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?

FIGURE P39.46

The uncertainty in knowing the position of the dust speck inside a hole of a width of $\left(10\mu \mathrm{m}\right)$is $(\Delta x=10\mu \mathrm{m})$. Now, the uncertainty in the velocity of the particle can be calculated using the Heisenberg uncertainty principle

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

Assuming the most accurate measurements possible, we can replace $(\ge )$with $(=)$. Moreover, we know that $\Delta p_x=m\Delta v_x$. Thus

$\Delta v_x=\frac{h}{2m\Delta x}=\frac{6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}}{2\left(1\times {10}^{-16}\mathrm{kg}\right)\left(10\times {10}^{-6}\mathrm{m}\right)}=3.31\times {10}^{-13}\mathrm{m}/\mathrm{s}$

which means that the range of possible velocities for the dust speck is from $\left(-\Delta v_x/2=-1.66\times {10}^{-13}\mathrm{m}/\mathrm{s}\right)$to $\left(\Delta v_x/2=1.66\times {10}^{-13}\mathrm{m}/\mathrm{s}\right)$. In terms of speed, the maximum possible speed for the dust speck is $1.66\times {10}^{-13}\mathrm{m}/\mathrm{s}$. Thus, the maximum kinetic energy of the speck is

$K=\frac{1}{2}m{v}^2$

$=\frac{1}{2}\left(1\times {10}^{-16}\mathrm{kg}\right){\left(1.66\times {10}^{-13}\mathrm{m}/\mathrm{s}\right)}^2$

$=1.38\times {10}^{-42}\mathrm{J}$

Now, in order for the speck to get out of the hole, it must gain potential energy that allows it to reach a height of $1\mu \mathrm{m}$, and this amount of potential energy can be calculated as follows

$U=mgh$

$=\left(1\times {10}^{-16}\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1\times {10}^{-6}\mathrm{m}\right)$

$=9.8\times {10}^{-22}\mathrm{J}$

$K=1.38\times {10}^{-42}\mathrm{J}=mgh$

$h=\frac{K}{mg}$

$=\frac{1.38\times {10}^{-42}\mathrm{J}}{\left(1\times {10}^{-16}\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}$

$=1.41\times {10}^{-27}\mathrm{m}$