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Identify the crystal structure if its density is 9.6 x 102 kg/m3, lattice constant is 4.3 A? and atomic weight is 23.
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written 3.5 years ago by
teamques10
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Given –
$\rho =9.6 \times 10^{2} kg/m^{3}$
$a=4.3A^{\circ} = 4.3 \times 10^{-10}m$
$M=23$
To find – $n$, (identify the crystal) Solution – According to, $n=\dfrac {\rho \cdot a^{3}}{\left(\frac{M}{N_A}\right)}$ $n=\dfrac {9.6 \times 10^{2}\times (4.3\times 10^{-10})^{3}}{\frac{23}{6.0238 \times 10^{23}}}=2$ $\because n=2$
∴ The crystal structure is BCC.
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