Answer
511 meters per second
Work Step by Step
We first find $\bar{K}$.
$\bar{K} = \frac{3}{2}kT = \frac{3}{2}(1.38\times10^{-23})(293)=6\times 10^{-21}\ J$
We now find the mass:
$m = 2(14)(1.66 \times10^{-27})=4.7 \times 10^{-26} \ kg$
Finally, we find v:
$v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(6\times 10^{-21}\ J)}{4.7 \times 10^{-26} \ kg}} \approx 511 \ m/s$
