Question

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 $1\mathrm{mm}$. The scientists have been able to cool the atoms in an atom trap to a temperature of approximately $1\mathrm{nK}$, 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 $v_{\text{rms}}$of the atoms in the trap is half the value you found in part a. Use this $v_{\mathrm{rms}}$to estimate the temperature of the atoms when they've been cooled to the limit set by the uncertainty principle.

Short answer

(a)The smallest range of speed for sodium atom is from$0\mathrm{m}/\mathrm{s}\text{to}4.3\times {10}^{-6}\mathrm{m}/\mathrm{s}$

(b)the lowest particle value of temperature is $4.0\times {10}^{-15}\mathrm{k}$is very much smaller (lower) than $1\mathrm{nk}$

Step-by-step solution

Part (a) Step 1: Given Information

Soot particles diameter $=15\mathrm{nm}$

$=15\mathrm{nm}\left(\frac{1\mathrm{m}}{{10}^9\mathrm{nm}}\right)$

$Density=1200\mathrm{kg}/{\mathrm{m}}^3$

localid="1650901525255" $Diameterofthinplate=0.50\mu \mathrm{m}$

localid="1650901546201" $Diameterofthecircle=2000\mathrm{nm}$

$=2000\mathrm{nm}\left(\frac{1\mathrm{m}}{{10}^9\mathrm{nm}}\right)$

Part (a) Step 2: Solution

Uncertainty velocity of sodium atom is

$\begin{array}{r}\Delta vx=\frac{h}{2m\Delta x}\\ h=6.63\times {10}^{-34}\mathrm{JS}\end{array}$

Applying $\mathrm{m}=23\left(1.67\times {10}^{-27}\mathrm{kg}\right)$

$$\begin{gathered} \Delta \mathrm{x}=1\times {10}^{-3}\mathrm{m} \\ \Delta v\mathrm{x}=\frac{6.63\times {10}^{-3}\mathrm{JS}}{2\left(23\left(1.67\times {10}^{-21}\mathrm{kg}\right)\right)\left(1\times {10}^{-3}\mathrm{m}\right)} \\ \Delta v\mathrm{v}=8.6\times {10}^{-6}\mathrm{m}/\mathrm{s} \end{gathered}$$

I herefore the possıble range of velocities (speed) Is from $-4.3\times {10}^6\mathrm{m}/\mathrm{s}$to $4.3\times {10}^6\mathrm{m}/\mathrm{s}$. Speed do not have negative values, the possible range of speed is from $0\mathrm{m}/\mathrm{s}$to $4.3\times {10}^{-6}\mathrm{m}/\mathrm{s}$

Conclusion: The smallest range of speed for sodium atom is from $-0\mathrm{m}/\mathrm{s}$to $4.3\times {10}^{-6}\mathrm{m}/\mathrm{s}$

Part (b) Step 1: Soluction

From the equation(2)

The rms speed of atom is

$\mathrm{Vrms}=\frac{1}{2}\mu \max$

Applying $\mu =4.3\times {10}^{-6}\mathrm{m}/\mathrm{s}$

$$\begin{gathered} V\mathrm{rms}=\frac{1}{2}\left(4.3\times {10}^{-6}\mathrm{m}/\mathrm{s}\right) \\ V\mathrm{rms}=2.15\times {10}^{-6}\mathrm{m}/\mathrm{s} \end{gathered}$$

From the equation (3), the lowest temperature is

$\mathrm{T}=\frac{{\mathrm{mv}}^2\mathrm{rms}}{3\mathrm{kB}}$

Applying values,

$$\begin{gathered} \mathrm{T}=\frac{23\left(1.67\times {10}^{-22}\mathrm{kg}\right){\left(2.15\times {10}^{-6}\mathrm{m}/\mathrm{s}\right)}^2}{3\left(1.38\times {10}^{-2}\mathrm{J}/\mathrm{K}\right)} \\ \left[\mathrm{kB}=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right] \\ \mathrm{T}=4.0\times {10}^{-15}\mathrm{k} \end{gathered}$$

Conclusion:Thus, the lowest particle value of temperature is $4.0\times {10}^{-15}\mathrm{k}$is very much smaller (lower) than 1 nk..