Answer
(a) $4.1\times 10^{-16}J$
(b) $7.0\times 10^5m/s$
Work Step by Step
(a) We can determine the required average kinetic energy as follows:
$K.E_{avg}=\frac{3}{2}k_B T$
We plug in the known values to obtain:
$K.E_{avg}=\frac{3}{2}(1.38\times 10^{-23})(2\times 10^7)$
$\implies K.E_{avg}=4.1\times 10^{-16}J$
(b) We can determine the required rms speed as follows:
$v_{rms}=\sqrt{\frac{3k_BT}{m}}$
We plug in the known values to obtain:
$v_{rms}=\sqrt{\frac{3\times 1.38\times 10^{-23}\times 2\times 10^7}{1.6726\times 10^{-27}}}$
This simplifies to:
$v_{rms}=7.0\times 10^5m/s$