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Chapter 37: Problem 65

An approximate one-dimensional quantum well can be formed by surrounding a layer of GaAs with layers of \(\mathrm{Al}_{x} \mathrm{Ga}_{1-x}\) As. The GaAs layers can be fabricated in thicknesses that are integral multiples of the single-layer thickness, \(0.28 \mathrm{nm}\). Some electrons in the GaAs layer behave as if they were trapped in a box. For simplicity, treat the box as an infinite one-dimensional well and ignore the interactions between the electrons and the Ga and As atoms (such interactions are often accounted for by replacing the actual electron mass with an effective electron mass). Calculate the energy of the ground state in this well for these cases: a) 2 GaAs layers b) 5 GaAs layers

Short Answer

Expert verified
Answer: The ground state energy for a GaAs well surrounded by two GaAs layers is around \(15.94 eV\), and for the well surrounded by five GaAs layers, it is around \(2.26 eV\).

Step by step solution

01

Calculate the size of the well for both cases

Since the GaAs layers can be fabricated at thicknesses that are integral multiples of the single-layer thickness (\(0.28nm\)), we calculate the size of the well for both cases: a) \(L_a = 2 \times 0.28 nm = 0.56 nm\) b) \(L_b = 5 \times 0.28 nm = 1.4 nm\) Convert these lengths to meters: a) \(L_a = 0.56 \times 10^{-9} m\) b) \(L_b = 1.4 \times 10^{-9} m\)
02

Calculate the effective mass of the electron in GaAs

To find out the effective mass of the electron in GaAs, we need to multiply the effective mass ratio (\(m^* \approx 0.067 m_e\)) by the electron mass (\(m_e \approx 9.11 \times 10^{-31} kg\)): \(m^* = 0.067 \times 9.11 \times 10^{-31} kg \approx 6.1 \times 10^{-32} kg\)
03

Calculate the ground state energy for both cases using the formula

Now we are ready to use the formula for the energy levels of an infinite potential well to calculate the ground state energy (\(n = 1\)) for both cases: a) \(E_{1a} = \frac{1^2 \pi^2 \hbar^2}{2 m^* L_a^2}\) b) \(E_{1b} = \frac{1^2 \pi^2 \hbar^2}{2 m^* L_b^2}\) a) \(E_{1a} = \frac{1^2 \pi^2 (6.58 \times 10^{-16} eV \cdot s)^2}{2 (6.1 \times 10^{-32} kg) (0.56 \times 10^{-9} m)^2} \approx 15.94 eV\) b) \(E_{1b} = \frac{1^2 \pi^2 (6.58 \times 10^{-16} eV \cdot s)^2}{2 (6.1 \times 10^{-32} kg) (1.4 \times 10^{-9} m)^2} \approx 2.26 eV\) Thus, the ground state energy for a GaAs well surrounded by two GaAs layers is around \(15.94 eV\), and for the well surrounded by five GaAs layers, it is around \(2.26 eV\).

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Most popular questions from this chapter

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.

Find the probability of finding an electron trapped in a one-dimensional infinite well of width \(2.00 \mathrm{nm}\) in the \(n=2\) state between 0.800 and \(0.900 \mathrm{nm}\) (assume that the left edge of the well is at \(x=0\) and the right edge is at \(x=2.00 \mathrm{nm}\) ).

Consider an attractive square-well potential, \(U(x)=0\) for \(x<-\alpha, U(x)=-U_{0}\) for \(-\alpha \leq x \leq \alpha\) where \(U_{0}\) is a positive constant, and \(U(x)=0\) for \(x>\alpha .\) For \(E>0,\) the solution of the Schrödinger equation in the 3 regions will be the following: For \(x<-\alpha, \psi(x)=e^{i \kappa x}+R e^{-i \kappa x}\) where \(\kappa^{2}=2 m E / \hbar^{2}\) and \(R\) is the amplitude of a reflected wave. For \(-\alpha \leq x \leq \alpha, \psi(x)=A e^{i \kappa^{\prime} x}+B e^{-i \kappa^{\prime} x}\) and \(\left(\kappa^{\prime}\right)^{2}=2 m\left(E+U_{0}\right) / \hbar^{2}\). For \(x>\alpha, \psi(x)=T e^{i \kappa x}\) where \(T\) is the amplitude of the transmitted wave. Match \(\psi(x)\) and \(d \psi(x) / d x\) at \(-\alpha\) and \(\alpha\) and find an expression for \(R\). What is the condition for which \(R=0\) (that is, there is no reflected wave)?

A beam of electrons moving in the positive \(x\) -direction encounters a potential barrier that is \(2.51 \mathrm{eV}\) high and \(1.00 \mathrm{nm}\) wide. Each electron has a kinetic energy of \(2.50 \mathrm{eV},\) and the electrons arrive at the barrier at a rate of 1000 electrons/s (1000. electrons every second). What is the rate \(\mathrm{I}_{\mathrm{T}}\) in electrons/s at which electrons pass through the barrier, on average? What is the rate \(\mathrm{I}_{\mathrm{R}}\) in electrons/s at which electrons reflect back from the barrier, on average? Determine and compare the wavelengths of the electrons before and after they pass through the barrier.

Determine the lowest 3 energies of the wave function of a proton in a box of width \(1.0 \cdot 10^{-10} \mathrm{~m}\).
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