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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

A particle of energy \(E=5 \mathrm{eV}\) approaches an energy barrier of height \(U=8 \mathrm{eV}\). Quantum mechanically there is a finite probability that the particle tunnels through the barrier. If the barrier height is slowly decreased, the probability that the particle will reflect from the barrier will a) decrease. b) increase. c) not change.

A positron and an electron annihilate, producing two 2.0 -MeV gamma rays moving in opposite directions. Calculate the kinetic energy of the electron when the kinetic energy of the positron is twice that of the electron.

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 neutrons in a parallel beam, each having kinetic energy \(1 / 40 \mathrm{eV}\) (which is approximately corresponding to 鈥渞oom temperature"), are directed through two slits \(0.50 \mathrm{~mm}\) apart. How far apart will the interference peaks be on a screen \(1.5 \mathrm{~m}\) away?

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.
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