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Chapter 5: Problem 140

The Chemistry in Action essay "Super Cold Atoms" in Section \(5.7 .\) describes the cooling of rubidium vapor to \(5.0 \times 10^{-8} \mathrm{~K}\). Calculate the root-mean-square speed and average kinetic energy of a \(\mathrm{Rb}\) atom at this temperature.

Short Answer

Expert verified
The root-mean-square speed of a Rubidium atom at \(5.0 \times 10^{-8} \mathrm{~K}\) is \(0.12 m/s\) and the average kinetic energy is \(1.035 \times 10^{-29} J\).

Step by step solution

01

Calculate the Mass of a Rubiuim Atom

Rubidium (Rb) has a molar mass of 85 g/mol. But the mass of one atom is needed here. Using Avogadro's number (\(N_A = 6.022 \times 10^{23} \, atoms/mol\)), the mass of one rubidium atom can be calculated as \(m = M/N_A = 85g/mol / 6.022 \times 10^{23} atoms/mol = 1.41 \times 10^{-25} kg/atom\).
02

Calculate the Root-Mean-Square Speed

With the mass of one Rb atom and the given temperature (T), substitute the values into the RMS speed equation \(v_{rms} = \sqrt{3kT/m}\) to obtain \(v_{rms} = \sqrt{3\cdot \, 1.38 \cdot 10^{-23} J/K \cdot 5 \cdot 10^{-8} K / 1.41 \cdot 10^{-25} kg} = 0.12 m/s.\)
03

Calculate the Average Kinetic Energy

Substitute the given temperature (T) into the average kinetic energy equation \(KE_{avg} =\frac{3}{2} kT\) to obtain \(KE_{avg} =\frac{3}{2} \cdot 1.38 \cdot 10^{-23} J/K \cdot 5 \cdot 10^{-8} K = 1.035 \times 10^{-29} J\).

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