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Chapter 37: Problem 68

A surface is examined using a scanning tunneling microscope (STM). For the range of the working gap, \(L\), between the tip and the sample surface, assume that the electron wave function for the atoms under investigation falls off exponentially as \(|\Psi|=e^{-\left(10.0 \mathrm{nm}^{-1}\right) a}\). The tunneling current through the STM tip is proportional to the tunneling probability. In this situation, what is the ratio of the current when the STM tip is \(0.400 \mathrm{nm}\) above a surface feature to the current when the tip is \(0.420 \mathrm{nm}\) above the surface?

Short Answer

Expert verified
Answer: The ratio of the tunneling current is approximately \(e^{0.4}\).

Step by step solution

01

Write down the probability density function at each distance

We know the probability density function is given by \(|\Psi|^2\). We'll calculate the probability densities, \(P1\) and \(P2\), for the STM tip at distances \(0.400 \mathrm{nm}\) and \(0.420 \mathrm{nm}\) respectively, using the formula: \(|\Psi_i|^2 = e^{-2 \times (10.0 \mathrm{nm}^{-1}) a_i}\), where i={1,2}
02

Calculate the probability density at a distance of \(0.400 \mathrm{nm}\)

We can substitute \(a_1 = 0.400 \mathrm{nm}\) into the formula for the probability density: \(P1 = |\Psi_1|^2 = e^{-2 \times (10.0 \mathrm{nm}^{-1}) \times 0.400 \mathrm{nm}}\) Now we can calculate \(P1\): \( P1 = e^{-2 \times (10.0 \mathrm{nm}^{-1}) \times 0.400 \mathrm{nm}}\) \( P1 = e^{- 8 }\)
03

Calculate the probability density at a distance of \(0.420 \mathrm{nm}\)

We can substitute \(a_2 = 0.420 \mathrm{nm}\) into the formula for the probability density: \(P2 = |\Psi_2|^2 = e^{-2 \times (10.0 \mathrm{nm}^{-1}) \times 0.420 \mathrm{nm}}\) Now we can calculate \(P2\): \( P2 = e^{-2 \times (10.0 \mathrm{nm}^{-1}) \times 0.420 \mathrm{nm}}\) \( P2 = e^{- 8.4 }\)
04

Calculate the ratio of the tunneling current

We are told that the tunneling current is proportional to the tunneling probability, which means the ratio of the tunneling current is equal to the ratio of the probability densities: \(\frac{I1}{I2} = \frac{P1}{P2}\) Substitute in the values for \(P1\) and \(P2\): \(\frac{I1}{I2} = \frac{e^{- 8}}{e^{- 8.4}}\) Now we can simplify the expression to get the ratio: \(\frac{I1}{I2} = e^{- 8 + 8.4}\) \(\frac{I1}{I2} = e^{0.4}\) So, the ratio of the tunneling current when the tip is \(0.400 \mathrm{nm}\) above the surface to the current when the tip is \(0.420 \mathrm{nm}\) above the surface is approximately \(e^{0.4}\).

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Most popular questions from this chapter

Is the superposition of two wave functions, which are solutions to the Schrödinger equation for the same potential energy, also a solution to the Schrödinger equation? a) no b) yes c) depends on potential energy d) only if \(\frac{d^{2} \psi(x)}{d x^{2}}=0\)

An electron is in an infinite square well of width \(a\) : \((U(x)=\infty\) for \(x<0\) and \(x>a\) ). If the electron is in the first excited state, \(\Psi(x)=A \sin (2 \pi x / a),\) at what position is the probability function a maximum? a) 0 b) \(a / 4\) c) \(a / 2\) d) \(3 a / 4\) and \(3 a / 4\) e) at both \(a / 4\)

Suppose \(\psi(x)\) is a properly normalized wave function describing the state of an electron. Consider a second wave function, \(\psi_{\text {new }}(x)=e^{i \phi} \psi(x),\) for some real number \(\phi .\) How does the probability density associated with \(\psi_{\text {new }}\) compare to that associated with \(\psi ?\)

Example 37.1 calculates the energy of the wave function with the lowest quantum number for an electron confined to a box of width \(2.00 \AA\) in the one-dimensional case. However, atoms are three-dimensional entities with a typical diameter of \(1.00 \AA=10^{-10} \mathrm{~m} .\) It would seem then that the next, better approximation would be that of an electron trapped in a three-dimensional infinite potential well (a potential cube with sides of \(1.00 \mathrm{~A}\) ). a) Derive an expression for the electron wave function and the corresponding energies for a particle in a three dimensional rectangular infinite potential well. b) Calculate the lowest energy allowed for the electron in this case.

For a finite square well, you have seen solutions for particle energies greater than and less than the well depth. Show that these solutions are equal outside the potential well if the particle energy is equal to the well depth. Explain your answer and the possible difficulty with it.
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