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(Solved): Aluminum (Al) has an FCC crystal structure and an atomic radius of 0.1431 nm. Compute the interplan ...




Aluminum (Al) has an FCC crystal structure and an atomic radius of 0.1431 nm. Compute the
interplanar spacing for the (110) s
Aluminum (Al) has an FCC crystal structure and an atomic radius of 0.1431 nm. Compute the interplanar spacing for the (110) set of planes. Hint: Use the top-down projection of the cubes. Draw the (110) plane. a b


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