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An element has atomic mass 93 g mol–1 and density 11.5 g cm–3 . If the edge length of its unit cell is 300 pm, identify the type of unit cell. |
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Answer» Given, M = 93 g mol-1 d = 11.5 g cm-3 a = 300 pm = 300 x 10-10 cm = 3 x 10-8 We know that, d = \(\frac{z\times M}{N_A\times a^3}\) z = \(\frac{d\times N_A\times a^3}{M}\) = \(\frac{11.5\times6.022\times10^{23}(3\times10^{-8})^3}{93}\) = 2.01 The number of atoms present in given unit cells is coming nearly equal to 2. Hence, the given unit cell is body centred cubic unit cell (BCC). |
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