de Broglie Waves in Bohr Orbits

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Remember Bohr's model? The electron can move only in certain orbits, where m v r equals n h by two pi. Only those orbits, nothing in between. But why? Why should nature allow only these orbits? Hmm, what is so special about them? Today, one picture, a wave wrapped round a circle, will give you Bohr's rule in one line. And look, JEE questions do ask for the electron's wavelength in an orbit, and there is a classic trap waiting for you. So, let's begin.

0:37Two rules you already know

So, first, two rules you already know. Rule one, from de Broglie: every moving particle has a wavelength. Lambda, the Greek letter on screen, equals h over m v, okay? Here h is Planck's constant, m is the mass, and v is the speed, okay? Rule two, from Bohr: m v r equals n h by two pi. m v r is the electron's angular momentum, and r is the radius of the orbit. And n is a whole number: one, two, three, and so on; the three dots mean the list continues. Bohr gave this as a postulate, a rule to accept. Clear? Now look at both lines together. One rule is about waves, the other is about orbits. Do you see any link between them?

1:25Wrap the electron's wave round the orbit

Yes, there is a link! de Broglie says the electron is also a wave, right? So imagine its wave, wrapped round the orbit. In the picture, the dashed circle is the orbit, the black dot in the centre is the nucleus, and the wavy line is the electron's wave, going round it, okay? Look at the left one, in dark blue. Go round once, and the wave joins up exactly with itself. Count the waves: four whole wavelengths fit in the circumference, that is, in two pi r, okay? So two pi r equals four lambda: a steady, standing wave. This orbit is allowed, with n equals four. Now the right one, in orange. Here two pi r is not a whole number of wavelengths; the crossed equals sign means not equal to. Go round once, and the wave does not join up with itself. So there is no steady wave, and this orbit is not allowed, okay? So that is the secret, and it is really simple: an orbit is allowed only when a whole number of waves fits round it. And the two lines on the right of the picture? That is where this idea leads. Let's build them now, step by step.

2:36From the wave to Bohr's condition

Let's put it in symbols. So, a whole number of wavelengths fits round the orbit, which means two pi r equals n lambda, where n is one, two, three, and so on, okay? Now use de Broglie's rule: lambda equals h over m v, remember? Put it in, and rearrange. What comes out? Let's see, did you get it? Two pi r equals n times h over m v, that is, n h over m v, okay? Now multiply both sides by m v, and divide by two pi. The double arrow means so: m v r equals n h by two pi. And look, this is exactly Bohr's condition! So, our puzzle from the start: why only certain orbits? Because only in those orbits does the electron's wave fit round a whole number of times. Got it? Now one quick question. Look at two pi r equals n lambda again. How many wavelengths fit round the nth orbit? Exactly n, yes! The first orbit holds one wavelength, the second holds two, and so on. Remember this; the next formula comes straight from here.

3:44Wavelength in the nth orbit

Alright, now the wavelength itself. n wavelengths fill the circumference, so one wavelength is two pi r n, divided by n, okay? The small n below lambda and r means in the nth orbit, clear? And r n, from Bohr's model, is n squared over Z, times a naught. The symbol a naught, the letter a with a small zero below, is the Bohr radius, the radius of the first orbit of hydrogen: zero point five two nine angstrom. Angstrom is the capital A with a small ring on top. And capital Z is the atomic number, okay? Now put this r n into lambda n. The radius grows as n squared. So the wavelength also grows as n squared, right? No, no, careful! This is the JEE trap. See, we still divide by n; the dot means times. n squared over n leaves just n, not n squared. So lambda n equals two pi a naught, times n over Z, okay? So remember the pair. The sign like a stretched alpha means proportional to. Lambda is proportional to n over Z, but r is proportional to n squared over Z, right? In ratio questions, students mix up these two all the time. Don't do that, got it?

5:03Example 1: de Broglie wavelength in H ground state

Right, let's try Example one. Find the de Broglie wavelength of the electron in the ground state of hydrogen. Pause the video, and try it first. Done? Okay, let's check. Ground state means the lowest orbit, so n equals one. And for hydrogen, H on screen, Z equals one, okay? So lambda one, with the small one below, equals two pi a naught, times one over one. Just two pi a naught: the whole first orbit is exactly one wavelength! That is two pi, times zero point five two nine. And that comes to about three point three two angstrom; the wavy equals sign means approximately. Really simple, right?

5:49Example 2: wavelength ratio, H (n = 2) to He+ (n = 1)

Now a slightly harder one, JEE style. Find the ratio of the de Broglie wavelengths in n equals two of hydrogen, and in n equals one of H e plus, the helium ion. Pause and try. Hint: you will not need the value of a naught at all. Done? Okay, let's check. In a ratio, two pi a naught sits on top and at the bottom, so it cancels. Lambda is just proportional to n over Z, okay? That is the topper's shortcut: compare n over Z, nothing else. Hydrogen, H, with n equals two and Z equals one: n over Z is two by one. The helium ion, H e plus, with n equals one and Z equals two: n over Z is one by two, okay? Now look, many students here grab n squared over Z, from the radius formula, and they get eight. So the answer is eight is to one, right? No, no, careful! That eight is the ratio of the radii, not of the wavelengths. The wavelength goes as n over Z, remember? Lambda H over lambda H e plus, where the small letters below name the atom, equals two by one, divided by one by two. That is two times two: four. So lambda H is to lambda H e plus is four is to one. Hydrogen's wave here is four times longer. See? We never needed the value of a naught. That's it, done!

7:16Quick revision: 2 pi r = n lambda, lambda varies as n/Z

So, quick revision. One: the electron's wave must fit a whole number of times round the orbit. Two pi r equals n lambda, and that gives Bohr's condition, m v r equals n h by two pi. Two: the nth orbit holds n de Broglie wavelengths. Three: lambda n equals two pi a naught, times n over Z, so lambda goes as n over Z, and not as n squared over Z, okay? Now go and try the practice problems on this page. You will find them easy, I promise. All the best!