Question

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

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

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

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