91影视

Potential well is a region surrounding a local minimum of potential energy. The energy inside that region is trapped that鈥檚 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,

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

Where, $v$= velocity of the particle,

$p$= momentum of the particle

$m$ = mass of the particle

$魔$= modified Plank鈥檚 constant

$k$= wave number

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

$\begin{array}{l}\mathrm{v}=\frac{\mathrm{鈩�}\mathrm{蟺}}{\mathrm{mW}}(\mathrm{位}=2\mathrm{W})\\ \mathrm{t}=\frac{{\mathrm{mW}}^2}{\mathrm{鈩�}\mathrm{蟺}}\end{array}$

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, $\mathrm{E}=\frac{1^2{\mathrm{鈩�}}^2{\mathrm{蟺}}^2}{2{\mathrm{mW}}^2}鈰�鈰�鈰�鈰�鈰�鈰�\left(1\right)$

If$\mathrm{E}<<{\mathrm{U}}_0$,Transmittance, $\mathrm{T}鈮�16\frac{\mathrm{E}}{{\mathrm{U}}_0}{\mathrm{e}}^{鈭�2\mathrm{L}\frac{\sqrt{2{\mathrm{mU}}_0}}{\mathrm{鈩�}}}鈰�鈰�鈰�鈰�鈰�鈰�\left(2\right)$

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

$\begin{array}{l}\mathrm{E}=\frac{{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{2{\mathrm{mW}}^2}\\ \mathrm{T}=\frac{8{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{{\mathrm{mW}}^2{\mathrm{U}}_0}{\mathrm{e}}^{鈭�2\mathrm{L}\frac{\sqrt{2{\mathrm{mU}}_0}}{\mathrm{鈩�}}}\end{array}$

Therefore, lifetime is:$\frac{{\mathrm{mW}}^2}{\mathrm{鈩�}\mathrm{蟺}}\frac{{\mathrm{mW}}^2{\mathrm{U}}_0}{8{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{\mathrm{e}}^{鈭�2\mathrm{L}\frac{\sqrt{2{\mathrm{mU}}_0}}{\mathrm{鈩�}}}=\frac{{\mathrm{mW}}^4}{\mathrm{鈩�}\mathrm{蟺}}\frac{{\mathrm{U}}_0}{64{\mathrm{蟺}}^2{\mathrm{L}}^2}{\mathrm{e}}^{\mathrm{蟽}}$

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

$\mathrm{蟿}=\frac{{\mathrm{mW}}^4}{2000{\mathrm{hL}}^2}{\mathrm{蟽}}^2{\mathrm{e}}^{\mathrm{伪}}$