Soot particles, from incomplete combustion in diesel engines, are typically $15\mathrm{nm}$in diameter and have a density of $1200\mathrm{kg}/{\mathrm{m}}^3$. FIGURE P39.45 shows soot particles released from rest, in vacuum, just above a thin plate with a $0.50-\mu \mathrm{m}$-diameter holeroughly the wavelength of visible light. After passing through the hole, the particles fall distance d and land on a detector. If soot particles were purely classical, they would fall straight down and, ideally, all land in a $0.50-\mu \mathrm{m}$-diameter circle. Allowing for some experimental imperfections, any quantum effects would be noticeable if the circle diameter were $2000\mathrm{nm}$. How far would the particles have to fall to fill a circle of this diameter?

Step 2 Calculation

volume of each particle is

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

The mass is then density times volume

$m=\rho V=\frac{1}{6}\rho a^3\pi$

Returning this into the expression for d we get

$d=\frac{g{w}^2{D}^2{\rho }^2{a}^6{\pi }^2}{72h^2}$

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

$\Delta x\Delta p_x\ge \frac{h}{2}$

$\Delta p_x\ge \frac{h}{2\Delta x}=\frac{h}{2D}$

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

$p_x=0\pm \Delta p_x=\pm \frac{h}{2D}$

Notice that we have taken the smallest possible uncertainty i.e. we have set $\Delta p_x=\frac{h}{2D}$, 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

$v_x=\pm \frac{h}{2Dm}$

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

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

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