JEE Advanced Questions (2009–2017): Bohr Model

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. An electron in a hydrogen atom jumps from one orbit to another, and you are told just one number: its potential energy before the jump, divided by its potential energy after, is six point two five. From that single number, can you tell which orbit it landed in? Sounds impossible, right? But JEE Advanced asked exactly this in two thousand seventeen, and by the end of this video it will take you less than a minute. So let us solve two of the JEE Advanced questions on the Bohr model from two thousand nine to two thousand seventeen. Bohr model questions do appear in JEE, so let's begin.

0:43The Bohr toolkit for one-electron atoms

So, first, let us pack our toolkit, because every question today uses it. Capital L is the angular momentum, and it equals m v r: the electron's mass, times its speed, times the radius of its orbit. Bohr's rule says L equals n times h over two pi, where h is Planck's constant, and pi is the Greek letter you know from circles. And n is the principal quantum number; the three dots mean it takes the values one, two, three, and so on. Okay? Next, let us look at the radius. r n, with the small n below, is the radius of orbit number n, and it equals a naught times n squared over capital Z, the atomic number, which counts the protons in the nucleus. And a naught, the small a with a zero below it, is the Bohr radius, zero point five two nine angstrom; the capital letter with a little circle on top is the angstrom symbol. With me? Now, the electron also behaves as a wave, with a de Broglie wavelength called lambda, the Greek letter in the box. So tell me, how many of these wavelengths fit round the nth orbit? Exactly n of them fit, a whole number. So two pi r n, the circumference of the orbit, equals n times lambda. Remember this line, because it turns a radius straight into a wavelength. Clear? And one condition: all these results hold only for atoms with just one electron. So hydrogen, the helium ion H e plus, and the lithium ion L i two plus, where the charge written up high counts the electrons removed. Got it?

2:19Energies in orbit n

Now let us look at the energies. E n, the total energy in orbit n, equals minus thirteen point six times Z squared over n squared, in electron volts, the e V on screen. The minus sign tells us the electron is bound to the nucleus. Okay? The total energy is capital K, the kinetic energy, plus capital U, the potential energy. Now, do you remember how K and U are related to the total energy? Yes, exactly, K equals minus E, and U equals two times E, so the potential energy is twice the total energy, and negative too. Know any one of the three, and you know all three. Right? So U is proportional to minus Z squared over n squared; the sign like a stretched alpha means proportional to. And here is a topper's trick: when a question gives you a ratio, write the proportionality, and the constants, like thirteen point six, cancel out. Only n and Z are left. With me?

3:20Example 1: smallest final orbit from PE ratio 6.25 (2017)

Now, remember our puzzle from the start? Here it is, example one, from JEE Advanced two thousand seventeen. An electron in a hydrogen atom goes from an orbit of quantum number n i to one of quantum number n f; the small i and f below stand for initial and final. The initial and final potential energies are V i and V f, and V i over V f equals six point two five. Find the smallest possible n f. Pause the video and try it first. Done? Okay, let's check. Careful here: capital V in this question means potential energy, not electric potential, so it plays the role of our U, okay? For hydrogen, Z equals one, so V is proportional to minus one over n squared. Right? Now take the ratio: minus one over n i squared, divided by minus one over n f squared. The minus signs cancel, and the two squares swap places, so n f squared over n i squared equals six point two five. Clear? Take the square root of both sides; that is the root sign on screen. The square root of six point two five is two point five, which is five over two. So n f over n i equals five over two. Okay? Now, here is a tempting shortcut. Some students take n i as one, and write n f equals two point five. Is that allowed? No, it is not allowed, because quantum numbers are whole numbers. So we need whole numbers in the ratio five to two, where the colon is read as to. The pairs are two and five, four and ten, and so on. With me? The smallest pair is n i equals two and n f equals five, so the smallest possible n f is five. And there is our puzzle solved: from just six point two five, the electron went from orbit two out to orbit five. Very good if you got it!

5:24Example 2: de Broglie wavelength in Li2+ with L = 3h/2pi

Now the next one, example two, from JEE Advanced two thousand fifteen. An electron in an excited state of L i two plus has angular momentum three h over two pi. Its de Broglie wavelength is p times pi times a naught, where a naught is the Bohr radius, and we have to find the number p. Pause the video and try it first. Done? Okay, let's check. The question does not tell us the orbit directly. So, from the angular momentum alone, which orbit is the electron in? Yes, exactly, it is in the third orbit. Bohr's rule says L equals n times h over two pi, and here L is three h over two pi, so n equals three; the double arrow means implies. Okay? Next we need Z for L i two plus. The ion has a charge of two plus, so is Z equal to two? No, and this is the classic trap, so be careful. Z is the atomic number, the number of protons in the nucleus, not the charge of the ion. Lithium has atomic number three, so Z equals three. Right? Now the radius: a naught times n squared over Z, with n equal to three and Z equal to three. Three squared is nine, and nine over three is three, so r equals three times a naught. With me? And now the de Broglie picture: n wavelengths fit round the orbit, so lambda equals two pi r over n. Two pi times three times a naught, over three, gives two pi times a naught. Do not forget to divide by n here; many students slip at this step. Clear? Compare this with p times pi times a naught, and p equals two. Three small steps: the orbit from the angular momentum, the radius from n and Z, and the wavelength from the circumference. Got it?

7:20Quick revision: Bohr toolkit, ratios and the Z trap

So, quick revision. L equals n times h over two pi, r n equals a naught times n squared over Z, and two pi r n equals n times lambda. E n equals minus thirteen point six times Z squared over n squared electron volts, with K equal to minus E, and U equal to two E, okay? In ratio questions the constants cancel, n is always a whole number, and Z is the atomic number, never the charge of the ion. Now try the other JEE Advanced questions on this page. All the best!