Question

In the atom interferometer experiment of Figure $38.13$, laser cooling techniques were used to cool a dilute vapor of sodium atoms to a temperature of $0.0010\mathrm{K}=1.0\mathrm{mK}$. The ultracold atoms passed through a series of collimating apertures to form the atomic beam you see circling the figure from the left. The standing light waves were created from a laser beam with a wavelength of $590\mathrm{nm}$.

a. What is the rms speed $v_{\text{me}}$of a sodium atom $(A-23)$in a gas at temperature $1.0\mathrm{mK}$?

b. By treating the laser beam as if it were a diffraction grating. Calculate the first-order diffraction angle of a sodium atom traveling at the rms speed of part a.

c. how far apart are the points $B$and $C$if the second sanding wave is $10\mathrm{cm}$from the first?

d. Because interference is observed between the two paths, each individual atom is apparently present at both points $B$and point $C$Describe, in your own words, what this experiment tells you about the nature of matter.

Short answer

(a).$v_{rms}=1.08\frac{\mathrm{m}}{\mathrm{s}}$

(b) $\theta =3.12°$

(c) $y=1.09\mathrm{cm}$

(d) We can conclude that the atoms have nonlocalized behavior like waves.

Step-by-step solution

Step: 1 Given information

(a) sodium atom $(A-23)$in a gas at a temperature $1.0\mathrm{mK}$?

Calculation

(a),We can begin by using the following speed equation from the kinetic theory of gases:

$$\begin{gathered} v_{rms}=\sqrt{\frac{3kT}{m}} \\ m=Au \\ {v}_{rms}=\sqrt{\frac{3kT}{Au}} \\ {v}_{\text{rms}}=\sqrt{\frac{3·1.38·{10}^{-23}·0.001}{23·1.67·{10}^{-27}}} \\ {v}_{rms}=1.08\frac{m}{s} \end{gathered}$$

Given information

(b) by treating the laser beam as if it were a diffraction grating. calculate the first-order diffraction angle of a sodium atom traveling at the rms speed of part a

Calculation

We must first locate de Broglie's wavelength.

$$\begin{gathered} \lambda =\frac{h}{p} \\ \lambda =\frac{h}{mv} \end{gathered}$$

$$\begin{gathered} d\sin \theta =n\lambda \\ d=\frac{{\lambda }_{\text{laser}}}{2} \end{gathered}$$

$$\begin{gathered} \frac{{\lambda }_{\text{laser}}}{2}\sin \theta =n\lambda \\ \theta =\arcsin \frac{2n\lambda }{{\lambda }_{\text{laser}}} \\ \theta =\arcsin \frac{2nh}{Auv{\lambda }_{\text{laser}}} \\ \theta =\arcsin \frac{2·1·h}{23·1.67·{10}^{-27}·1.08·590·{10}^{-9}} \\ \theta =3.12° \end{gathered}$$

Step:5 Given information

(c).How far apart are the points $B$and $C$if the second sanding wave is $10\mathrm{cm}$from the first?

Step: 6 Calculation

(c). From Figure $38.13$and geometry, we can find the distance between $B$and $C$

$$\begin{gathered} y=2L\tan \theta \\ y=2·0.1·\tan 3.2° \\ y=1.09\mathrm{cm} \end{gathered}$$

Given information

The individual atom is apparently present at both point $B$and point$C$. Describe, in your own words, what this experiment tells you about the nature of matter.

Step: 8 Calculation

(d). We can deduce that atoms behave like waves in terms of nonlocality.