IIT-JEE Questions (1981–2003): Bohr Model

Transcript

0:00Introduction

Hello everyone! Here is a quick question first. Hydrogen holds its single electron, in the ground state, with thirteen point six electron volts. Now take lithium, and strip away two of its three electrons, so that only one is left, just like hydrogen. How much energy pulls out that last electron? The same thirteen point six? Or three times as much, because lithium has three protons? The answer is neither, as you will see in one line. Today we solve two past IITJEE questions on the Bohr model. Questions like these come up often, so let's begin.

0:36Bohr's model for hydrogen-like atoms

So, first, our tools. Bohr's model works for hydrogen-like atoms: one electron going round a nucleus of charge plus Z e, where Z is the atomic number and e is the electronic charge. So hydrogen; helium plus, where the small plus up top means one electron removed; and lithium two plus, also called doubly ionised lithium. Bohr's results hold only for these one-electron atoms, okay? Now, Bohr allowed the electron only certain orbits, where it does not radiate. So which quantity did he allow only in whole-number steps? Yes, exactly, it is the angular momentum. Capital L equals m v r, mass times speed times radius, and it must equal n times h, over two pi, with pi the Greek letter. Here h is Planck's constant, and n, the principal quantum number, is one, two, three, and so on, as the three dots say. Clear? This rule fixes the radius of orbit number n, written r with a small n below. It is a naught times n squared over Z, where the small two up top means squared. And a naught, the letter a with a small zero below, is the Bohr radius, zero point five two nine times ten to the power minus ten metres. With me? Now the energy of level n, written E with a small n below. It is minus thirteen point six times Z squared over n squared, in electron volts, the e V on screen. The minus sign means the electron is bound. Okay? Let us try it on hydrogen, where Z is one. At n equals one, the energy is minus thirteen point six. So tell me, what are E two and E three? Thirteen point six over four gives three point four, and thirteen point six over nine gives about one point five one. So E one is minus thirteen point six, E two is minus three point four zero, and E three is minus one point five one electron volts.

2:41Excited states, ionisation and photons

Got it? Now, here is a word trap that catches many students. Suppose a question says the second excited state. So that means n equals two, right? That is exactly the trap, so be careful. The ground state is n equals one, the first excited state is n equals two, so the second excited state is n equals three. With me? Next comes ionisation. How much energy must we give to pull the electron out completely, starting from level n? Yes, just enough to lift its energy up to zero, where it is free. So the ionisation energy from level n is thirteen point six times Z squared over n squared electron volts, the binding energy of the electron in that level. Clear? Thirteen point six electron volts has its own name, and we call it one rydberg of energy. It equals capital R times h times c, where R is the Rydberg constant and c is the speed of light. Okay? Now, when the electron jumps down from level E i to level E f, a photon comes out; the small i and f stand for initial and final. Its energy, h nu, equals E i minus E f, where nu, the Greek letter that looks like a v, is the frequency. Right? For the wavelength, use h c equals twelve hundred and forty electron volt nanometre, the e V n m on screen. So lambda, the wavelength in nanometres, equals twelve hundred and forty over delta E in electron volts, where the little triangle, delta, means the energy difference.

4:19Example 1: energy to remove the electron from Li2+ (1997)

Got it? Example one, from IITJEE nineteen ninety seven, and it is our opening question. Find the minimum energy needed to remove the electron from the ground state of doubly ionised lithium, where Z equals three. Pause the video and try it first. Done? Okay, let's check. Doubly ionised lithium has lost two of its three electrons, so one electron is left, and it is hydrogen-like. Right? The minimum energy to remove it from the ground state is just the ionisation energy, with n equals one and Z equals three. Okay? Now, lithium's nucleus has three protons, three times the charge of hydrogen's nucleus. So the energy is three times thirteen point six, right? That is exactly the trap, so be careful, because the energy goes as Z squared, not as Z itself. So the factor is nine, not three. Clear? With n squared equal to one, the energy needed is thirteen point six times nine, and that gives one hundred and twenty two point four electron volts. So our opening answer is nine times hydrogen's value! Got it?

5:30Example 2: 4 rydberg gives He+, 30.4 nm photon, first radius

Now a harder one, from IITJEE nineteen eighty four. The ionisation energy of a hydrogen-like Bohr atom is four rydberg. First, find the wavelength emitted when the electron jumps from the first excited state to the ground state. Then find the radius of its first orbit; the letters in brackets mark these two parts. Pause the video and try it first. Done? Okay, let's check. We are not told the atom, so we find Z first. The ionisation energy from the ground state is thirteen point six times Z squared, so what is Z here? Four rydberg is four times thirteen point six, which is fifty four point four electron volts. That equals thirteen point six times Z squared, so Z squared is four, and the double arrow says Z equals two, which is helium plus. Clear? Now the first part, from the first excited state to the ground state. Remember our word trap? So which n, to which n? Yes, from n equals two down to n equals one, as the arrow shows. With the ground level at minus fifty four point four, the energy difference is fifty four point four times, one minus one quarter. That is fifty four point four times three quarters, which gives forty point eight electron volts. Right? So the wavelength is twelve hundred and forty over forty point eight, in nanometres, which gives about thirty point four nanometres. The wavy equals sign means approximately. With me? Now the second part. With n equal to one, r one is a naught over Z, so zero point five two nine times ten to the power minus ten, over two, in metres. Okay? That gives two point six four five times ten to the power minus eleven metres, which is about two point six five times ten to the power minus eleven metres. It is half the radius for hydrogen, because the doubled charge pulls the electron closer. Got it?

7:37Quick revision: r and E formulas, n = 3 trap, 1240/delta E

So, quick revision. For hydrogen-like atoms, the radius is a naught times n squared over Z, and the energy is minus thirteen point six times Z squared over n squared electron volts. The second excited state is n equals three, and the ionisation energy is thirteen point six times Z squared over n squared. The wavelength in nanometres is twelve hundred and forty over the energy difference in electron volts, and one rydberg is thirteen point six electron volts. Now try the other twelve past questions on this page before you open their solutions. All the best!