7.6 The quantum tunneling of particles through potential barriers  (Page 5/22)

 Page 5 / 22
${\beta }_{1}=\sqrt{26.254\left(10.00\phantom{\rule{0.2em}{0ex}}\text{eV}-{E}_{1}\right)\text{/}\text{eV}}\frac{1}{\text{nm}}=\sqrt{26.254\left(10.00-7.00\right)}\frac{1}{\text{nm}}=\frac{8.875}{\text{nm}},$
$T\left(L,{E}_{1}\right)=16\frac{{E}_{1}}{{U}_{0}}\left(1-\frac{{E}_{1}}{{U}_{0}}\right){e}^{\text{−}2{\beta }_{1}L}=16\frac{7}{10}\left(1-\frac{7}{10}\right){e}^{\text{−}17.75\phantom{\rule{0.2em}{0ex}}L\text{/}\text{nm}}=3.36{e}^{\text{−}17.75\phantom{\rule{0.2em}{0ex}}L\text{/}\text{nm}}.$

For a higher-energy electron with ${E}_{2}=9.00\phantom{\rule{0.2em}{0ex}}\text{eV}$ :

${\beta }_{2}=\sqrt{26.254\left(10.00\phantom{\rule{0.2em}{0ex}}\text{eV}-{E}_{2}\right)\text{/}\text{eV}}\frac{1}{\text{nm}}=\sqrt{26.254\left(10.00-9.00\right)}\frac{1}{\text{nm}}=\frac{5.124}{\text{nm}},$
$T\left(L,{E}_{2}\right)=16\frac{{E}_{2}}{{U}_{0}}\left(1-\frac{{E}_{2}}{{U}_{0}}\right){e}^{\text{−}2{\beta }_{2}\phantom{\rule{0.2em}{0ex}}L}=16\frac{9}{10}\left(1-\frac{9}{10}\right){e}^{\text{−}5.12\phantom{\rule{0.2em}{0ex}}L\text{/}\text{nm}}=1.44{e}^{\text{−}5.12\phantom{\rule{0.2em}{0ex}}L\text{/}\text{nm}}.$

For a broad barrier with ${L}_{1}=5.00\phantom{\rule{0.2em}{0ex}}\text{nm}$ :

$T\left({L}_{1},{E}_{1}\right)=3.36{e}^{\text{−}17.75\phantom{\rule{0.2em}{0ex}}{L}_{1}\text{/}\text{nm}}=3.36{e}^{\text{−}17.75·5.00\phantom{\rule{0.2em}{0ex}}\text{nm}\text{/}\text{nm}}=3.36{e}^{-88}=3.36\left(6.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-39}\right)=2.1\text{%}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-36},$
$T\left({L}_{1},{E}_{2}\right)=1.44{e}^{\text{−}5.12\phantom{\rule{0.2em}{0ex}}{L}_{1}\text{/}\text{nm}}=1.44{e}^{\text{−}5.12·5.00\phantom{\rule{0.2em}{0ex}}\text{nm}\text{/}\text{nm}}=1.44{e}^{\text{−}25.6}=1.44\left(7.62\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-12}\right)=1.1\text{%}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}.$

For a narrower barrier with ${L}_{2}=1.00\phantom{\rule{0.2em}{0ex}}\text{nm}$ :

$T\left({L}_{2},{E}_{1}\right)=3.36{e}^{\text{−}17.75\phantom{\rule{0.2em}{0ex}}{L}_{2}\text{/}\text{nm}}=3.36{e}^{\text{−}17.75·1.00\phantom{\rule{0.2em}{0ex}}\text{nm}\text{/}\text{nm}}=3.36{e}^{-17.75}=3.36\left(5.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\right)=1.7\text{%}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4},$
$T\left({L}_{2},{E}_{2}\right)=1.44{e}^{\text{−}5.12\phantom{\rule{0.2em}{0ex}}{L}_{2}\text{/}\text{nm}}=1.44{e}^{\text{−}5.12·1.00\phantom{\rule{0.2em}{0ex}}\text{nm}\text{/}\text{nm}}=1.44{e}^{\text{−}5.12}=1.44\left(5.98\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\right)=0.86\text{%}.$

Significance

We see from these estimates that the probability of tunneling is affected more by the width of the potential barrier than by the energy of an incident particle. In today’s technologies, we can manipulate individual atoms on metal surfaces to create potential barriers that are fractions of a nanometer, giving rise to measurable tunneling currents. One of many applications of this technology is the scanning tunneling microscope (STM), which we discuss later in this section.

Check Your Understanding A proton with kinetic energy 1.00 eV is incident on a square potential barrier with height 10.00 eV. If the proton is to have the same transmission probability as an electron of the same energy, what must the width of the barrier be relative to the barrier width encountered by an electron?

${L}_{\text{proton}}\text{/}{L}_{\text{electron}}=\sqrt{{m}_{e}\text{/}{m}_{p}}=2.3\text{%}$

In 1928, Gamow identified quantum tunneling as the mechanism responsible for the radioactive decay of atomic nuclei. He observed that some isotopes of thorium, uranium, and bismuth disintegrate by emitting $\text{α}$ -particles (which are doubly ionized helium atoms or, simply speaking, helium nuclei). In the process of emitting an $\text{α}$ -particle, the original nucleus is transformed into a new nucleus that has two fewer neutrons and two fewer protons than the original nucleus. The $\text{α}$ -particles emitted by one isotope have approximately the same kinetic energies. When we look at variations of these energies among isotopes of various elements, the lowest kinetic energy is about 4 MeV and the highest is about 9 MeV, so these energies are of the same order of magnitude. This is about where the similarities between various isotopes end.

When we inspect half-lives (a half-life is the time in which a radioactive sample loses half of its nuclei due to decay), different isotopes differ widely. For example, the half-life of polonium-214 is $160\phantom{\rule{0.2em}{0ex}}µ\text{s}$ and the half-life of uranium is 4.5 billion years. Gamow explained this variation by considering a ‘spherical-box’ model of the nucleus, where $\text{α}$ -particles can bounce back and forth between the walls as free particles. The confinement is provided by a strong nuclear potential at a spherical wall of the box. The thickness of this wall, however, is not infinite but finite, so in principle, a nuclear particle has a chance to escape this nuclear confinement. On the inside wall of the confining barrier is a high nuclear potential that keeps the $\text{α}$ -particle in a small confinement. But when an $\text{α}$ -particle gets out to the other side of this wall, it is subject to electrostatic Coulomb repulsion and moves away from the nucleus. This idea is illustrated in [link] . The width L of the potential barrier that separates an $\text{α}$ -particle from the outside world depends on the particle’s kinetic energy E . This width is the distance between the point marked by the nuclear radius R and the point ${R}_{0}$ where an $\text{α}$ -particle emerges on the other side of the barrier, $L={R}_{0}-R$ . At the distance ${R}_{0}$ , its kinetic energy must at least match the electrostatic energy of repulsion, $E={\left(4\pi {\epsilon }_{0}\right)}^{-1}Z{e}^{2}\text{/}{R}_{0}$ (where $+Ze$ is the charge of the nucleus). In this way we can estimate the width of the nuclear barrier,

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