# Calculating the de Broglie Wavelength

1. Jul 8, 2010

### Jenkz

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

Calculate the de Broglie wavelength of an oxygen molecule (molecular mass 32)in air at room temperture.

2. Relevant equations
P = h/ $$\lambda$$
P= mv
E= hv = 1/2mv$$^{2}$$ = P$$^{2}$$/(2m)

3. The attempt at a solution

hv = 1/2mv$$^{2}$$

h = 1/2mv

If v = P/m then h = mp/2m

so p= 2h

This doesnt seem correct to me... help?

Last edited: Jul 8, 2010
2. Jul 8, 2010

### dulrich

E = hf, where f is the frequency of the photon. Sometimes also written as $E = h\nu$.

But $\frac12 mv^2 = p^2 / 2m$ is correct and $p = h/\lambda$ is too. What you need is a way of calculating the (average) kinetic energy of the oxygen molecules. Note that you are told the situation involves room temperature.

3. Jul 8, 2010

### Jenkz

Would i use KE = 3/2 RT where R = Gas constant 8.31 JK^-1mol^-1 ?

And equate this to p$$^{2}$$/2m

4. Jul 8, 2010

### dulrich

That will get you the internal kinetic energy of one mole of a monatomic ideal gas. You want a similar formula for one molecule of the gas. In addition, the oxygen molecule is diatomic, so the 3/2 is not correct.

5. Jul 9, 2010

### Jenkz

it has 6 degrees of freedom? 3 for movement and 3 for momentum?

Would I use KE=6/2 KT where K is Boltzmann's constant: 1.380 x 10^-23 J/K.

6. Jul 9, 2010

### RoyalCat

First, the relation E=hf is only valid for a photon.

Second, a diatomic molecule has 5 degrees of freedom.
http://en.wikipedia.org/wiki/Degrees_of_freedom_(physics_and_chemistry)
Let's imagine you lay out your molecule along a Cartesian coordinate system (x, y and z axes). We'll have the axis of the molecule point along the x direction. Imagine one oxygen atom at position (d,0,0) and the other at position (-d,0,0).

The 5 degrees of freedom are linear motion along the x, y and z axes (3 degrees of freedom), rotation about the y or z axis (Note how these two types of motion are completely equivalent, since any axes perpendicular to that of the molecule's axis are indistinguishable (1 degree of freedom), and oscillations along the x axis (Like two masses on a spring, the two atoms can move closer and further apart) (1 degree of freedom).

We don't count the rotation about the x axis since it incorporates very little energy, just like we disregard the rotation of a mono-atomic gas. (I'm not quite sure why this is valid and how it sits with the equipartition principle, but it's just the way it is. Maybe it has something to do with how two snapshots of the rotation would be indistinguishable from each other)

The whole degrees of freedom thing is a bit tricky, I may have mislead you here a bit, so if someone better informed comes along, feel free to take their word over mine.
The gist of the matter is that you're supposed to remember the following, mono-atomic - 3 degrees of freedom, diatomic - 5 degrees of freedom.
If you're dealing with a polyatomic molecule, you'll either be asked for a qualitative description, or simply given the number of degrees of freedom, since counting degrees of freedom gets complicated.

Oh, and on a final note, movement and momentum are the same thing. What you might have meant is position and momentum, but position does not constitute a degree of freedom separate from that axis' associated momentum/KE.

Last edited: Jul 9, 2010
7. Jul 9, 2010

### Jenkz

Oh ok, thank you. I'll remember that in the future.

I got an answer of 2.6 x10^-11 m ??

8. Jul 9, 2010