What is the minumum acceleration potential (V) that must be applied to a beam of free electrons for diffraction to occur in a crystal of niobium (Nb)? Express your answer in volts.
Any ideas?
2.6 *10^3 Volts
2.7*10^3, sorry
Dude its wrong again
Nb is BCC and if you solve it, it gives you 2667 Volts but when you put it as your answer it gives X mark :(
My solution
if BCC first plane to give maximum wavelenth is (110) plane.
Then use lambda = 2dsin(theta) but since you want to maximize the lamdba assume that (theta) = 90 degrees
So lambda = 2d only
d = a(lattice constant) / sqrt (h^2 + k+2 + l^2)
solve for a, plug in (110) for (hkl)
then use lambda = hc/eV
to solve for Volts