Heisenberg's Uncertainty Principle

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Remember Bohr's model? He gave the electron an exact orbit, and an exact speed. So tell me: at one single moment, can we know exactly where an electron is, and exactly how fast it is moving? Sounds easy, right? Just measure both! Hmm, not so fast. And look, in JEE, this can come as quick numericals and as a theory question: why the modern atom has orbitals, not orbits. So, let's begin.

0:32Position and momentum: never both exact

So, here is the answer, from Heisenberg, in nineteen twenty seven. It is impossible to know both the exact position and the exact momentum of a particle at the same time. Not difficult, impossible. Okay? Now in symbols. Delta x, that is the Greek letter delta, the little triangle, with x, is the uncertainty in position. Delta p is the uncertainty in momentum, and the dot between them means times. Their product is greater than or equal to h over four pi; that is the sign with a line below it. Here h is Planck's constant, and pi is this Greek letter. Suppose I locate a particle very, very precisely, so delta x is tiny. But the product can never go below h over four pi. So what must happen to delta p? Yes, it must become large! Small delta x, so large delta p, and the other way round too. The double arrow means so. Like a seesaw: one side down, the other side up. Got it?

1:34The same law with speed

Alright, but JEE numericals usually ask about speed, not momentum. Momentum p is m v, mass times velocity. The mass is fixed, so delta p equals m delta v, okay? Now put m delta v in place of delta p. The m comes along with delta v. So where should that m go? Yes, exactly, divide both sides by m! So delta x times delta v is greater than or equal to h over four pi m. Same law, now written with speed. Clear? And here is a trick for every numerical. When the question says least uncertainty, just use the equals sign, and solve. So the least delta v equals h, over four pi m delta x. Simple, right? With me? Now careful, a classic JEE trap. Look at this line, lit up again: the speed form has m in the bottom. The momentum form had no m at all. Mix them up, and the answer goes wrong. Keep that in mind, okay?

2:40Heavy objects and the electron

Right, now the big question. Look at h over four pi m. Planck's constant is tiny, and m sits in the bottom. So for a ball, or any everyday object with a large mass, is this big, or small? Small, yes! For everyday objects, h over four pi m is negligible. That is why you never notice it with a cricket ball. But an electron is extremely light. So for an electron inside an atom, the uncertainty in its speed is enormous. We will calculate it in Example one, okay? Now connect this to Bohr. An orbit means an exact position and an exact speed at every moment. If we cannot know both, can the electron have a definite orbit? No! So, a definite orbit has no meaning. That neat circle we drew in Bohr's model is not a real path for the electron at all. So what does the modern model use? Orbitals. An orbital is a region of probability, not a fixed path. Notice, orbit and orbital: two letters more, a completely different idea. Do not mix up these two words.

3:54Energy and time

One more pair, and then we solve. Energy and time follow a similar relation. Delta E, the uncertainty in energy, times delta t, the uncertainty in time, is at least h over four pi, okay? And for the least uncertainty, the same trick: use the equals sign. So the least delta E equals h over four pi delta t. That's it, that's the whole theory. Now let's use it.

4:20Example 1: speed uncertainty of an electron in an atom

So, remember our puzzle from the start? Let's finish it. The uncertainty in the position of an electron is one angstrom, about the size of an atom. Angstrom is the capital A with a small ring on top. Find the least uncertainty in its speed. Pause the video, and try it first. Done? Okay, let's check. Delta x is one angstrom. So I just put one in the formula for delta x, right? No, no, careful! Units first. One angstrom is ten to the power minus ten metre. And the electron's mass, m, is nine point one times ten to the power minus thirty one kilogram, the k g on screen. Least uncertainty, so the equals sign. Delta v equals h over four pi m delta x. That is six point six two six times ten to the power minus thirty four, over four pi, times nine point one times ten to the power minus thirty one, times ten to the power minus ten. The bottom works out to one point one four times ten to the power minus thirty nine. Divide, and delta v is about five point eight times ten to the power five metres per second. The wavy equals sign means approximately, and m s to the power minus one is metres per second. Now the real question. Is this uncertainty in speed small enough to ignore? Not at all! It is comparable with the electron's own speed. So we cannot say how fast it is moving, and a definite orbit has no meaning. There you go, puzzle solved! That is why orbitals, not orbits. Crystal clear, right?

6:04Example 2: why no electron can live in a nucleus

Now a harder one, JEE level. Use the uncertainty principle to show that an electron cannot be confined inside a nucleus. The nucleus is tiny, so delta x is about ten to the power minus fourteen metre. Pause and try. Hint: the same formula as Example one. Done? Okay, first a quick guess. The nucleus is much smaller than the atom. Smaller box, so a smaller uncertainty in speed, right? No, no, careful! This is the JEE trap. Look where delta x sits: in the bottom of the fraction. Smaller delta x, bigger delta v. So delta v is at least h over four pi m delta x. Put in six point six three times ten to the power minus thirty four, the electron's mass, and ten to the power minus fourteen; the brackets just mean multiply. Work it out, and delta v is at least five point eight times ten to the power nine metres per second. Wow, that is huge! Now a topper's quick check. Ten to the power minus ten, over ten to the power minus fourteen, is ten to the power four. So delta x is ten to the power four times smaller than in Example one, and delta v that much larger: five point eight times ten to the power five becomes five point eight times ten to the power nine. It matches perfectly! So, the electron just moves very, very fast inside the nucleus, and that is fine? Oh, no way! That speed is more than the speed of light itself. No electron can move like that. So an electron cannot be confined inside a nucleus. Proved, in two lines! Got it?

7:52Quick revision: position-momentum and energy-time forms

So, quick revision. Delta x times delta p is at least h over four pi; with speed, it is h over four pi m. Delta E times delta t is at least h over four pi, and for the least uncertainty, use the equals sign. And for an electron in an atom, delta v is enormous, so orbitals, not orbits. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!