= Solution
At the same <temperature>, momenta increase as the <square root> of <mass>, so wavelengths decrease as its inverse <square root>. For a nucleus of <mass> $M_N\simeq A m_p$,
$$
\boxed{\frac{\lambda_N}{\lambda_e}=\sqrt{\frac{m_e}{M_N}}
\simeq\frac1{\sqrt{1836A}}.}
$$
At the <electron> overlap threshold, $\lambda_e=d$, the nuclear wavelength is about $7.0\,\mathrm{pm}/\sqrt A$, far below the $0.3$ nm spacing. Nuclear exchange effects are consequently much less important than <electron> exchange effects. This statement about nuclear overlap does not mean lattice vibrations are always classical.
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