In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is $356$ pm, then the radius (in pm) of carbon atom is:

77.07

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154.14

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251.70

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89.20

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Solution

The correct option is B $77.07$
In a diamond structure, there are atoms at each corner and alternate tetrahedral voids. Consider the body diagonal of unit cell (cube with side a). It corresponds to 8R, and length of body diagonal is $\sqrt{3} \times a$ .
Thus $R = \frac{\sqrt{3} \times a}{8} = \frac{356 \times 1.732}{8} = 77.07$ pm

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In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is 356 pm, then radius of carbon atom is

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In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is 356 pm, then the radius (in pm) of carbon atom is:

The correct option is B 77.07In a diamond structure, there are atoms at each corner and alternate tetrahedral voids. Consider the body diagonal of unit cell (cube with side a). It corresponds to 8R, and length of body diagonal is √3×a.Thus R=√3×a8=356×1.7328=77.07 pm

In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is $356$ pm, then the radius (in pm) of carbon atom is: