A conventional <body-centred cubic> cell contains two atoms, so its edge $a$ obeys $\rho a^3=2(23m_u)$. Taking $m_u=1.66054\times10^{-27}\,\mathrm{kg}$ gives $a\simeq4.29\times10^{-10}\,\mathrm m$. The <neighbour shells of a body-centred cubic lattice> have eight nearest neighbours that are at $\sqrt3a/2$; the six next-nearest neighbours lie along the cell axes. Thus \b[the next-nearest-neighbour distance is $a\simeq0.429\,\mathrm{nm}$].

Solution

Past exam of the physics A course of the University of Cambridge / 2011 / Paper 1 / Section A / 4 / Solution

 Solution (source code)