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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?

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jorgezarate

Answer:

[tex]\lambda=1.23[/tex]Å

Explanation:

Bragg's Law refers to the simple equation:

[tex]n\lambda = 2d sin(\theta)[/tex]

In this case:

n=1

θ = 12.062°

d=2.9344 Å

[tex]\lambda = 2*2.9344sin(12.062)=1.23[/tex]Å

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