Energy of the Stationary States

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. When the electron of hydrogen moves up to a higher orbit, it moves more slowly, so its kinetic energy goes down. And yet everybody says its energy goes up. So which is it, up or down? Both sound right, don't they? Soon you will see why both are true. Energy levels come up often in JEE, so let's begin.

0:26Kinetic, potential and total energy

So, here is a hydrogen-like atom: one electron round a nucleus of charge Z times e, where capital Z is the atomic number and e is the charge on a proton. The kinetic energy, capital K, equals Z e squared over eight pi epsilon naught r n. Epsilon naught, the Greek letter with a small zero, is the permittivity of free space, and r n, with the small n below, is the radius of orbit number n. Okay? Next, the potential energy, capital U, equals minus Z e squared over four pi epsilon naught r n, negative because the force is attractive. Look at the denominators: four here, but eight for the kinetic energy. Right? Add K and U to get the total energy, capital E, okay? So how is the total energy related to the kinetic energy? Yes, exactly, the total comes out as minus K, because U is minus two times K, right? So K equals minus E and U equals two times E, and the negative total energy means the electron is bound. With me? Now write it as a ratio, where each colon is read as is to. K is to U is to E as one is to minus two is to minus one, in every orbit. Here is a topper's trick: know any one of the three, and you know all three.

1:51The energy of level n

Clear? Now put in the allowed radius from Bohr's rule. E n, with n as a subscript, equals minus m e to the power four, over eight epsilon naught squared h squared, times Z squared over n squared, where m is the mass of the electron and h is Planck's constant. Okay? Put in all the constants, and E n equals minus thirteen point six, times Z squared over n squared, in electron volts, the e V on screen. This is the formula you will use most. Got it? That thirteen point six electron volts equals capital R times h times c, where R is the Rydberg constant and c is the speed of light. It is called one rydberg. Okay? And one condition: all this is for hydrogen-like atoms only, with just one electron. So hydrogen, the helium ion H e plus, and the lithium ion L i two plus, where the plus signs up top count the electrons removed. Right?

2:50Hydrogen's energy levels

Now let us use it for hydrogen, where Z is one. What is the energy for n equals one, and for n equals two? For n equals one, the ground state, E one is minus thirteen point six electron volts. For n equals two, minus thirteen point six over four gives minus three point four zero. For n equals three, minus thirteen point six over nine gives minus one point five one. Okay? Then E four is minus zero point eight five, and E five is minus zero point five four. Learn these five by heart. Right? Now look at the gaps. The first gap is big, but the gap between E four and E five is tiny. The levels crowd together, and as n tends to infinity, shown by the arrow and the infinity sign, the energy tends to zero. Clear? Now back to our puzzle. As n goes up, the total energy climbs towards zero, so it increases. But what does the kinetic energy do? Remember that K equals minus E. Yes, it decreases, because E is becoming less negative. So, as the up arrow says, when n increases the total and potential energies increase, while the kinetic energy decreases. Both halves of our puzzle were true. With me?

4:12Excitation and ionisation energy

Now, two important words. Excitation energy is the energy needed to raise the electron from the ground state to a higher state, and the arrow means goes to. Okay? So, how much energy lifts the electron of hydrogen from level one to level two? From minus thirteen point six up to minus three point four is a rise of ten point two electron volts, to the first excited state. Up to level three, the second excited state, the rise is twelve point zero nine. Careful, n equals two is the first excited state, not the second. Right? The matching potential differences are the excitation potentials, ten point two and twelve point zero nine volts, with capital V for volts. Got it? Next, the ionisation energy: the energy needed to remove the electron completely, to infinity, where the energy is zero. From level n, it is thirteen point six Z squared over n squared electron volts. Okay? This is also the binding energy of the electron in that level. And here is a trap: thirteen point six electron volts is hydrogen's ionisation energy only from the ground state. From a higher level, use that level's n. Clear?

5:29Example 1: K, U and orbit n when E is -0.85 eV

Now, example one, an easy one. The total energy of an electron in a certain orbit of hydrogen is minus zero point eight five electron volts. Find its kinetic energy, its potential energy, and the quantum number of the orbit. Pause the video and try it first. Done? Okay, let's check. K equals minus E, so the kinetic energy is plus zero point eight five electron volts, positive as it must be. Right? And U equals two times E, so the potential energy is two times minus zero point eight five, which is minus one point seven zero electron volts. With me? Now the orbit. For hydrogen, Z is one, so thirteen point six over n squared equals zero point eight five. Thirteen point six over zero point eight five gives sixteen, so n squared is sixteen, and n is four, where the double arrow means implies. Okay?

6:29Example 2: n = 2 energy with PE zero at the ground state

Now a harder one, with a trap inside. Suppose the zero of potential energy is taken at the ground state of hydrogen, instead of at infinity. What would be the total energy of the electron in the first excited state? Pause the video and try it first. Done? Okay, let's check. With the usual zero at infinity, what is the potential energy in the ground state? U equals two times E, so U one, with a subscript one, is two times minus thirteen point six, which is minus twenty seven point two electron volts. Right? Now we want this potential energy to become zero. Many students add thirteen point six to everything here. So which is it, thirteen point six or twenty seven point two? It is twenty seven point two, because we moved the zero of potential energy, not of total energy. So we add twenty seven point two electron volts to every potential energy, and so to every total energy. Clear? The first excited state is n equals two, with the usual total energy minus three point four. Add twenty seven point two, and E two prime, the little dash meaning the new value, equals twenty three point eight electron volts. It is positive now, as our new zero sits much lower. Got it?

7:55Quick revision: En formula, K:U:E ratio, ionisation energy

So, quick revision. E n equals minus thirteen point six Z squared over n squared electron volts, for hydrogen-like atoms only. K is to U is to E as one is to minus two is to minus one, in every orbit. And the ionisation energy from level n is thirteen point six Z squared over n squared. Now go and try the practice problems on this page. All the best!