Bohr's Postulates and the Orbits

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a hydrogen atom, with its electron in the first orbit. Now a helium ion, H e plus, with its electron in the second orbit. Different atoms, different orbits, but exactly the same energy! How? Just a coincidence? Hmm, strange, right? Today, with Bohr's model, we will see exactly why. And look, in JEE, Bohr's formulas give quick, sure marks, if you remember a few powers. So, let's begin.

0:35Bohr's three postulates (1913)

So, remember Rutherford's problem? An electron moving in a circle should radiate energy, and spiral into the nucleus. But atoms are stable, right? In nineteen thirteen, Bohr's first postulate answered this. The electron moves only in certain circular orbits, called stationary states, and in these it does not radiate energy at all. That's it, no spiralling in! Second postulate: the angular momentum, m v r. Here m is the mass of the electron, v its speed, and r the radius of its orbit. m v r equals n h over two pi, where h is Planck's constant, and n labels the orbit. Okay? Now tell me: can n be any number? Say, two point five? No! n can only be one, two, three, and so on; the three dots mean the list goes on. Whole numbers only: that is what quantised means, okay? Alright, postulate three. Energy is emitted or absorbed only when the electron jumps between orbits. Delta, the Greek triangle, means change. Delta E equals E n two minus E n one; the small n one and n two below are the two orbit numbers. And it equals h nu, the energy of one photon; nu, the Greek letter like a v, is the frequency. Got it?

1:54Radius and speed in orbit n

Now Bohr's results. They hold for any one-electron species: hydrogen, the helium ion H e plus, and the lithium ion L i two plus. Capital Z, the nuclear charge, is in brackets: one, two and three. Clear? The radius of orbit number n, r with a small n below it, equals zero point five two nine, times n squared over Z, in angstrom, the capital A with a small ring on top, okay? Look at the formula. More charge on the nucleus, a bigger Z: does the orbit get bigger, or smaller? Smaller, exactly! Z sits at the bottom. And n is squared on top, so the orbits grow fast as n goes up. Remember this n squared for example one. Next, the speed. v n, again with a small n below, equals two point one eight times ten to the power six, times Z over n, in metres per second, written m s to the minus one. See, Z on top, n at the bottom: more charge, a faster electron; a higher orbit, a slower one.

2:59Energy of orbit n

Now, the energy. E n, the energy in orbit n, equals minus thirteen point six, times Z squared over n squared, in electron volts, e V on screen. In the picture, the blue ladder on the left is hydrogen: its level n equals one sits at minus thirteen point six. Every level lies below the dashed zero line, so all these energies are negative. With me? In joules: minus two point one eight times ten to the power minus eighteen, times Z squared over n squared, joule, the capital J on screen. And look at the blue ladder going up: the levels crowd together as n grows. Now the helium ion, the orange ladder on the right. Z is two, so Z squared is four: every level is four times deeper. Its first level sits at minus fifty four point four. Really deep! Now see the double arrow marked equal: hydrogen's first level, and the helium ion's second. Our puzzle! Using the formula, tell me: why are they equal? Yes, exactly! Only Z squared over n squared matters. Helium ion, n equals two: four over four, which is one. Hydrogen, n equals one: also one. Same Z squared over n squared, same energy. Puzzle solved!

4:19Kinetic and potential energy

Now, this E is the total energy. From it come two more. Kinetic energy, capital K, equals minus E, okay? E is negative, so K comes out positive. And the potential energy, capital U, equals two times E, okay? So tell me: is U positive, or negative? Negative, yes! Two times a negative number. And twice as deep as the total energy. So, K is to U is to E equals one is to minus two is to minus one. Look, know any one energy, and you know all three.

4:54Topper's trick: remember the powers

Now a trick that toppers use. For ratios, forget the constants; just remember the powers. The sign like a stretched alpha means proportional to. r goes as n squared over Z, v as Z over n, and E as Z squared over n squared. The time period, capital T, the time for one revolution, goes as n cubed over Z squared. And the frequency of revolution is just the flip: Z squared over n cubed. Got it? And the angular momentum, n h over two pi, goes as n only. No Z at all! So in the same orbit, hydrogen and the helium ion have the same angular momentum. That's all; now write these powers down.

5:39Example 1: radius of H second orbit and Li2+ first orbit

Now, example one. Find the radius of the second Bohr orbit of hydrogen, and of the first orbit of L i two plus. Pause the video, and try it first. Done? Okay, let's check. Second orbit, so n is two, and the radius is zero point five two nine times two, right? No, no, careful! This is the JEE trap: n is squared. Z is one, so r equals zero point five two nine, times two squared over one: two point one one six, approximately two point one two angstrom. The wavy equals sign means approximately. Clear? Now L i two plus: n is one, and Z is three. r equals zero point five two nine, times one squared over three: approximately zero point one seven six angstrom. More charge, a smaller orbit. Just as we said!

6:33Example 2: find the orbit from KE, then the PE

Now a harder one, JEE level. The kinetic energy of the electron in an orbit of hydrogen is zero point eight five electron volt. Which orbit is it, and what is the potential energy? Pause the video, and try it. Hint: find the total energy first. Done? Okay, let's check. E equals minus K, so the total energy is minus zero point eight five electron volt. Negative, like every level in our picture. Hydrogen, so Z is one, and E equals minus thirteen point six over n squared. Put it equal to minus zero point eight five. The double arrow means so: n squared is thirteen point six over zero point eight five, exactly sixteen. So n is four, the fourth orbit. Not sixteen, okay? Sixteen is n squared. Now the potential energy. U is two times zero point eight five, so plus one point seven electron volt, right? No, no, careful! This is the trap. U equals two E, not two times K, okay? E is minus zero point eight five, so U is two times minus zero point eight five: minus one point seven electron volt. Perfect!

7:51Quick revision: postulates, radius, speed and energy

So, quick revision. Bohr's postulates: stationary orbits, m v r equals n h over two pi, and delta E equals h nu. Radius goes as n squared over Z, speed as Z over n, and energy is minus thirteen point six, times Z squared over n squared, electron volt. And K equals minus E, U equals two E: one is to minus two is to minus one. Now go and try the practice problems on this page. All the best!