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Homework Statement
What is the Bragg Scattering angle (i.e. the angle, in degrees, between the incoming incident beam and the reflected beam) for electrons scattered from a nickel crystal if their energy is 63 eV? The spacing between Ni Bragg planes is 0.215 nm. Give your result to 3 significant figures.
so the energy of the electrons is 63eV and d = 0.215nm
Homework Equations
n[tex]\lambda[/tex] = 2d sin [tex]\theta[/tex]
E = hc/[tex]\lambda[/tex]
The Attempt at a Solution
a condition for bragg diffraction is that [tex]\lambda[/tex] [tex]\leq[/tex] 2d so when you use E = hc/[tex]\lambda[/tex] to find the wavelength of the electrons you get ~19.72nm which is much greater than 2d. also the question doesn't say anything about an angle. am i missing something here?? not only will this not work with that wavelength being so much bigger than 2d but there's no information about an angle?