Question

Exercise 54 gives a rough lifetime for a particle trapped particle to escape an enclosure by tunneling.

(a) Consider an electron. Given that$\mathbf{W}\mathbf{=}\mathbf{100}\mathbf{\text{\hspace{0.17em}nm}}\mathbf{,}\mathbf{L}\mathbf{=}\mathbf{1}\mathbf{\text{\hspace{0.17em}nm\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}}{\mathbf{U}}_{\mathbf{0}}\mathbf{=}\mathbf{5}\mathbf{\text{\hspace{0.17em}eV}}$, first verify that the${\mathbf{E}}_{\mathbf{GS}}\mathbf{<}\mathbf{<}{\mathbf{U}}_{\mathbf{0}}$assumption holds, then evaluate the lifetime.

(b) Repeat part (a), but for a$\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\mu }\mathbf{g}$particle, with$\mathbf{W}\mathbf{=}\mathbf{1}\mathbf{nm}\mathbf{,}\mathbf{L}\mathbf{=}\mathbf{1}\mathbf{\mu }\mathbf{m}$, and a barrier height${\mathbf{U}}_{\mathbf{0}}$that equals the energy the particle would have if its speed were just$\mathbf{1}\mathbf{\text{\hspace{0.17em}mmperyear}}$.

Short answer

1. The assumption$E_{GS}<<U_0$ holds. The lifetime is $\mathrm{\tau }\simeq 33\min$.
2. The assumption $E_{GS}<<U_0$holds.

The relaxation time is very large ( $\cong {10}^{20}{\sigma }^2{e}^{\sigma }$) for such a huge classical particle.

Step-by-step solution

Concepts involved

Tunneling is a phenomena in which a particle is able to tunnel through a potential barrier when its kinetic energy

The ground state energy of infinite well is given by,

${\mathrm{E}}_{\mathrm{GS}}=\frac{{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{2{\mathrm{mL}}^2}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Where, $\mathrm{\hslash }$= Modified Plank’s Constant

$m$= mass of the particle

$L$ = Barrier width

Lifetime of the particle is given by,

$\mathrm{\tau }\cong \frac{{\mathrm{mW}}^4{\mathrm{\sigma }}^2}{2000\mathrm{\hslash }{\mathrm{L}}^2}{\mathrm{e}}^{\mathrm{\sigma }}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Where, =Wavelength/2

$\sigma =\frac{2L\sqrt{2m{U}_0}}{\hslash }\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)$

Step 2(a): Determine the lifetime of the particles

The infinite well ground state Energy from Equation (1):

$\begin{array}{c}{\mathrm{E}}_{\mathrm{GS}}=\frac{{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{2{\mathrm{mL}}^2}=\frac{{\mathrm{\pi }}^2{(1.055\times {10}^{-34}\text{\hspace{0.17em}J.s})}^2}{2(9.11\times {10}^{-31}\text{\hspace{0.17em}kg}){\left({10}^{-7}\text{\hspace{0.17em}m}\right)}^2}\\ =6\times {10}^{-24}\text{\hspace{0.17em}J}\\ {\mathrm{E}}_{\mathrm{GS}}=3.8\times {10}^{-5}\mathrm{eV}<<{\mathrm{U}}_0\end{array}$

Hence, the assumption holds, the value of $E_{GS}$is found to be much smaller than $U_0$.

Now, if you put the values in equation (3), and then in equation (2) you get,

$\begin{array}{c}\mathrm{\sigma }=\frac{\left({10}^{-9}\right)\sqrt{8(9.11\times {10}^{-31}\text{\hspace{0.17em}kg})(8\times {10}^{-19}\text{\hspace{0.17em}J})}}{(1.055\times {10}^{-34}\text{\hspace{0.17em}J.s})}\\ =22.9\\ \mathrm{\tau }=\frac{(9.11\times {10}^{-31}\text{\hspace{0.17em}kg}){\left({10}^{-7}\text{m}\right)}^4}{2000(1.055\times {10}^{-34}\text{J.s}){\left({10}^{-9}\text{\hspace{0.17em}m}\right)}^2}{\left(22.9\right)}^2{\mathrm{e}}^{22.9}\\ \cong 2000\text{\hspace{0.17em}s}\\ \cong \text{33min}\end{array}$

Hence, lifetime of the particle is 33 min.

Step 3(b): verify that the  EGS<<U0

Now, for new parameters:

From Equation (1), you get,

$\begin{array}{c}{\mathrm{E}}_1=\frac{{\mathrm{\pi }}^2(1.055\times {10}^{-34}\text{\hspace{0.17em}J.s})}{2\left({10}^{-10}\right){\left({10}^{-3}\right)}^2}\\ {\mathrm{E}}_1=5\times {10}^{-52}\text{J}\\ {\text{U}}_0=\frac{1}{2}\left({10}^{-10}\text{kg}\right)\left(\frac{{10}^{-2}\text{m/s}}{3.16\times {10}^7\text{s}}\right)\\ {\mathrm{U}}_0=5\times {10}^{-32}\text{J}\end{array}$

Now, if you put above obtained values in equation (3) and then in equation (2), you get,

$\begin{array}{c}\mathrm{\sigma }=\frac{\left({10}^{-6}\text{\hspace{0.17em}m}\right)\sqrt{8\left({10}^{-10}\text{\hspace{0.17em}kg}\right)(5\times {10}^{-32}\text{\hspace{0.17em}J})}}{(1.055\times {10}^{-34}\text{\hspace{0.17em}J.s})}=6\times {10}^7\\ \mathrm{\tau }=\frac{\left({10}^{-10}\text{kg}\right){\left({10}^{-3}\text{m}\right)}^4}{2000(1.05\times {10}^{-34}\text{J.s}){\left({10}^{-6}\text{m}\right)}^2}{\mathrm{\sigma }}^2{\mathrm{e}}^{\mathrm{\sigma }}\\ \cong {10}^{20}{\mathrm{\sigma }}^2{\mathrm{e}}^{\mathrm{\sigma }}\end{array}$

$\tau$would be very large for such a huge, classical particle.

The assumption of${\mathrm{E}}_{\mathrm{GS}}<<{\mathrm{U}}_0$holds good.

For such a huge classical particle, relaxation time would be very large.