The metal gold crystallizes in a face centered cubic unit cell with one atom per lattice point. when x-rays with λ = 1.436 å are used, the second-order bragg reflection from a set of parallel planes in a(n) gold crystal is observed at an angle θ = 20.62°. if the spacing between these planes corresponds to the unit cell length (d = a), calculate the radius of a(n) gold atom.

 r =  1.45 Å

Explanation:

given,

λ = 1.436 Å

θ = 20.62°

d = a

n = 2

metal gold crystallizes in a face centered cubic unit cell

Radius of the gold atom = ?

using Bragg's Law

 n λ = 2 d sin θ

 2 x 1.436 Å = 2 a sin 20.62°

 a = 4.077 Å

We know relation of radius for face centered cubic unit cell

 r =  1.45 Å

the radius of a(n) gold atom. is equal to 1.45 Å

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)