Question

A small speck of dust with mass $1.0\times {10}^{-13}\mathrm{g}$has fallen into the hole shown in FIGURE P39.46 and appears to be at rest. According to the uncertainty principle, could this particle have enough energy to get out of the hole? If not, what is the deepest hole of this width from which it would have a good chance to escape?

Short answer

The value of $\Delta \mathrm{E}$less than the value of $\mathrm{V}$, so the particle will have enough energy to get out the hole. The deepest hole is $0.17\times {10}^{-28}\mathrm{m}$for which, particle have a good chance to escape from the hole..

Step-by-step solution

Given Information

Dust mass $=1.0\times {10}^{-13}\mathrm{g}$

Width of the box $\Delta x=10\mu \mathrm{m}$

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

The uncertainity principle is given by,

$\Delta xpx\ge \frac{h}{2}$

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

Step 2:Solution

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

The energy and momenturm relation of particle is egven by

$\Delta T=\frac{{\Delta }_{n^1}}{=}$

AlH is uncertainity in measurem ent of energy of particle.

Applying values,

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

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

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" $\mathrm{AI}\mathrm{shend}\mathrm{be}\mathrm{greater}\mathrm{or}\mathrm{c्यeal}$

localid="1650898394321" $\Vert =\frac{A\mathrm{U}}{\pi t}$

Apphing values,

localid="1650898410368" $\Vert =0.17\times {10}^{-23}\mathrm{m}$

Concluiton: The value of localid="1650898415592" $∆\mathrm{E}$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" $0.17\times {10}^{-2}$"a for which, particle have a good chance to escape from the hole.