Question

Consider the symmetrical finite square well potential shown below. U(x) = 46 eV for xs-L/2 U(x) 0 eV for-L/2 < x < L/2 U(x) 46 eV for x 2 L/2 L-0.27mm Note: 46 ev 1. the width L is unchanged from the infinite well you previously considered 2, the potential outside x-±L/2 is finite with U-46 eV. 3. you found the three lowest energy levels for that infinite -8.135 0.135 potential well were: 5.16 ev, 20.64 ev, and46.45 ev. 1) Here are four candidates for part of the wave functions for bound states of the finite square well in the region around x L/2, which of these candidates have the proper behavior? a. b. C. d. 2 (b)

Answer 1

1] is correct since the wavefunction must be continuous, single-valued, and should have continuous derivative (continuous slope).

2] is correct since B=0 (because the wavefunction should be finite valued) and thus the value of the wavefunction should exponentially drop beyond x = L/2

3]

Wavelength of the fundamental energy energy in an inifinite potential well will be:

$\lambda = \frac{2L}{1} = 2[0.27]nm = 0.54 nm$

where the wavefunction terminates at the boundaries

in the finite potential case, the wavefunction penetrates the walls of the barrier and so the wavelength in this case will be 'stretched' out and should thus be longer than 0.54 nm.

Also, since the wavelength will be longer, the energy of the ground state will be smaller than the ground state energy of the infinite potential case [which is 5.16eV] and so the correct choices are 2nd option and 3rd option.