A given metal crystallizes out with a cubic structure having edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom?

108 pm

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40 pm

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127 pm

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80 pm

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Solution

The correct option is C
127 pm

Since there are 4 atoms in a unit cell, we can assume that it is an FCC structure.

$4 r = \sqrt{2} a$

$r = \frac{\sqrt{2} a}{4} = \frac{\sqrt{2} \times 361}{4} = 127 p m$

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A given metal crystallizes out with a cubic structure having edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom?

The correct option is C 127 pm Since there are 4 atoms in a unit cell, we can assume that it is an FCC structure. 4r=√2a r=√2a4=√2×3614=127pm

The correct option is C
127 pm

Since there are 4 atoms in a unit cell, we can assume that it is an FCC structure.

$4 r = \sqrt{2} a$

$r = \frac{\sqrt{2} a}{4} = \frac{\sqrt{2} \times 361}{4} = 127 p m$