## Essential University Physics: Volume 1 (4th Edition)

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: $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(6\times 10^{-21}\ J)}{4.7 \times 10^{-26} \ kg}} \approx 511 \ m/s$