Show that if you attempt to detect a particle while tunneling, your experiment must render its kinetic energy so uncertain that it might well be "over the top."

Tunnelingis a phenomenon when particles penetrate through a potential barrier which is higher than their kinetic energy.

According to the uncertainty principle, it’s impossible to calculate the momentum of a particle together with its position simultaneously with absolute accuracy.

$\mathrm{\Delta x}.\mathrm{\Delta p}\ge \mathrm{\hslash }/2$

Where, $\mathrm{\Delta x}$= Change in position of the particle

$\mathrm{\Delta p}$ = Change in momentum of the particle

$\hslash$ = modified form of plank’s constant

From the above equation, you get that, uncertainty in $x$ and $P$are inversely proportional.

If the particle lies in the barrier, the uncertainty in $x$needs to be greater than the wave number- $\delta$ such that $∆p$can be no less than $\hslash /\delta$.

Kinetic energy is given by ( $p^2/2m$), it needs to be greater than or equal to$\frac{{\mathrm{\hslash }}^2}{2{\mathrm{m\delta }}^2}=\frac{{\mathrm{\hslash }}^2}{2\mathrm{m}}\frac{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}{{\mathrm{\hslash }}^2}={\mathrm{U}}_0-\mathrm{E}$ .

Thus, the order of magnitude of uncertainty in the Kinetic Energy obtained by the experiment, makes up for difference between the potential energy barrier - $U_0$and the total energy - $E$.