Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is:
A
3(√3)4(√2)
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B
4(√3)3(√2)
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C
(√3)(√2)
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D
12
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Solution
The correct option is A3(√3)4(√2) For BCC, z = 2 and a = 4r(√3) For FCC, z = 4 and a = 2 √2r ∴d25∘Cd900∘C=(ZMNAa3)BCC(ZMNAa3)FCC =24(2√2r4r√3)3 =(3√34√2)