Question

Suppose that in the infinite square well problem, you set the potential energy at the bottom of the w ell to be equal to some constant V_0, rather than zero. Which of the following is the correct Schrodinger Equation inside the well? -h^2 partial differential^2 psi (x, t)/2 m partial differential x^2 = ih partial differential psi (x, t)/partial differential t -h^2 partial differential^2 psi (x, t)/2m partial differential x^2 + V_0 = ih partial differential psi (x, t)/partial differential t -h^2 partial differential^2 psi (x, t)/2 m partial differential x^2 - V_0 = ih partial differential psi (x, t)/partial differential t -h^2 partial differential^2 psi (x, t)/2m partial differential x^2 + V_0 psi (x, t) = ih partial differential psi (x, t)/partial differential t -*h^2 partial differential^2 psi (x, t)/2 m partial differential x^2 - V_0 psi (x, t) = ih partial differential psi (x, t)/partial differential t Which of the following are the correct energy levels for the infinite square well of width L with potential energy equal to V_0 at the bottom of the well? N^2 pi^2 h^2/2 mL^2 n^2pi^2h^2*/2mL^2 + V_0 n^2 pi^2 h^2/2mL^2 - V_0 V_) - n^2 pi^2 h^2/2 mL^2 n^2 pi^2 h^2/2 ml^2 V_0 V_0 n^2 pi^2 h^2/2 mL^2 None of the above - the potential energy has to be zero at the bottom of an infinite square well.

Answer 1

Second equation, second energy are corrects

Step-by-step explanation:

Schrödinger's equation is

-h’²/2m d²ψ/ dx² + V ψ = ih dψ / dt

Where h’= h/2π, h is the Planck constant, ψ the time-dependent wave function.

If we apply this equation to a well of infinite potential, there is only a solution within the well since, because the walls are infinite, electrons cannot pass to the other side,

In general we can place the origin of the regency system at any point, one of the most common to locate it at the bottom of the potential well so that V = 0, in this case it is requested that we place it lower so that V = V₀ , as this is an additive constant does not change the form of the solutions of the equation that is as follows

-h’²/2m d²ψ/dx² + Vo ψ = ih dψ / dt

Proposed Equations

First equation wrong missing the potential

Second equation correct

Third equation incorrect the power must be positive

Fifth. Incorrect is h ’Noel complex conjugate (* h’)

To find the potential well energy levels, we solve the independent equation of time

-h’² /2m d²φi /dx² + Vo φ = E φ

d²φ/ dx² = - 2m/h’² (E-Vo)φ

d²φ/dx² = k² φ

With immediate solution for being a second degree equation (harmonic oscillator), to be correct the solution must be zero in the well wall

φ = A sin k x

kL = √2m(E-Vo) /h’² = n pi

E- Vo = (h’²2 / 2mL²) n²

E = (h’² π² / 2 m L²) n² + Vo

Proposed Energies

First. wrong missing Vo

Second. correct

Third. Incorrect potential is positive, not a subtraction

Quarter. incorrect the potential is added energy is not a product