrogen, where one over lambda equals R times, one minus one quarter, which is three R over four. Okay? For lithium two plus, Z squared is nine. So nine R, times the bracket, one over n one squared, minus one over n two squared, must equal three R over four. Right? Cancel R and divide by nine: three quarters over nine gives one over twelve. With me? Now, which whole numbers n one and n two give one over twelve? Try a few values yourself. If you found n one equal to three, and n two equal to six, very good! One over nine, minus one over thirty six, is three over thirty six, which is one over twelve, exactly. Clear? So the answer is the transition six to three. And look at the pattern: hydrogen's levels, two and one, each multiplied by Z, which is three, give exactly six and three. Very good!

7:59Quick revision: scattering count, U = 2E, series limits

So, quick revision. First, the scattered count goes as one over sine to the power four of theta over two, so one hundred and twenty degrees gets one ninth of the sixty degree count. Second, in a Bohr orbit, K equals minus E, and U equals two times E, so read which energy is asked. Third, the series limit is n one squared over R Z squared, and for the same jump in an ion, the wavelength is divided by Z squared. Now try the other problems of Set three on this page. All the best!