Chapters 2 and 3: Problems 2-9, 3-1, 3-3, 3-4, 3-11,3-12, 3-26. An electron in a
ID: 980649 • Letter: C
Question
Chapters 2 and 3: Problems 2-9, 3-1, 3-3, 3-4, 3-11,3-12, 3-26. An electron in a one-dimensional box makes a transition from n=4 to n=2 and emits a photon of wavelength 539 nm. What is the size of the box What wavelength is emitted when the electron makes a transition from n=3 to n= 1 in this one-dimensional box A particle is confined in a one-dimensional box between times = 0 and times = a, with a wave function given by phi_n(X), where n is the quantum number. Show that the wave functions for a particle in a box are orthonormal, such that integral 0 to a phi_i ^* (x) phi_j(x) dx = delta_ij where delta_ij = 1 if i = j and = 0 if i not equal to j. What are the average values for (a) (x cap p cap ^2_x) and (b) (x cap^2 p cap_x) for a particle in a one-dimensional box of length a in a quantum level nExplanation / Answer
Let us consider two wave function for particle in one dimensional box are
(2/a)1/2 Sin (n*pi*x/a) and (2/a)1/2 Sin (m*pi*x/a)
Case 1: m=n=1
when we do intergral inbetween 0 to a and apply n=1 in both wave functions, we will get
(2/a) { Integral (0 to a) sin2 (pi*x/a) dx}
we know that, Sin2x = (1-Cos2x) / 2, so the above equation transformed into
(2/a) Integral (0 to a) dx/2 - integral (0 to a) Cos(2*pi*x/2a) dx.
The cosine term is become Zero because Interal cosx = - sin x. when we applied to limits 0 to a, sin 0 and sin pi will become zero. so we can neglect this term.
Therefore, m=n=1, Integral (wave function)2 dx = (2/a) [x/2] (limits 0 to a) = (2/a)*(a/2) = 1.
So when m=n, wave function is normalised each other.
Case 2: m=1 and n = 2
(2/a) integral (0 to a) sin (pi*x/a) sin(2*pi*x/a) dx
= (2/a) [ {sin (pi/a - 2pi/a)/2(pi/a - 2pi/a)} - {sin (pi/a + 2pi/a)/2(pi/a + 2pi/a)}] (limits 0 to a)
substituting limits 0 and a into this expression and evaluating,
= (2/a) [ {sin (-pi) / (-2pi/a) - sin 3pi / (6pi/a)} - (2/a) [ {sin0/ (-2pi/a)} - {sin0 / (6pi/a)}] = 0 = orthogonal.
Since sin pi and sin 0 has Zero value and this expression becomes zero. Therefore, when m is not equal to n, those wavefunctions are orthogonal to each other.