Iron has body centered cubic lattice. If the edge length of the unit cell is $286$ pm, the radius (in pm) of iron atom is:

80

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62

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160

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124

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Solution

The correct option is B $124$
In BCC lattice,

a= 286 pm

$\sqrt{3} a = 4 r$

$∴ r = \frac{\sqrt{3} \times 286}{4} = 124$ pm

Option D is correct.

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Iron has body centered cubic lattice. If the edge length of the unit cell is 286 pm, the radius (in pm) of iron atom is:

The correct option is B 124In BCC lattice,a= 286 pm √3a=4r∴r=√3×2864=124 pmOption D is correct.

Iron has body centered cubic lattice. If the edge length of the unit cell is $286$ pm, the radius (in pm) of iron atom is:

The correct option is B $124$
In BCC lattice,

a= 286 pm

$\sqrt{3} a = 4 r$

$∴ r = \frac{\sqrt{3} \times 286}{4} = 124$ pm

Option D is correct.