Learning Goal: To understand, qualitatively and quantitatively, the uncertainty principle.
This idea is difficult to reconcile with common experience. To understand it better, you must consider the properties of a wave. According to the de Broglie equation, the momentum of a wave is directly related to its wavelength. For the wave in the first figure, the wavelength is clearly well defined. However, the position is not well defined at all. The question, "Where is the wave?" does not have a welldefined answer, as we expect for a particle. This is the essence of the indeterminacy principle. We could just as easily draw a single sharp point at some particular x coordinate. This could be considered a wave with a very well determined position. However, any notion of wavelength for such a wave seems strange.
A wave like the one shown in the second figure can be built up by adding together waves with different wavelengths. Recall that if two waves with similar frequencies, and , are added together, a wave with a beat frequency of is produced . This gives a wave with somewhat welldefined position and wavelength. If you add contributions from all of the frequencies between and , then you get a wave packet, which looks essentially like a single isolated beat cycle. In this problem, you will consider such a wave packet as simply being one beat cycle of this wave. While not exactly correct, this will give a useful approxmation.
Let the distance between the two nodes of the wave be the uncertainty in position . Since the beat frequency is given by , and the wave travels at speed , the uncertainty in position is given by
.
Part A  

The de Broglie relation can be rewritten in terms of the wave number as . Recall that wave number is defined by . Using the fact that , find the wave numbers and corresponding to frequencies and . Express your answer as two expressions separated by a comma. Use , , , and .

Part B  

Find an expression for the uncertainty in the wave number. Use your results from Part A. Express your answer in terms of quantities given in Part A.

Part C  

What is the value of the product ? Use to find the uncertainty in the momentum of the particle. Express your answer in terms of quantities given in Part A and fundamental constants.

Part D  

In an atom, an electron is confined to a space of roughly meters. If we take this to be the uncertainty in the electron's position, what is the minimum uncertainty in its momentum? Express your answer in kilogram meters per second to two significant figures.

Part E  

What is the kinetic energy of an electron with momentum kilogram meters per second? Express your answer in electron volts to two significant figures.
Notice that this energy is similar to the energy scale for electrons in an atom, which typically ranges from a bit less than an electron volt up to a few dozen electron volts. A good estimate for the energy scale of a particle can often be found by calculating the energy the particle would have if you set the momentum equal to the minimum uncertainty in momentum. The justification for this sort of estimation lies in the rigorous statistical definition of the uncertainty; it is sufficient now for you to know that this will give a reasonably good orderofmagnitude estimate of the energy for a variety of quantum systems. 
Part F  

Suppose that you know the position of a 100gram pebble to within the width of an atomic nucleus ( meters). What is the minimum uncertainty in the momentum of the pebble? Express your answer in kilogram meters per second to one significant figure.
For a 100gram pebble, this corresponds to an uncertainty in the speed of about meters per second. Such tiny values are the reason that you are unaware of the uncertainty principle in everyday situations. In practice, it would be impossible to measure the position of a pebble to such accuracy, much less its speed 
That was really helpful, thanks :)
ReplyDeleteThat was really helpful...if I just wanted the answers. Having an explanation would be nice too.
ReplyDeleteThen feel free to ask questions regarding specific parts. I'll help if I can.
ReplyDeleteAnswer to Part D is wrong?
ReplyDeleteI believe you are correct, DisaFear. If we take (dx)(dp) = h, then (dp) = h / (dx) ~= 6.626*10^24. I can't check these answers anymore, but you could try that.
ReplyDeletePart D should be 3.3*10^24 kgm/s using the uncertainty formula:
ReplyDeletedx.dp>=h/2
I got 5.3*10^25 for part D. I'm getting part F wrong, too.
ReplyDeletePart F: deltX*deltP>=hbar/2, therefore deltp>=hbar/(2*deltX)=5*10^20 kg*m/s
ReplyDeletePart D is correct. If you're using Mastering Physics then you'll in the comments above this portion of the question a different uncertainty formula; (dx)(dp)>h(bar) and not h(bar)/2. I do not understand why it is different but it is and 1.05E24 (or 1.1E24 for 2 sig fig's) gives a correct answer. It would be awesome if anyone could give a little more detail to Part F.
ReplyDeleteDisregard my issue with Part F, "... one sig fig."
ReplyDelete3*10^19 will be the answer for F
ReplyDelete