If the edge length of $\ce{NaH}$ unit cell is $\pu{488 pm}$, what is the length in $\pu{pm}$ of an $\ce{NaH}$ bond? The answer is $\pu{244 pm}$, I thought that $d = 2 \sqrt{2}\times r$ , do you not need to know the radius for this?

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    $\begingroup$ Please elaborate what you mean by radius. Also note the edits made to format number as equations "\$ \$" for equations, "\$ \ce{} \$" for chemical formulas and equations, and "\$ \pu{} \$" for units. $\endgroup$ – A.K. Jul 25 '18 at 3:15

$\ce{NaH}$ has a rock salt type structure meaning each edge has two $\ce{Na-H}$ bonds. When working out bond lengths in a unit cell, you assume the atoms are point charges (i.e. they take up no volume) and so you do not need to know the radius, only the unit cell parameter which as you have said is $\pu{488 pm}$. Since there are two bonds along each edge, the bond length is simply half of the unit cell parameter, therefore $\pu{244 pm}$.

Finally it dawned on me, why you may have been concerned about knowing atomic radii, yes, you need to know the radii to determine whether $T_\mathrm{d}$ or $O_\mathrm{h}$ sites are preferred. However, given the context of the question, there must have been something previously alluding to the octahedral sites being occupied (and hence rock salt structure) rather than the tetrahedral sites (which would give you a ccp structure with half of $T_\mathrm{d}$ sites occupied). However, you still wouldn't need to use the ionic radii in any calculations pertaining to bond length. I have also included a drawn picture of the unit cell of Sodium Hydride to help you visualise how to calculate the bond length.

Sodium Hydride unit cell


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