Calculating the de Broglie Wavelength

In summary, an oxygen molecule has a de Broglie wavelength of 2.6 x10^-11 meters in air at room temperature.
  • #1
Jenkz
59
0

Homework Statement



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

Homework Equations


P = h/ [tex]\lambda[/tex]
P= mv
E= hv = 1/2mv[tex]^{2}[/tex] = P[tex]^{2}[/tex]/(2m)

The Attempt at a Solution



hv = 1/2mv[tex]^{2}[/tex]

h = 1/2mv

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

so p= 2h

This doesn't seem correct to me... help?
 
Last edited:
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  • #2
E = hf, where f is the frequency of the photon. Sometimes also written as [itex]E = h\nu[/itex].

But [itex]\frac12 mv^2 = p^2 / 2m[/itex] is correct and [itex]p = h/\lambda[/itex] 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
Would i use KE = 3/2 RT where R = Gas constant 8.31 JK^-1mol^-1 ?

And equate this to p[tex]^{2}[/tex]/2m
 
  • #4
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
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
I have several comments,
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.
 
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  • #7
Oh ok, thank you. I'll remember that in the future.

I got an answer of 2.6 x10^-11 m ??
 
  • #8
Yep, same answer I got.
 

Related to Calculating the de Broglie Wavelength

What is the de Broglie wavelength?

The de Broglie wavelength is a concept in quantum mechanics that describes the wave-like nature of matter. It is the wavelength associated with a particle and is calculated using the particle's momentum.

How do you calculate the de Broglie wavelength?

The de Broglie wavelength (λ) can be calculated using the following formula: λ = h/mv, where h is Planck's constant, m is the mass of the particle, and v is the velocity of the particle.

What is the significance of the de Broglie wavelength?

The de Broglie wavelength helps to explain the wave-particle duality of matter. It suggests that all particles have both wave-like and particle-like properties, and the wavelength can be used to describe the behavior of particles on a quantum level.

What units are used for the de Broglie wavelength?

The de Broglie wavelength is typically measured in meters (m). However, it can also be expressed in other units such as nanometers (nm) or angstroms (Å).

Can the de Broglie wavelength be observed?

No, the de Broglie wavelength is a theoretical concept and cannot be directly observed. However, its effects can be observed in experiments such as the double-slit experiment, where particles exhibit wave-like behavior.

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