Question

You irradiate a crystalline sample that has 2.9344 Å between atoms with electrons to do an electron diffraction experiment in reflection mode. You observe a first order (m = 1) diffraction peak at θ = 12.062°. What is the wavelength of the electrons?

Answer 1

`\lambda=1.23`Å

Step-by-step explanation:

Bragg's Law refers to the simple equation:

`n\lambda = 2d sin(\theta)`

In this case:

n=1

θ = 12.062°

d=2.9344 Å

`\lambda = 2*2.9344sin(12.062)=1.23`Å