Double slit experiment with smoke particles

[DODGEVIPER13]

Homework Statement

Suppose we wish to do a double slit experiment with a beam of the smoke particles of Example 4.1c (which is lamda=6.6e-20m). Assume we can construct a double slit whose separation is about the same size of the particles. Estimate the separation between the fringes if the double slit and the screen were on opposite coasts of the United States.

Homework Equations

2dsin(theta)=n(lamda)
lamda=h/p

The Attempt at a Solution

Well I see how they got lamda and I understand that I am looking for the separation between the fringes. And that the separation between the slits is 6.6e-20 meters. Honestly I would show work if I understood what was happening here but I am confused. I am confused on how to find theta or un less I assume n=1 but I still don't know lamda.

 

[NascentOxygen]

I still don't know lamda.

I believe you wrote ...

(which is lamda=6.6e-20m)[/size]

It seems this question is about the de Broglie wavelength associated with particles and their momentum.

 

[DODGEVIPER13]

Yah sorry and you are correct it is de broglies wavelength but how should I start this?

 

[NascentOxygen]

Can you explain the terms in this formula you provided?

2. Homework Equations
2dsin(theta)=n(lamda)

 

[DODGEVIPER13]

D is slit width is the angl found from the incident beam, the 2 comes from the distance from the second level of atoms and n stands for the number of maximums and finally lambda stands for the wavelength

 

[NascentOxygen]

D is slit width is the angl found from the incident beam, the 2 comes from the distance from the second level of atoms and n stands for the number of maximums and finally lambda stands for the wavelength

You sound a little confused. See http://en.wikipedia.org/wiki/Double-slit_experiment
The formula for maxima in the bands is explained under Classical wave-optics formulation
The formula enables you to find sinθ, the apex angle of a long thin triangle of height equalling the width of north America.