Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle to start. Hydrogen's red line, H alpha, comes from the jump from level three to level two. Now helium plus, a helium ion with one electron, gives a line of exactly the same wavelength. Which jump is that? Three to two again, right? Keep your guess; it is problem nine of today's set. This is Part A of the unit problem bank, sixteen problems at JEE Main level. So, let us pack three tools first.

0:32Tool 1: one-electron atoms (Bohr model)

Tool number one is the Bohr model, for one-electron atoms only: hydrogen, helium plus, lithium two plus and beryllium three plus. The signs up top give the charge on the ion, and capital Z, the atomic number, counts the protons. Okay? In orbit n, n being 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. The speed, v sub n, is two point one nine times ten to the power six, times Z over n, in metres per second, the m s to the minus one; the cross means times. With me? The energy, E sub n, is minus thirteen point six times Z squared over n squared, in electron volts, the e V on screen. For hydrogen: minus thirteen point six, minus three point four zero, minus one point five one, and minus zero point eight five. Right? Now let me ask you something. Lithium two plus has Z equal to three. In the same orbit, is its radius three times or nine times smaller than hydrogen's? Three times smaller, because the radius has Z, not Z squared. So the radius is divided by three, the line with two dots being division, the speed is multiplied by three, and only the energy is multiplied by Z squared, which is nine. Clear?

1:54Tool 2: spectral lines

Tool number two is the spectrum. In the figure, each downward arrow is one jump, giving one spectral line. Its wavelength is lambda, the Greek letter like an upside down y, and one over lambda equals capital R, times Z squared, times the bracket, one over n one squared, minus one over n two squared. Here n one and n two, with small one and two below, are the lower and upper levels, and capital R is the Rydberg constant; 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 level two, mostly in the visible, with its limit in the near ultraviolet, and Paschen on level three, in the infrared, as the figure shows. Right? The series limit, the shortest wavelength of a series, is the jump from n two equal to infinity, the sideways eight, the dashed line at the top. Then one over n two squared is zero, so the limit is n one squared over R Z squared. Got it?

2:54Problem 9: which He+ jump gives the H-alpha line

Back to our puzzle, problem nine. H alpha, the capital H with a small alpha below, is the first Balmer line, the jump three to two; the arrow means the jump. So one over lambda is R times, one quarter minus one ninth. Okay? So for helium plus, it is three to two again, is that right? That is exactly the trap, so be careful here. Helium plus has Z equal to two, so the same jump gives four times the energy, and the wavelength is divided by four. So that line is not H alpha at all. With me? In the figure, hydrogen is on the left and helium plus on the right. Level two of helium plus sits at minus thirteen point six, level with hydrogen's ground state; see the arrow marked equal. At level two n, Z squared over, two n whole squared, is four over four n squared, which is one over n squared, just as for hydrogen at level n. So which helium plus jump copies three to two? Yes, exactly, we double both quantum numbers and get six to four. Level six of helium plus matches hydrogen's level three, and level four matches level two. Clear? As a check, four R times, one sixteenth minus one thirty sixth, equals R times, one quarter minus one ninth, the same as H alpha. So the puzzle is solved: six to four, not three to two. Got it?

4:17Tool 3: mass defect and binding energy

Tool number three is for the nucleus: the mass defect, delta m, the little triangle being delta. It is Z times m H, plus A minus Z in brackets, times m n, minus M atom, where capital A is the mass number. Here m H, m n and M atom, with the small H, n and atom written below, are the masses of a hydrogen atom, a neutron, and the whole atom. These are atomic masses, so the electron masses cancel. Right? The binding energy, E b, with a small b below for binding, 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 mega electron volts, the M e V on screen, so multiply the mass defect in u by it. With me?

5:01Example 1: first excitation energy of Be3+

Alright, Example one is problem four of the set. Find the first excitation energy of beryllium three plus. Pause the video and try it first. Done? Okay, let's check. Beryllium has Z equal to four, and three plus means three of its four electrons are gone; one is left, so it is hydrogen-like. Now, the first excitation takes the electron from the ground state to which level? To n equal to two, the first excited state. So delta E, the energy needed, is thirteen point six, times Z squared, sixteen, times the bracket, one minus one quarter. Okay? Thirteen point six times sixteen gives two hundred and seventeen point six, and three quarters of that gives one hundred and sixty three point two electron volts. That is sixteen times ten point two, the hydrogen value. Right?

5:55Example 2: binding energy of the triton, H-3

Example two, problem thirteen of the set, is a step harder. Find the binding energy of the triton, the H with mass number three up top and atomic number one below. Its atomic mass is three point zero one six zero four nine u. Pause the video and try it first. Done? Okay, let's check. Z is one, so one proton, and A minus Z is two, so two neutrons. The given mass includes the atom's electron. So for the proton, do we use m p, a bare proton, or m H, a hydrogen atom? We use m H, one point zero zero seven eight two five u, so that its electron cancels the electron in the atomic mass. And m n, the neutron mass, is one point zero zero eight six six five u. Clear? So delta m is one point zero zero seven eight two five, plus two times one point zero zero eight six six five, minus three point zero one six zero four nine, which gives zero point zero zero nine one zero six u. With me? Multiply by nine hundred and thirty one point five, and the binding energy, E b, is about eight point four eight mega electron volts; the wavy equals sign means approximately. Got it?

7:14Quick revision: Z squared rule, He+ 6 to 4, 931.5 MeV per u

So, quick revision. For one-electron atoms, E sub n is minus thirteen point six times Z squared over n squared electron volts, so the radius is divided by Z, but the energy is multiplied by Z squared. Helium plus level two n matches hydrogen level n, so six to four copies H alpha. And for the nucleus, find delta m using m H, then multiply by nine hundred and thirty one point five M e V per u. Now attempt all sixteen problems of Part A as a test, then check the solutions. You have all the tools now, so you can do this. All the best!