Consider a particle of mass m inside the well as shown in the figure. If bound, its lowest energy state would of course be the ground state, but would it be bound? Assume that for a while, it at least occupies the ground state, which is much lower than${\mathbf{U}}_0$, and the barriers qualify as wide. Show that a rough average time it would remain bound is given by:$\mathbf{\tau }\mathbf{=}\frac{{\mathbf{mW}}^{\mathbf{4}}}{\mathbf{2000}{\mathbf{hL}}^{\mathbf{2}}}{\mathbf{\sigma }}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{\alpha }}$ where$\mathbf{\sigma }\mathbf{=}\mathbf{L}\frac{\sqrt{\mathbf{8}{\mathbf{mU}}_{\mathbf{0}}}}{\mathbf{h}}$.

Potential well is a region surrounding a local minimum of potential energy. The energy inside that region is trapped that’s why it is unable to convert to another form of energy.

The quantum particle trapped has finite quantized energy levels and hence it is capable oftunnelingthrough the energy barrier.

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

You know that,

$\mathrm{v}=\mathrm{p}/\mathrm{m}=\mathrm{ħk}/\mathrm{m}$

Where, $v$= velocity of the particle,

$p$= momentum of the particle

$m$ = mass of the particle

$ħ$= modified Plank’s constant

$k$= wave number

If you put value of $k$in the above obtained expression, it will be

$\begin{array}{l}\mathrm{v}=\frac{\mathrm{\hslash }\mathrm{\pi }}{\mathrm{mW}}(\mathrm{\lambda }=2\mathrm{W})\\ \mathrm{t}=\frac{{\mathrm{mW}}^2}{\mathrm{\hslash }\mathrm{\pi }}\end{array}$

If time,$t=\text{distance/speed}$

The time it would last would be the time obtained in the previous step divided by the tunneling probability.

You know that, Total energy, $\mathrm{E}=\frac{1^2{\mathrm{\hslash }}^2{\mathrm{\pi }}^2}{2{\mathrm{mW}}^2}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

If$\mathrm{E}<<{\mathrm{U}}_0$,Transmittance, $\mathrm{T}\cong 16\frac{\mathrm{E}}{{\mathrm{U}}_0}{\mathrm{e}}^{-2\mathrm{L}\frac{\sqrt{2{\mathrm{mU}}_0}}{\mathrm{\hslash }}}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Now, if you use equation (1) in equation (2), you get,

$\begin{array}{l}\mathrm{E}=\frac{{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{2{\mathrm{mW}}^2}\\ \mathrm{T}=\frac{8{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{{\mathrm{mW}}^2{\mathrm{U}}_0}{\mathrm{e}}^{-2\mathrm{L}\frac{\sqrt{2{\mathrm{mU}}_0}}{\mathrm{\hslash }}}\end{array}$

Therefore, lifetime is:$\frac{{\mathrm{mW}}^2}{\mathrm{\hslash }\mathrm{\pi }}\frac{{\mathrm{mW}}^2{\mathrm{U}}_0}{8{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{\mathrm{e}}^{-2\mathrm{L}\frac{\sqrt{2{\mathrm{mU}}_0}}{\mathrm{\hslash }}}=\frac{{\mathrm{mW}}^4}{\mathrm{\hslash }\mathrm{\pi }}\frac{{\mathrm{U}}_0}{64{\mathrm{\pi }}^2{\mathrm{L}}^2}{\mathrm{e}}^{\mathrm{\sigma }}$

Hence, We can also rewrite the above equation after simplifying it as,

$\mathrm{\tau }=\frac{{\mathrm{mW}}^4}{2000{\mathrm{hL}}^2}{\mathrm{\sigma }}^2{\mathrm{e}}^{\mathrm{\alpha }}$