JEE Advance - Chemistry (2008 - Paper 2 Offline - No. 20)
The empty space in this HCP unit cell is :
74%
47.6%
32%
26%
Explanation
Packing fraction = $$\frac{\mathrm{Total~volume~of~unit~cell}}{\mathrm{Volume~occupied}}$$
Substituting the values of occupied sphere and volume of the unit cell we have:
$$ \Rightarrow {{6 \times {4 \over 3} \times \pi {r^3}} \over {24\sqrt 2 {r^3}}} = 0.74$$
Rearranging and solving % occupied space = 74%
The space in the HCP unit cell = 100 $$-$$ 74 = 26%
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