Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the rms speed is given by
$$v_{\rm rms}=\sqrt{\dfrac{3k_BT}{m}}$$
So for the given sodium atom,
$$v_{\rm rms}=\sqrt{\dfrac{3(1.38\times 10^{-23})(1\times 10^{-3})}{(23\times 1.67\times 10^{-27})}}=\color{red}{\bf 1.04}\;\rm m/s$$
$$\color{blue}{\bf [b]}$$
First, we need to find the de Broglie wavelength of the atoms
$$\lambda_{\rm atom}=\dfrac{h}{mv_{\rm rms}}\tag 1$$
For diffraction grating, we know that
$$d\sin\theta_m=m\lambda_{\rm atom}$$
And for the first-order diffraction angle where $m=1$,
$$d\sin\theta_1= \lambda_{\rm atom}$$
Hence,
$$\theta_1=\sin^{-1}\left[\dfrac{\lambda_{\rm atom}}{d}\right]\tag 2$$
And since the laser beam works as a diffraction gratings, the antinodes in the standing wave of the laser beam represent the grating.
This means that $d=\frac{1}{2}\lambda_{\rm beam}$
Plug into (2);
$$\theta_1=\sin^{-1}\left[\dfrac{2\lambda_{\rm atom}}{\lambda_{\rm beam}}\right] $$
Plug from (1);
$$\theta_1=\sin^{-1}\left[\dfrac{2h}{\lambda_{\rm beam}mv_{\rm rms}}\right] $$
Plug the known;
$$\theta_1=\sin^{-1}\left[\dfrac{2(6.63\times 10^{-34})}{(590\times 10^{-9})(23\times 1.67\times 10^{-27})(1.04)}\right]$$
$$\theta=\color{red}{\bf 3.23}^\circ$$
$$\color{blue}{\bf [c]}$$
From the geometry of Figure 38.14, the distance between B and C is given by
$$\tan\theta_1=\dfrac{x}{L}$$
where $\overline{BC}=2x$, and $x=L\tan\theta_1$.
Thus,
$$\overline{BC}=2L\tan\theta_1$$
Plug the known;
$$\overline{BC}=2(10\times 10^{-2})\tan3.23^\circ= \color{red}{\bf 1.13} \;\rm cm$$
$$\color{blue}{\bf [d]}$$
When the atoms come together at point D, they create an interference pattern. This only happens if each atom somehow travels through both paths of the interferometer at the same time.
So, midway through, each atom is effectively in two places at once- both at point B and point C- even though these points are more than a centimeter apart. This shows us that, at the atomic level, atoms don’t behave like small, solid particles that are in just one place.
Instead, they behave more like waves, spreading out and existing in multiple places at the same time, something we usually only think of with waves.
In other words, in the quantum world, atoms are perfect waves.
