Given the following wavefunction for the ground state of a finite quantum well of width L=2nm, ground state energy of E1=0.05eV.
wavefunction 1= Acos(kx) and wavefunction 2 = Be^(alpha.x)
Determine the barrier energy from the decay constant alpha?
Given the following wavefunction for the ground state of a finite quantum well of width L=2nm,...
1. Given the following wavefunction for the ground state of a finite quantum well of width 2nm, ground state energy of E1=0.05eV -A cos(kx) and ψ,-Beax A.) Find the values of k and a (remember to keep the wavefunction continuous and smooth)[10ptsl B.) Find the normalization constants A and B (you will need to find k first of course) [10pts] C.) Determine the barrier energy from the decay constant a? [5pts D.)If the well were replaced with a semi-infinite well...
(III) Quantum Tunneling Consider an electron in 1D in presence of a potential barrier of width L represented by a step function ſo I<0 or 1>L V U. r>0 and 2<L The total wavefunction is subject to the time-independent Schrödinger equation = EV (2) 2m ar2 +V where E is the energy of the quantum particle in question and m is the mass of the quantum particle. A The total wavefunction of a free particle that enters the barrier from...
A particle in an infinite well of width L is in its ground state. a) If L is 30 cm, what is the ground state energy? (3 marks) Where is the particle most likely to be found? Use sketching to further explain. (4 marks)
Special Problem (20 pts) Consider an undoped AljGa7As/GaAs/ Al3Ga7As quantum well (QW) of width W-15 nm. (a) Due the quantum mechanical confinement in the quantum well, the lowest energy states of in the conduction band is no longer the conduction band edge, but the CB edge plus the confined state energy (particle in the box problem), where the confinement energy relative to the CB edge is given by the solutions for infinite barriers where n-1,2,.is the quantum number, n-1 is...
An infinite square well and a finite square well in 1D with
equal width. The potential energies of these wells are
Infinite square well: V(x)=0, from 0 < x < a, also V(x) =
, elsewhere
Finite square well: V(x)= 0, from 0 < x < a, also V(x) =
,
elsewhere
The ground state of both systems have identical particles.
Without solving the energies of ground states, determine which
particle has the higher energy and explain why?
Problem 6 Electron with mass mo-9.11x103 kg is placed in a quantum well of width L-70nm with finite barriers Vi-0.35eV Calculate and plot the electron wave function and probability density of finding the electron depending on its coordinate z for two lowest energy levels Ei and E 4/2/
Problem 6 Electron with mass mo-9.11x103 kg is placed in a quantum well of width L-70nm with finite barriers Vi-0.35eV Calculate and plot the electron wave function and probability density of finding...
An electron is bound in the ground state of a finite square well with U0 = 73 eV. (a) How much energy is required to free the electron from the well if the ground-state energy is 2.6 eV? eV (b) If this transition is accomplished through the absorption of one photon of light, what is the maximum wavelength of that photon? m
2. Goal of this problem is to study how tunnelling in a two-well system emerges. In particular, we are interested in determining how the tunnelling rate T' of a particle with mass m scales as a function of the (effective) height Vo - E and width b of an energy barrier separating the two wells. The following graphics illustrates the set-up. Initially the particle may be trapped on the left side corresponding to the state |L〉, we are now interested...
An electron in a one-dimensional infinite potential well of length L has ground-state energy E1. The length is changed to L' so that the new ground-state energy is E1' = 0.234E1. What is the ratio L'/L?
Consider a particle of mass in a 10 finite potential well of height V. the domain – a < x < a. a) Show that solutions for – a < x < a take the form on (x) = A cos(knx) for odd n, and on (x) = A sin(knx) for even n. . Show a) Match the boundary conditions at x = a to prove that cos(ka) = Bk where k is the wave vector for -a < x...