Answer
(a) $1.24\times 10^{-19}J$
(b) $1.2\times 10^4m/s$
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:
$K.E_{avg}=\frac{3}{2}(1.38\times 10^{-23})(6\times 10^4)$
This simplifies to:
$K.E_{avg}=1.24\times 10^{-19}J$
(b) We know that
$v_{rms}=\sqrt{\frac{3k_B T}{m}}$
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:
$v_{rms}=1.2\times 10^4m/s$
