Question

What is the ratio of Burgers vector magnitude in FCC crystals of Cu and Al.

Answer 1

burger vectors of FCC form is

b= a/2 * (uvw)

where a = unit edge length

magnitude of b = a/2 *(u^2 + v^2 + v^2 ) ^1/2

For Cu

R = 0.1278 nm     (from internet )

    a= R(2)^1/2   =   0.1278*(2)^1/2 = 0.3614

For FCC Burger vector is b = a/2 (110)

magnitude of b = a/2 (u^2 + v^2 + =w^2) ^1/2

                       = 0.3614/2 (2^2 +1^2 +0^2)^1/2 = 0.2555 nm

For Al

R = 0.125nm

so magnitude of b = 0.12592)^1/2 *1/2   (1^2 +1 ^2 +0^2)^1/2   = 0.

                             = 0.08839nm

ratio of Burgers vector magnitude in FCC crystals of Cu and Al. =0.2555/0.08839 = 2.89