91Ó°ÊÓ

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

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

Assuming the most accurate measurements possible, we can replace $(â‰\yen )$with $(=)$. Moreover, we know that $\mathrm{Δ}{p}_x=m\mathrm{Δ}{v}_x$. Thus

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

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

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

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

$=1.38×{10}^{-42}\mathrm{J}$

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 $1\mathrm{μ}\mathrm{m}$, and this amount of potential energy can be calculated as follows

$U=mgh$

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

$=9.8×{10}^{-22}\mathrm{J}$

$K=1.38×{10}^{-42}\mathrm{J}=mgh$

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

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

$=1.41×{10}^{-27}\mathrm{m}$