91影视

Dust mass $=1.0脳{10}^{-13}\mathrm{g}$

Width of the box $螖x=10渭\mathrm{m}$

$=\left(0渭\mathrm{m}\left(\frac{1\mathrm{m}}{{10}^6渭\mathrm{m}}\right)\right)$

The uncertainity principle is given by,

$螖xpx鈮�\frac{h}{2}$

Where, $螖\mathrm{x}$is measurement of position, $螖\mathrm{px}$is measurement of the momentum of the particle and $\mathrm{h}$is plank's constant.

$A\left[1=3.3脳{10}^{-23}\lg /5^{-1}\right]$

The energy and momenturm relation of particle is egven by

$螖T=\frac{{螖}_{n^1}}{=}$

AlH is uncertainity in measurem ent of energy of particle.

Applying values,

$螖\mathrm{H}-1.{65}^{+{10}^{-3}}\mathrm{J}$

The height of potertial barrier is $g$ven by,

$V=mnh$

Where v is height af potental barrier, th is mass of particle, $\mathrm{B}$ts acceleratien due to graify

Apply ing values,

localid="1650898379149" $$\begin{gathered} V=\left(1.0脳{10}^{-13}\mathrm{R}\right)\left(9{\mathrm{xms}}^{-3}\right)\left(1\mathrm{ma}\right) \\ =\left(1.0脳{10}^{-13}\mathrm{e}\right)\left(9{\mathrm{Nms}}^{-2}\right)\left(1渭\mathrm{m}\right)\left(\frac{1\mathrm{m}}{1\mathrm{m}}\right) \\ \left.\mathrm{V}=0脳8脳{10}^{-2\mathrm{I}}\right) \end{gathered}$$

Hence, localid="1650898382670" $AEcV$the oarticle all not have enough energy to get out the hole

To find the masimum height, ursig the formula.

localid="1650898390302" $\mathrm{m}t\mathrm{H}=\mathrm{A}$EI localid="1650898386256" $\mathrm{AI}\mathrm{shend}\mathrm{be}\mathrm{greater}\mathrm{or}\mathrm{肠啷嵿く别补濒}$

localid="1650898394321" $鈥�=\frac{A\mathrm{U}}{蟺t}$

Apphing values,

localid="1650898410368" $鈥�=0.17脳{10}^{-23}\mathrm{m}$

Concluiton: The value of localid="1650898415592" $鈭�\mathrm{E}$less than the value of localid="1650898452767" $V_2$so the particle aill have enough energy to get eut the hole. The deepest hole is localid="1650898424914" $0.17脳{10}^{-2}$"a for which, particle have a good chance to escape from the hole.