How many peaks in the interference pattern?

[Abdul.119]

Homework Statement

In a diffraction experiment in which electrons of kinetic energy 110 eV are scattered from a crystal, a first maximum in the intensity of the scattered electrons occurs at an angle θ=10.7
a) How many peaks will there be in the interference pattern?
b) What is the spacing between the atom planes?

Homework Equations

2d sinθ = nλ
λ = h/p = h/√(2m KE)

The Attempt at a Solution

From the kinetic energy of the electron I found the wavelength to be 1.17*10^-10 m , then I don't understand how to use it in that equation, I believe n is the number of peaks, and d is the spacing, so how would I solve this equation while two variables are missing?

 

[gneill]

n is the number of a given peak (well, the pair of peaks placed symmetrically about the centerline, where the central peak is for n = 0).

You're given the angle of the first peak for which n = 1. That should get you going. For more detail, take a look at the Hyperphysics entry on diffraction gratings.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/grating.html

 

[Abdul.119]

n is the number of a given peak (well, the pair of peaks placed symmetrically about the centerline, where the central peak is for n = 0).

You're given the angle of the first peak for which n = 1. That should get you going. For more detail, take a look at the Hyperphysics entry on diffraction gratings.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/grating.html

Oh ok, so having n=1 will help me solve for the d, but how do I find how many peaks there are then?

 

[gneill]

Oh ok, so having n=1 will help me solve for the d, but how do I find how many peaks there are then?

Usually you'd look for values of n where the equations start giving un-physical results (like imaginary angles, or impossible (real) values for a trig function).

 

[Abdul.119]

Usually you'd look for values of n where the equations start giving un-physical results (like imaginary angles, or impossible (real) values for a trig function).

Solving for the angle θ, I start getting imaginary numbers at n=6, so that means there are 5 peaks?

 

[gneill]

Solving for the angle θ, I start getting imaginary numbers at n=6, so that means there are 5 peaks?

Sounds reasonable. Those would be pairs of peaks, and don't forget symmetry and the central peak.

 

[Abdul.119]

Sounds reasonable. Those would be pairs of peaks, and don't forget symmetry and the central peak.

Okay, thank you very much for the help