University:
Santa Fe College
Course:
PHY 2048 | General Physics 1 with Calculus
Academic year:

2024
Views:

85

Pages:

6
Author:

Jabin F.

Bound state 0 < R < pi/2 => 0 < beta < pi/2 => 0 < K II L / 2 < pi/2 => 0 < K II^2 L^2 / 4 < pi^2 / 4 => 0 < K II^2 < pi^2 / L^2 => 0 < E < pi^2 * hbar^2 / 2 * m * L^2 => 0 < 2 * m * E / hbar^2 < pi^2 / L^2 Radius Energy No. of Even Parity States No. of Odd Parity States No. of Bound States 0 < R < pi/2 0 < E1 < pi^2 * hbar^2 / 2 * m * L^2 1 0 1 pi/2 <= R < pi pi^2 * hbar^2 / 2 * m * L^2 <= E2 < 4 * pi^2 * hbar^2 / 2 * m * L^2 1 1 2 pi <= R < 3 * pi/2 4 * pi^2 * hbar^2 / 2 * m * L^2 <= E3 < 9 * pi^2 * hbar^2 / 2 * m * L^2 2 1 3 3 * pi/2 <= R < 2 * pi 9 * pi^2 * hbar^2 / 2 * m * L^2 <= E4 < 16 * pi^2 * hbar^2 / 2 * m * L^2 2 2 4 ... ... ... ... ... (n - 1) * pi/2 <= R < n * pi/2 ((n - 1)^2 * pi^2 * hbar^2 / 2 * m * L^2) <= En < (n^2 * pi^2 * hbar^2 / 2 * m * L^2) n/2 (Even) (n - 1)/2 (Odd) n (total) ... ... ... ... ... (n - 1) * pi/2 < R < n * pi/2 2 0 < beta1 < pi/2, pi/2 <= beta2 < pi Calculation for Table pi^2 / L^2 <= (k_II)2^2 <= 4 * pi^2 / L^2^2 pi^2 * hbar^2 / 2mL^2 <= E2 <= 4 * pi^2 * hbar^2 / 2mL^2 Similar to for E3, E4, ... For infinite pot well The no of bound states are infinite. E1 Infinite <= Enfin <= En Infinite. 1/nfin (n-1)^2 * pi^2 * hbar^2 / 2mL^2 <= En < n^2 * pi^2 * hbar^2 / 2mL^2 For finite potential well there are finite No. of Bound states. V(x) Vo n=2 n=1 = For State psi1 psi1(x) = { A e^(Kx), x <= -L/2 { C cos K_II x, -L/2 <= x <= L/2 { A e^(Kx), x >= L/2 psi1(x) Compare for finite pot well for infinite pot well 20 Novero for psi_2(x) psi_2 = {A e^(Kx), x <= -L/2 {C sin K_II x, -L/2 <= x <= L/2 {-A e^(-Kx), x >= L/2 A = F V(x) V=0 x=-L/2 x=L/2 x=0 x -Vo 3 JULY 2020 @ue1-13 Page-31 V(x) = V(x) = { 0, |x| < a/2 { Vo, |x| > a/2 Hence V(x) = { 0, |x| < a/2 {-Vo, |x| > a/2 K_I^2 = 2mE / hbar^2, K_II^2 = 2m(E + Vo) / hbar^2 K_I^2 + K_II^2 = R^2 = 2mV_o / hbar^2 d²ψ / dx² + (2m/ħ²) (E - Vm) ψ = 0 for x -> -∞ d²ψ / dx² + (2m/ħ²) (E + Vo) ψ = 0 K₁² = (2m/ħ²) (E + Vo) This is right. for ex. Vo = 4 eV -Vo = +4 eV E = -2 eV E + Vo = 2 eV + ve value then. K₁ as x -> +∞ ce ψ₁ = A sin K₁x + B cos K₁x ψ₂ = Ce^(Kx) + De^(-Kx) ψ₂ = De^(-Kx) ψ₂ | x=0 => B = 0 K I 2 +K II 2 = ℏ 2 2mV o α=k I l β=k II l =>[α 2 +β 2 =⋅ ℏ 2 α m v o l 2 =R 2 ] α=−β cot β=+β tan( 2 π +β) only tre plot allowed for Ground state or one Bound state 2 π ≤R<Π R min = 2 π R min 2 = 4 π 2 ℏ 2 2 mv o m! e 2 = 4 π 2 V omin = ℏ 2 8m 2 l 2 (option-a)