Answer
$0.43cm/s$
Work Step by Step
We can determine the required rms speed of cesium atoms as follows:
$v_{rms}=\sqrt{\frac{3k_B TN_A}{M}}$
We plug in the known values to obtain:
$v_{rms}=\sqrt{\frac{3(1.38\times 10^{-23})1\times 10^{-7}\times 6.02\times 10^{23}}{0.123}}$
This simplifies to:
$v_{rms}=4.3\times 10^{-3}m/s=0.43cm/s$
