The de Broglie wavelength of molecules

[roam]

Homework Statement

http://img580.imageshack.us/img580/7492/50177994.jpg

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:

\lambda = \frac{h}{p} =\frac{h}{mv}

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

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

So I used the following relationship to work out v

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})}}

\therefore \ v=516.96 \ m/s

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

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]

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

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]

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?

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]

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?

that's it. Your method is fine too.

now try the ultra cool question (part b)

 

[roam]

that's it. Your method is fine too.

now try the ultra cool question (part b)

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

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

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

\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

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/intro/whatbec/what%20is%20BEC.gif All the particles start to behave like a single particle - its beautiful and strange. Gotta love physics.