Question

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?

Short answer

Hence, the particle would have to fall$0.34\times {10}^{-15}\mathrm{m}$to fill a circle of diameter of$2000\mathrm{nm}\text{.}$

Step-by-step solution

 Step 1: Given Informaiton

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

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

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

Diameter of thin plate$=\mathbf{0}.50\mu \mathrm{m}$

Diameter of the circle=2000 nm

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

 Step 2: solution

Formula to be used:

The uncertainity principle is given by

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

Where, $\Delta \mathrm{x}$is measurement of position, $∆\mathrm{px}$is measurement of the momentum of the particle and $\mathrm{h}$is planks constant.

Calculation:

The mass of the particle is given by

$\mathrm{m}=\ell \left(\frac{1}{6}\pi D^3\right)$[considering the shape of particle is sphere]

Where $\mathrm{m}$is mass, $\ell$is density and $\mathrm{D}$is diameter

Applying values,

$$\begin{gathered} m=1200{\mathrm{kgm}}^3\left[\frac{1}{6}\pi {\left(15\mathrm{nm}\left(\frac{1\mathrm{m}}{{10}^9\mathrm{nm}}\right)\right)}^3\right. \\ \mathrm{m}=2.1\times {10}^{-21}\mathrm{kg} \end{gathered}$$

The minimum uncertainity momentum is given by,

$\Delta \mathrm{px}=\frac{\mathrm{h}}{2\Delta \mathrm{x}}$

Applying values, $∆px=\frac{6.6\times {10}^{-7/s}}{2(2000\mathrm{nm})\left(\frac{1\mathrm{m}}{{10}^9\mathrm{mm}}\right)}$

$\Delta \mathrm{px}=1.65\times {10}^{-2\mathrm{s}}\mathrm{kgm}/\mathrm{s}$

The momentum and energy relation is given by

$\Delta \mathit{E}=\frac{\Delta {\mathrm{px}}^2}{2\mathrm{m}}$[where $\Delta \mathrm{E}$is measurement of energy of particle]

Applying values,

$$\begin{gathered} \Delta E=\frac{1.65\times {10}^{-2\mathrm{kgm}/\mathrm{s}}}{2\left(2.1\times {10}^{-21}\mathrm{kg}\right)} \\ \Delta E=6.5\times {10}^{-36\mathrm{J}} \end{gathered}$$

The potential energy of the falling particle is given by,

$\mathrm{V}=\mathrm{mgd}$

Where v is height of potential barrier, m is mass, g is acceleration due to gravity and d is distance of falling particle.

The relation between potential energy and uncertainty energy is given by $\mathrm{mgd}=\Delta \mathrm{E}$[P.E of particle should be more or equal to $∆\mathrm{E}$]

$\mathrm{d}=\frac{\Delta \mathrm{E}}{\mathrm{mg}}$

Applying values,

$$\begin{gathered} d=\frac{6.5\times {10}^{-36}\mathrm{J}}{\left(2.1\times {10}^{-2}\mathrm{kg}\right)(9.8\mathrm{m}/\mathrm{s})} \\ d=H=0.34\times {10}^{-15}\mathrm{m} \end{gathered}$$

Conclusion:Hence, the particle would have to fall $0.34\times {10}^{-15}\mathrm{m}$to fill a circle of diameter of $2000\mathrm{nm}$.