91Ó°ÊÓ

Tunnelling is a phenomenon when a particle propagates through a potential barrier when the potential energy of the barrier is higher than the kinetic energy of the particle.

Tunneling Probability is the ratio of squared amplitudes of the waves after crossing the barrier to the incident waves.

The potential energy is modelled as,

$U\left(x\right)-E=\mathrm{Ï•}-Mx$

If the particle enters where x = 0 and exits where E = U, or $x=\mathrm{Ï•}/M$.

Hence, by method 1, tunnelling probability is given by

role="math" localid="1660047829428" $\begin{array}{rcl}{T}_1& â‰\dots & e^{\frac{-2}{\sout{h}}\left\{{∫}_0^{\mathrm{Ï•}/M}\sqrt{2m\left(\mathrm{Ï•}/Mx\right)}dx\right\}}\\ & =& e^{\frac{-2}{\sout{h}}\left\{{\overline{)\frac{{\left(2m\left(\mathrm{Ï•}/Mx\right)\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3Mm}}}_0^{\mathrm{Ï•}/M}\right\}}\\ & =& e^{\frac{-2}{\sout{h}}\left\{\frac{\sqrt{8m{\mathrm{Ï•}}^3}}{3M}\right\}}\end{array}$

Also, by method 2, tunnelling probability is given by

$\begin{array}{rcl}{T}_2& =& e^{\frac{-2}{\sout{h}}\left\{\frac{\sqrt{2m{\mathrm{Ï•}}^3}}{M}\right\}}\\ & =& e^{\left\{\frac{\sqrt{8m{\mathrm{Ï•}}^3}}{\sout{h}M}\right\}}\end{array}$

Now the required ratio, T₁/T₂ can be calculated as,

$\begin{array}{rcl}{T}_1/T_2& =& e^{\left\{\frac{-2}{\sout{h}}\frac{\sqrt{8m{\mathrm{Ï•}}^3}}{3M}+\frac{\sqrt{8m{\mathrm{Ï•}}^3}}{\sout{h}M}\right\}}\\ & =& e^{\frac{\sqrt{8m{\mathrm{Ï•}}^3}}{3M}}\end{array}$

Hence, the ratio T₁/T₂ is obtained as $\begin{array}{rcl}& & e^{\frac{\sqrt{8m{\mathrm{Ï•}}^3}}{3M}}\end{array}$.

As it can be clearly seen from the ratio obtained in the previous step, the ratio will have the largest value, i.e., the probabilities differ the most when the argument of the exponential is large or when the tunnelling probability is small.