Overview

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Suppose electrons of thirteen electron volts each hit hydrogen atoms in the ground state. Freeing hydrogen's electron needs thirteen point six electron volts, so thirteen is not even enough for that. So how many spectral lines come out? None at all? Keep your guess. This chapter is mixed JEE Main-pattern problems on atoms and nuclei, so let's begin.

0:29How this chapter works

So, how does this chapter work? There are three timed sets of fifteen original problems each, in the style of recent JEE Main papers. Okay? The topics are mixed: scattering, the Bohr model, spectra, the nucleus, fission and fusion. So first, spot the topic. Right? Give yourself twenty five minutes per set, then check the solutions. Now let us pack the tools.

0:55Tools 1: scattering and Bohr orbits

First, scattering. An alpha particle, the Greek letter like a rounded a, comes head-on at a nucleus of atomic number capital Z, okay? Its distance of closest approach, r naught, the r with a small zero below, is two point eight eight times Z, over capital K, its kinetic energy in mega electron volts, M e V, and the answer comes out in femtometres, f m. Clear? Next, Bohr's model, for one-electron atoms only: hydrogen, helium plus and lithium two plus, the signs up top give the charge. With n the principal quantum number, the radius r sub n is zero point five two nine times n squared over Z, in angstroms, the capital A with a small circle on top. And the speed v sub n is proportional to Z over n; the open alpha sign means proportional to. Right? The energy, E sub n, is minus thirteen point six times Z squared over n squared, in e V, electron volts. Okay? Now, the electron moves to a higher orbit, so its total energy rises. Does its kinetic energy rise too? It actually falls, and many students get this wrong. Capital K equals minus E, so as E becomes less negative, K becomes smaller. And the potential energy, capital U, is two times E, so it rises too. Got it?

2:25Tools 2: spectra

Now spectra. In the figure, each line is a level n with its energy on the right, and each arrow is a jump down, giving one photon. The Rydberg formula gives the wavelength lambda, the Greek letter like an upside down y. 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; the small one and two below n mark the lower and upper levels. And one over R is ninety one point two nanometres, n m. Okay? Each series is named by its lower level, n one. Lyman ends on level one, in the ultraviolet, Balmer on two, in the visible and near ultraviolet, and Paschen on three, in the infrared, as the figure shows. Right? Now a quick one. Which level is the second excited state of hydrogen? It is n equals three, not two, so be careful. Level one is the ground state, level two the first excited state, and level three the second. With me? And when many atoms are excited to level n, every jump between two levels happens somewhere. So the number of lines is n times n minus one, over two, with n minus one in brackets.

3:46Tools 3: the nucleus

Clear? Now the nucleus. Its radius, capital R, is R naught times A to the power one third, the cube root of the mass number, capital A, okay? R naught, the R with a small zero below, is about one point two femtometres, and the wavy equals sign means approximately. Right? Next, the mass defect, delta m, with delta the little triangle. It is Z times m H, plus A minus Z in brackets, times m n, minus M atom; the small labels below give the masses of a hydrogen atom, a neutron, and the whole atom. With atomic masses, the electron masses cancel. Okay? The binding energy, E sub b, is delta m times c squared, with c the speed of light. In practice, one atomic mass unit, u, is worth nine hundred and thirty one point five M e V, right? But stability is measured by the binding energy per nucleon, E sub b n, which is E sub b over capital A, not by the total. In the graph, it is largest, about eight point seven five M e V, near iron fifty six, the F e with fifty six up top. Clear? Now, a uranium nucleus, far right on the graph, splits into two middle-mass nuclei. Is energy released, or must we supply it? Yes, exactly, energy is released. The pieces sit higher on the curve, as the fission arrow shows. And in fusion, two very light nuclei join and also move up the curve. Both go towards larger E sub b n. Got it? And how much comes out? That is the Q value, capital Q: the binding energy of the products, minus that of the reactants. Okay?

5:35Example 1: number of nucleons in a nucleus of radius 6.0 fm

Alright, Example one, from Set two. How many nucleons are there in a nucleus of radius six point zero femtometres, with R naught equal to one point two femtometres? Pause the video and try it first. Done? Okay, let's check. Start from capital R equals R naught times A to the power one third. The double arrow means it follows that A to the power one third is six point zero over one point two, which is five. Right? So the nucleus has five nucleons, is that correct? That is not correct, and it is exactly the trap, because five is only the cube root of capital A, the mass number. Cube it, and capital A equals five to the power three, which is one hundred and twenty five nucleons.

6:27Example 2: lines emitted when 13 eV electrons hit hydrogen

Clear? Now Example two, also from Set two, and it is our puzzle. An electron beam of energy thirteen point zero electron volts bombards hydrogen atoms in the ground state. How many lines can be emitted? Pause the video and try it first. Done? Okay, let's check. Remember, a photon is absorbed only if its energy matches a gap exactly, and no gap from the ground state is exactly thirteen. So photons would do nothing. But do electrons also need an exact match? They do not, and that is the key. An electron can give up any part of its kinetic energy, so it needs only at least the gap, while a photon needs the exact gap. Right? So, level four is at minus zero point eight five electron volts, as in the figure. The gap from the ground state is thirteen point six minus zero point eight five, which gives twelve point seven five, and the less-than sign shows it is below thirteen. That means level four is reached. With me? Level five is at minus zero point five four four electron volts. Its gap is thirteen point six minus zero point five four four, which gives thirteen point zero six, and the greater-than sign shows it is above thirteen. So level five is out of reach. Okay? So the atoms reach level four, and there are many of them. The number of lines is four times three, over two, which gives six; the cross means times. Puzzle solved: not zero lines, but six. Got it?

8:05Quick revision: Bohr, photon vs electron gap, nuclear radius

So, quick revision. For one-electron atoms, the radius goes as n squared over Z, the energy is minus thirteen point six times Z squared over n squared electron volts, and K equals minus E, okay? Photons need the exact gap; electrons need at least the gap. And for the nucleus, R equals R naught times A to the power one third, and energy comes out when E sub b n goes up. Now open Set one, give it twenty five minutes, then check the solutions. All the best!