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Chapter 7: Problem 70

A 6.0-eV electron impacts on a barrier with height \(11.0 \mathrm{eV} .\) Find the probability of the electron to tunnel through the barrier if the barrier width is (a) \(0.80 \mathrm{nm}\) and (b) \(0.40 \mathrm{nm}\)

Short Answer

Expert verified
The probability of the electron tunneling through the barrier is approximately \(3.90 \times 10^{-6}\) when the barrier width is 0.80 nm and approximately \(2.58 \times 10^{-3}\) when the barrier width is 0.40 nm.

Step by step solution

01

Calculate the wave vector inside the barrier

First, we must determine the wave vector inside the barrier, using the formula: \[k_2 = \sqrt{\frac{2m(V - E)}{\hbar^2}}\] where m is the mass of the electron, which is approximately \(9.11 \times 10^{-31} kg\), and \(\hbar\), the reduced Planck constant, is approximately \(1.055 \times 10^{-34} J鈰卻\). We are given E and V, so we can plug in the values: \[k_2 = \sqrt{\frac{2(9.11 \times 10^{-31} kg)(11.0 eV - 6.0 eV)}{(1.055 \times 10^{-34} J鈰卻)^2}}\] Note that we must convert eV to J in order to use the formula correctly: \[1 eV = 1.602 \times 10^{-19} J\] Therefore, \[k_2 = \sqrt{\frac{2(9.11 \times 10^{-31} kg)(5.0 \times 1.602 \times 10^{-19} J)}{(1.055 \times 10^{-34} J鈰卻)^2}} = 1.241 \times 10^{10} m^{-1}\] Now we have the wave vector, k_2.
02

Calculate the transmission probability for d = 0.80 nm

Next, we calculate the transmission probability using the given formula when the barrier width, d, is 0.80 nm: \[T(E) = \frac{1}{1 + \frac{V^2 \sinh^2(k_2 d)}{4E(V - E)}}\] Plugging in the values and converting d to meters: \[T(E) = \frac{1}{1 + \frac{(11.0 \times 1.602 \times 10^{-19} J)^2 \sinh^2(1.241 \times 10^{10} m^{-1} \times 0.8 \times 10^{-9} m)}{4(6.0 \times 1.602 \times 10^{-19} J)(5.0 \times 1.602 \times 10^{-19} J)}}\] \[T(E) 鈮� 3.90 \times 10^{-6}\] Thus, the probability of the electron tunneling through the barrier with a width of 0.80 nm is approximately \(3.90 \times 10^{-6}\).
03

Calculate the transmission probability for d = 0.40 nm

Finally, we calculate the transmission probability when the barrier width, d, is 0.40 nm: \[T(E) = \frac{1}{1 + \frac{V^2 \sinh^2(k_2 d)}{4E(V - E)}}\] Plugging in the values and converting d to meters: \[T(E) = \frac{1}{1 + \frac{(11.0 \times 1.602 \times 10^{-19} J)^2 \sinh^2(1.241 \times 10^{10} m^{-1} \times 0.4 \times 10^{-9} m)}{4(6.0 \times 1.602 \times 10^{-19} J)(5.0 \times 1.602 \times 10^{-19} J)}}\] \[T(E) 鈮� 2.58 \times 10^{-3}\] Thus, the probability of the electron tunneling through the barrier with a width of 0.40 nm is approximately \(2.58 \times 10^{-3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quantum Mechanics
Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. It challenges classical mechanics with its concepts of quantization of energy, wave-particle duality, and the uncertainty principle.

At the heart of quantum mechanics is the idea that particles don't have exact positions or velocities but rather probabilities of being in various places and states. This probabilistic nature extends to phenomena such as electron tunneling, which we see in the exercise where an electron has a chance to cross a barrier that would be unsurmountable according to classical physics.
Wave-Particle Duality
Wave-particle duality is a core concept in quantum mechanics, describing how every particle or quantic entity exhibits both particle and wave-like properties. For instance, electrons can be diffracted like light waves, and photons can collide like billiard balls.

Evidence of this duality is observed in experiments like the double-slit experiment. This duality is crucial in understanding how an electron behaves when encountering a potential barrier. It doesn't just 'hit' the barrier; instead, its wave-like properties allow it to have a probability of 'tunneling' through.
Transmission Probability
Transmission probability in the context of quantum tunneling refers to the likelihood that a particle will penetrate and pass through a potential barrier. The calculation is not straightforward, as it incorporates the properties of the barrier and the particle's wave function.

In our case, the transmission probability changes significantly as the width of the barrier changes鈥攈ighlighted by the calculations for a 0.80 nm and 0.40 nm barrier where the probabilities differ greatly. This illustrates how even small changes in a quantum system can have large effects on the outcome.
Electron Tunneling
Electron tunneling is a quantum mechanical phenomenon where a particle moves through a barrier that it classically shouldn't be able to pass. This behavior happens because of the wave-like nature of electrons that allows them to penetrate potential barriers to a certain extent.

In the textbook exercise, we see this concept at work when calculating the probability for an electron to 'magically' appear on the other side of an energy barrier. While traditional physics would say the electron should bounce back, quantum mechanics allows for the chance, albeit small, that the electron will tunnel through the seemingly impassable barrier.

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Most popular questions from this chapter

Suppose an electron is confined to a region of length \(0.1 \mathrm{nm}\) (of the order of the size of a hydrogen atom) and its kinetic energy is equal to the ground state energy of the hydrogen atom in Bohr's model (13.6 eV). (a) What is the minimum uncertainty of its momentum? What fraction of its momentum is it? (b) What would the uncertainty in kinetic energy of this electron be if its momentum were equal to your answer in part (a)? What fraction of its kinetic energy is it?

Consider an infinite square well with wall boundaries \(\begin{array}{llllll}x=0 & \text { and } & x=L & \text { Show } & \text { that } & \text { the function }\end{array}\) \(\psi(x)=A \sin k x \quad\) is the solution to the stationary Schr枚dinger equation for the particle in a box only if \(k=\sqrt{2 m E} / \hbar .\) Explain why this is an acceptable wave function only if \(k\) is an integer multiple of \(\pi / L\)

What is the difference between a wave function \(\psi(x, y, z)\) and a wave function \(\Psi(x, y, z, t)\) for the same particle?

A wave function of a particle with mass \(m\) is given by $$\psi(x)=\left\\{\begin{array}{cl} A \cos \alpha x, & -\frac{\pi}{2 \alpha} \leq x \leq+\frac{\pi}{2 \alpha} \\ 0, & \text { otherwise } \end{array}\right.$$ where \(\alpha=1.00 \times 10^{10} / \mathrm{m} .\) (a) Find the normalization constant. (b) Find the probability that the particle can be found on the interval \(0 \leq x \leq 0.5 \times 10^{-10} \mathrm{m}\). (c) Find the particle's average position. (d) Find its average momentum. (e) Find its average kinetic energy \(-0.5 \times 10^{-10} \mathrm{m} \leq x \leq+0.5 \times 10^{-10} \mathrm{m}\)

In STM, an elevation of the tip above the surface being scanned can be determined with a great precision, because the tunneling-electron current between surface atoms and the atoms of the tip is extremely sensitive to the variation of the separation gap between them from point to point along the surface. Assuming that the tunneling-electron current is in direct proportion to the tunneling probability and that the tunneling probability is to a good approximation expressed by the exponential function \(e^{-2 \beta L}\) with \(\beta=10.0 / \mathrm{nm}\) determine the ratio of the tunneling current when the tip is \(0.500 \mathrm{nm}\) above the surface to the current when the tip is \(0.515 \mathrm{nm}\) above the surface.
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