Chapter 39: Q. 44 Exercises and Problems (page 1139)

Physicists use laser beams to create an atom trap in which atoms are confined within a spherical region of space with a diameter of about $1 mm$ . The scientists have been able to cool the atoms in an atom trap to a temperature of approximately $1 nK$ , which is extremely close to absolute zero, but it would be interesting to know if this temperature is close to any limit set by quantum physics. We can explore this issue with a onedimensional model of a sodium atom in a $1.0 - mm$ -long box.
a. Estimate the smallest range of speeds you might find for a sodium atom in this box.
b. Even if we do our best to bring a group of sodium atoms to rest, individual atoms will have speeds within the range you found in part a. Because there's a distribution of speeds, suppose we estimate that the root-mean-square speed $v_{ms}$ of the atoms in the trap is half the value you found in part a. Use this $v_{rms}$ to estimate the temperature of the atoms when they've been cooled to the limit set by the uncertainty principle.

Short Answer

Expert verified

a.) Smallest range of speeds is $(0 m / s) to (4.34 \times 10^{- 6} m / s)$

b.) Estimated the temperature is $4.34 \times 10^{- 15} K$

Step by step solution

a.) Step 1 Given InformationDiameter of spherical region is 1 mm and the temperature of an atom is 1nK, size of the one-dimensional model is 1.0-mm-long box.

a.) The uncertainty in knowing the position of the atom inside a one-dimensional box of a length of ( $1 mm$ ) is ( $Δ x = 1 mm$ ). Now, the uncertainty in the velocity can be calculated using the Heisenberg uncertainty principle $Δ x Δ p_{x} \geq \frac{h}{2}$

Assuming the most accurate measurements possible, we can replace $(\geq)$ with ( = ) . Moreover, we know that $Δ p_{x} = m Δ v_{x}$ and the mass of the sodium atom is $3.817 \times 10^{- 26} kg$ . Thus

$Δ v_{x} = \frac{h}{2 m Δ x} = \frac{6.626 \times 10^{- 34} J \cdot s}{2 (3.817 \times 10^{- 26} kg) (1 \times 10^{- 3} m)} = 8.68 \times 10^{- 6} m / s$

which means that the range of possible velocities for the electron is from $(- Δ v_{x} / 2 = - 4.34 \times 10^{- 6} m / s)$ to $(Δ v_{x} / 2 = 4.34 \times 10^{- 6} m / s)$ . In terms of the speed, the smallest range of speeds might be found for a sodium atom in the given box is from $(0 m / s)$ to $(4.34 \times 10^{- 6} m / s)$ .

b.) Given InformationThe root-mean-square speed of the atoms in the trap is half the value you found in part a

We can find an expression for the temperature by using the fact that the kinetic energy of the atom can be written as $\frac{3}{2} k T$ , and also as $\frac{1}{2} m v_{rms}^{2}$ Hence $\frac{3}{2} k T = \frac{1}{2} m v_{rms}^{2}$