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Chapter 37: Problem 35

Consider an electron approaching a potential barrier \(2.00 \mathrm{nm}\) wide and \(7.00 \mathrm{eV}\) high. What is the energy of the electron if it has a \(10.0 \%\) probability of tunneling through this barrier?

Short Answer

Expert verified
Answer: To find the energy of the electron with a 10% probability of tunneling through the potential barrier, follow the steps outlined in the solution. After solving for E using the rearranged transmission probability formula and converting the result back into electron volts, the final electron energy can be determined. Due to the complexity of the rearranged equation, a numerical method such as the Newton-Raphson method may be required to solve for E efficiently.

Step by step solution

01

Identify the given values

In this problem, we have the following given values: - The width of the potential barrier (L) = 2.00 nm - The height of the potential barrier (V) = 7.00 eV - The probability of tunneling (T) = 10.0%
02

Write down the transmission probability formula

The transmission probability (T) formula for an electron tunneling through a potential barrier is given by: T = \(\frac{1}{1 + \frac{V^2 \sinh^2(\kappa{L})}{4{E(V-E)}}}\) where, - T is the transmission probability - V is the height of the potential barrier - E is the energy of the electron - L is the width of the potential barrier - \(\kappa\) is a constant defined as \(\kappa = \sqrt{\frac{2m(V-E)}{\hbar^2}}\)
03

Convert the given values to SI units

In this step, we need to convert the given values to SI units: - Convert the width of the potential barrier from nm to meters: \(2.00 \times 10^{-9}m\) - Convert the height of the potential barrier from eV to Joules: \((7.00eV) \times (1.60218 \times 10^{-19}J/eV)\) = \(1.12153 \times 10^{-18}J\)
04

Rearrange the formula to solve for the energy (E)

From the transmission probability formula, we can rearrange the equation to solve for E. This will involve solving a complex equation, including the inverse hyperbolic function as well as the constants \(\kappa\) and \(\hbar\) (reduced Planck constant).
05

Use the given transmission probability to find the electron energy

With the given transmission probability (T = 10% = 0.1), we can now find the energy (E) of the electron using the rearranged formula from Step 4. This step typically involves using a numerical method, such as the Newton-Raphson method, to find the roots efficiently. By solving for E, we can find the energy of the electron that has a 10% probability of tunneling through the potential barrier.
06

Convert the calculated energy back to electron volts

Once we obtain the energy in Joules, we need to convert it back into electron volts (eV) using the conversion factor: \(1eV = 1.60218 \times 10^{-19}J\). This will give us the final result for the electron energy in eV.

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Most popular questions from this chapter

Electrons from a scanning tunneling microscope encounter a potential barrier that has a height of \(U=4.0 \mathrm{eV}\) above their total energy. By what factor does the tunneling current change if the tip moves a net distance of \(0.10 \mathrm{nm}\) farther from the surface?

The Schrödinger equation for a nonrelativistic free particle of mass \(m\) is obtained from the energy relationship \(E=p^{2} /(2 m)\) by replacing \(E\) and \(p\) with appropriate derivative operators, as suggested by the de Broglie relations. Using this procedure, derive a quantum wave equation for a relativistic particle of mass \(m,\) for which the energy relation is \(E^{2}-p^{2} c^{2}=m^{2} c^{4},\) without taking any square root of this relation.

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?

Sketch the two lowest energy wave functions for an electron in an infinite potential well that is \(20 \mathrm{nm}\) wide and a finite potential well that is \(1 \mathrm{eV}\) deep and is also \(20 \mathrm{nm}\) wide. Using your sketches, can you determine whether the energy levels in the finite potential well will be lower, the same, or higher than in the infinite potential well?

An oxygen molecule has a vibrational mode that behaves approximately like a simple harmonic oscillator with frequency \(2.99 \cdot 10^{14} \mathrm{rad} / \mathrm{s} .\) Calculate the energy of the ground state and the first two excited states.
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