91Ó°ÊÓ

Step 2 Calculation

volume of each particle is

$V=\frac{4}{3}{\left(\frac{a}{2}\right)}^3\mathrm{Ï€}=\frac{1}{6}{a}^3\mathrm{Ï€}.$

The mass is then density times volume

$m=\mathrm{ÒÏ}V=\frac{1}{6}\mathrm{ÒÏ}{a}^3\mathrm{Ï€}$

Returning this into the expression for d we get

$d=\frac{g{w}^2{D}^2{\mathrm{ÒÏ}}^2{a}^6{\mathrm{Ï€}}^2}{72h^2}$

Plugging in $w=2000\mathrm{nm}$and the rest of the given numerical values we find

The particles would have to fall through localid="1651038700218" $d=50\mathrm{m}$to notice quantum effects.

Therefore, from Heisenberg's uncertainty principle, we have that

$\mathrm{Δ}x\mathrm{Δ}{p}_xâ‰\yen \frac{h}{2}$

$\mathrm{Δ}{p}_xâ‰\yen \frac{h}{2\mathrm{Δ}x}=\frac{h}{2D}$

In the classical case, the particles wouldn't have any momentum in x direction and they would fall straight down forming a circle of diameter D. However, due to the quantum uncertainty, our particles have the momentum of

$p_x=0Â\pm \mathrm{Δ}{p}_x=Â\pm \frac{h}{2D}$

Notice that we have taken the smallest possible uncertainty i.e. we have set $\mathrm{Δ}{p}_x=\frac{h}{2D}$, because this would require bigger d minimum and we want to find the distance d that would always do the job, even for the smallest uncertainty in x direction.This yields for the speed in x direction

$v_x=Â\pm \frac{h}{2Dm}$

where m is the mass of the particle. The time necessary for the particle to fall down is

$t=\sqrt{\frac{2d}{g}}$

Therefore, the beam would diverge in x direction by $v_{x}t$towards each side, i.e. it would form the circle of diameter

$w=2v_{x}t=\frac{h}{Dm}\sqrt{\frac{2d}{g}}$

$d=\frac{g{w}^2{D}^2{m}^2}{2h^2}$