91Ó°ÊÓ

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$=\mathbf{0}.50\mathrm{μ}\mathrm{m}$

Diameter of the circle=2000 nm

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

Formula to be used:

The uncertainity principle is given by

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

Where, $\mathrm{Δ}\mathrm{x}$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

$\mathrm{m}=\mathrm{â„“}\left(\frac{1}{6}\mathrm{Ï€}{D}^3\right)$[considering the shape of particle is sphere]

Where $\mathrm{m}$is mass, $\mathrm{â„“}$is density and $\mathrm{D}$is diameter

Applying values,

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

The minimum uncertainity momentum is given by,

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

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

$\mathrm{Δ}\mathrm{px}=1.65×{10}^{-2\mathrm{s}}\mathrm{kgm}/\mathrm{s}$

The momentum and energy relation is given by

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

Applying values,

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

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 $\mathrm{mgd}=\mathrm{Δ}\mathrm{E}$[P.E of particle should be more or equal to $∆\mathrm{E}$]

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

Applying values,

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

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