Answer
(a) $310nm$
(b) $250m/s$
(c) $2.1\times 10^{-22}J$
Work Step by Step
(a) The mean free path can be determined as follows:
$\lambda=\frac{k_BT}{4\sqrt{2}\pi p r^2}$
We plug in the known values to obtain:
$\lambda=\frac{1.38\times 10^{-23}\times 10}{4\sqrt{2}\pi \times 1\times 10^4(5\times 10^{-11})^2}$
This simplifies to:
$\lambda=310nm$
(b) The rms speed of the atoms can be calculated as
$v_{rms}=\sqrt{\frac{3k_B TN_A}{M}}$
We plug in the known values to obtain:
$v_{rms}=\sqrt{\frac{3(1.38\times 10^{-23})\times 10\times 6.02\times 10^{23}}{0.004}}$
$v_{rms}=250m/s$
(c) The average energy per atom can be calculated as
$E_{avg}=\frac{3}{2} k_B T$
$E_{avg}=\frac{3}{2}(1.38\times 10^{-23})(10)$
This simplifies to:
$E_{avg}=2.1\times 10^{-22}J$
