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Iron Packs in a Body-Centered Cubic Structure 2(126pm)=178pm\sqrt{2}(126 \mathrm{pm})=178 \mathrm{pm} B) 2(126 Pm)= 252 Pm
C)

Question 1

Multiple Choice

Iron packs in a body-centered cubic structure.If an iron atom has a radius of 126 pm,what is the distance between lattice points at two opposite corners of the unit cell? (Note: a line drawn between the points would go through the center of the unit cell)


A) 2(126pm) =178pm\sqrt{2}(126 \mathrm{pm}) =178 \mathrm{pm}
B) 2(126 pm) = 252 pm
C) 32(126pm) =109pm\frac{\sqrt{3}}{2}(126 \mathrm{pm}) =109 \mathrm{pm}
D) 4(126 pm) = 504 pm
E) 3(126pm) =218pm\sqrt{3}(126 \mathrm{pm}) =218 \mathrm{pm}

Correct Answer:

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