The de Broglie wavelength of molecules

roam

1. The problem statement, all variables and given/known data



3. The attempt at a solution

I first attempt part (a) since both parts are similar:

(a) I need to know "v" in order to use the following equation to find the de Broglie wavelength:

[itex]\lambda = \frac{h}{p} =\frac{h}{mv}[/itex]

So I first found the kinetic energy of a N₂ molecule:

[itex]K=3(1.38065 \times 10^{-23})(300)/2 = 6.21292 \times 10^{-21} \ J[/itex]

So I used the following relationship to work out v

[itex]K=\frac{1}{2} mv^2 \implies v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(6.213 \times 10^{-21})}{28(1.6605 \times 10^{-27})}}[/itex]

[itex]\therefore \ v=516.96 \ m/s[/itex]

[itex]\lambda = \frac{6.63 \times 10^{-34}}{(28(1.6605\times 10^{-27})\times 516.96)} = 2.7593 \times 10^{-11}[/itex]

My answer seems wrong because this is 0.0275 nanometers (or 27.589 picometers) which is in the X-ray region of the spectrum! So what did I do wrong?

Also the question says "compare this with the diameter (less than 1 nm) of the molecule", what does that mean? My value for λ is shorter than the diameter of the molecule which doesn't sound right.

Any help is greatly appreciated.

NoPoke

What does the de Broglie wavelength tell you about the behaviour of a particle in a physics experiment?

roam

It tells us about its momentum, the higher the speed (and therefore momentum) the shorter the wavelength. But my working is not right, is it? Because I think it takes a lot more energy to produce x-rays...

Rooted

Your working looks good to me, I think. Think about it this way - what would be the de Broglie wavelength of, say, a lead atom or an electron with the same kinetic energy as the sodium atom above? Doe that help?

NoPoke

The de Broglie wavelength is not associated with an electromagnetic wave like an X-ray. Does that help?

So if it isn't an electromagnetic wave what use is it? What insight into the behaviour of large things and small things is it giving you?

roam

Alright, I see. I think it has to do with the wave and particle duality of particles. But I was wondering if my method and working were correct or not?

NoPoke

that's it. Your method is fine too.

now try the ultra cool question (part b)

roam

Okay thank you. Here's what I did for part (b):

[itex]K=3(1.38065 \times 10^{-23})(450\times10^{-12}) /2 = 9.3194 \times 10^{-33} \ J[/itex]

[itex]v= \sqrt{\frac{2(9.3194 \times 10^{-33})}{23(1.6605 \times 10^{-27})}} = 6.98595 \times 10^{-4} \ m/s[/itex]

[itex]\lambda = \frac{6.63 \times 10^{-34}}{(23(1.6605 \times 10^{-27}))\times (6.9859 \times 10^{-4})} = 2.48497 \times 10^{-5} \ m[/itex]

So the de Broglie wavelength is 24.85 μm. Is this correct as well? I'm not sure if it was valid to apply the equation K=1/2mv² to work out the velocity of the particle, since the kinetic energy of the particles were already given by another formula K=3k_BT/2. I'm not sure if that equation is appropriate for this situation of gas particles.

NoPoke

temperature reduced by 12 orders of magnitude so the de Broglie wavelength increases by 6 orders of magnitude. ( I haven't checked your actual calculation )

yes I believe that both energy equations can still be applied even at such low temperatures , you are right to be cautious of what velocity means when the particle is no longer behaving like an independent lump but like a wave. I know basically nothing about Bose Einstein condensates which is what this ultra cool part b is leading towards. This picture is basically the limit of my knowledge http://cua.mit.edu/ketterle_group/in...20is%20BEC.gif All the particles start to behave like a single particle - its beautiful and strange. Gotta love physics.