Answer
$0.124nm$
Work Step by Step
We can find the required separation as follows:
$d=\frac{\lambda}{sin\theta_{max}}$
We plug in the known values to obtain:
$d=\frac{0.030\times 10^{-9}m}{sin(14^{\circ})}$
$d=\frac{0.030\times 10^{-9}m}{0.2419}$
$d=0.124\times 10^{-9}m$
$d=0.124nm$
