Beyond the Simple Model

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take hydrogen, and its heavy cousin deuterium, whose nucleus is about twice as heavy. Both have one electron and Z equal to one, and Bohr's energy formula, minus thirteen point six Z squared over n squared electron volts, depends only on Z and n. So their spectra should be identical, right? And yet, the lines of deuterium are slightly shifted from those of hydrogen. Where does that tiny difference come from? Today we go beyond the simple Bohr model: a moving nucleus, exotic atoms, other forces, and the model's limits. These ideas do appear in JEE questions, so let's begin.

0:45The nucleus moves too: reduced mass

So, the simple model pretended the nucleus sits perfectly still, as if it were infinitely heavy. But a real nucleus has a finite mass, so the electron and the nucleus both revolve about their common centre of mass. Okay? The fix is really simple: the reduced mass, mu, the Greek letter that looks like a u with a tail. Mu equals small m times capital M, over small m plus capital M, where small m is the electron's mass and capital M is the nucleus's mass. So mu is always smaller than the electron's mass, and for a real atom, with its heavy nucleus, only a little smaller. Right? Now the rule: every Bohr formula still holds if you replace m by mu. So E n, the energy of orbit number n, with n as a small subscript, is proportional to mu. The symbol like a stretched alpha means proportional to. The radius r n goes as one over mu, and capital R, the Rydberg constant, goes as mu. Clear? Now back to our puzzle. Deuterium's nucleus is about twice as heavy as hydrogen's. So is its reduced mass larger or smaller than hydrogen's? Yes, it is larger, because a heavier nucleus brings mu even closer to the electron's mass. So deuterium has a slightly larger Rydberg constant, and its lines shift slightly. That solves our puzzle: the nucleus moves! Got it?

2:14Swap the electron for a heavier particle

Next, exotic atoms. Replace the electron by a particle with the same charge but a different mass, m prime. The little dash is read as prime. And keep the nucleus fixed for now. With me? Then the radii are divided by m prime over m, and the energies are multiplied by it. So a heavier particle sits in a smaller orbit, and is bound more tightly. Okay? Now tell me, what happens to its speed in each orbit? The speed stays exactly the same, which surprises many students. Remember, v n is proportional to Z over n, where Z is the atomic number. There is no mass in it. Right? Here is a toppers' trick: never derive the formulas again. Start from hydrogen's minus thirteen point six electron volts, the e V on screen, and zero point five two nine angstrom, the capital letter with a small circle on top. Then multiply the energy by the mass ratio, and divide the radius by it. Clear?

3:17Positronium: an electron and a positron

Now the most exotic one, positronium: an electron and a positron bound together. The positron has the same mass as the electron, but a positive charge. So this partner certainly does not stay still. Got it? So tell me, with capital M equal to m, what is the reduced mass of positronium? Put capital M equal to small m in the formula. Mu equals m times m, over m plus m, which is m over two, exactly half the electron's mass. With me? Now here is where many students slip. The reduced mass is halved, so they halve the orbit too. So tell me, does the separation of the two particles become half, or double? Oh no, careful, it doubles, because the radius goes as one over mu. Half the mu means double the distance. So the separation is two times a naught, times n squared, where a naught, the small a with a zero below it, is the Bohr radius. Okay?

4:20Other force laws: the same two equations

Now, suppose the force is not electric at all. The recipe stays exactly the same. First, m v squared over r equals F of r, the force at a distance r, with r in brackets. Right? Second, Bohr's condition: m v r equals n h over two pi, where h is Planck's constant. Solve the two together for the radius and the speed of every orbit. Clear? For the electric force, the radius came out proportional to n squared. Will that be true for every force? No, it depends on the force. For a constant force, for example, r turns out proportional to n to the power two thirds. So never copy n squared blindly. Got it?

5:06Where the Bohr model fails

Finally, the limits of the model. Let me ask you this: can Bohr's formulas describe a neutral helium atom? No, they cannot, because neutral helium has two electrons. The model works only for one-electron systems. Okay? Next, it cannot give the relative intensities of the lines, their fine structure, or their splitting in a magnetic field. Right? And last, it conflicts with the uncertainty principle. Remember all of these. With me?

5:39Example 1: positronium ground-state and ionisation energy

Now, example one. Find the ground-state energy and the ionisation energy of positronium. Pause the video and try it first. Done? Okay, let's check. For positronium, mu is m over two, and the energy is proportional to mu. So, does the energy become half, or double? It becomes half, because the energy goes as mu itself; only the radius goes as one over mu. So E one is minus thirteen point six times one half, which is minus six point eight electron volts. Clear? Now, ionisation takes the atom from the ground state up to zero energy, with the two particles pulled apart. So the ionisation energy is zero minus E one, which is plus six point eight electron volts. Never write it with a minus sign. Got it? And in general, E n equals minus six point eight over n squared electron volts, half of hydrogen's value at every level. Right?

6:44Example 2: muon bound to a proton, energy and first radius

Now a harder one. Here a muon is bound to a proton. The muon's mass, m mu, with a small mu below for muon, is two hundred and seven times m e, the electron's mass. The proton's mass, m p, is one thousand eight hundred and thirty six times m e. Taking the proton's motion into account, find the ground-state energy and the radius of the first orbit. Pause the video and try it first. Done? Okay, let's check. Many students simply use two hundred and seven times m e as the mass. Is that enough here? No, because the question asks us to include the proton's motion. So the reduced mass, mu, equals two hundred and seven times one thousand eight hundred and thirty six, over their sum, two thousand and forty three, all times m e. That is about one hundred and eighty six times m e. The wavy equals sign means approximately. With me? Now use our trick. E one equals minus thirteen point six times one hundred and eighty six electron volts. That is about minus two point five three k e V, kilo electron volts. Okay? The radius goes as one over mu, so this time we divide. r one equals zero point five two nine angstrom, divided by one hundred and eighty six. That gives about two point eight four times ten to the power minus thirteen metres. Right?

8:14Quick revision: reduced mass, positronium, Bohr limits

So, quick revision. When the nucleus moves, replace m by the reduced mass, mu, which is m times capital M over m plus capital M, okay? For a heavier particle with the nucleus fixed, divide the radii by m prime over m, multiply the energies by it, and the speed stays the same. Positronium has mu equal to m over two, energies of minus six point eight over n squared electron volts, and a separation of two times a naught, times n squared. And the Bohr model works for one-electron systems only. Now go and try the practice problems on this page. All the best!