Imagine that two infinitely thin walls are added to the middle of a box at x = L
ID: 475667 • Letter: I
Question
Imagine that two infinitely thin walls are added to the middle of a box at x = L/3 and x = 2L/3. At these locations, the potential is infinite just as it is at x = 0 and L but the particle can tunnel through the walls because they are infinitely thin. write the allowed values of the wavefunction at psi (x = 0), psi (x = L), psi (x = L/3), and psi (x = 2L/3); write an equation for the normalized wavefunction that satisfies all of these boundary conditions; substitute the new wavefunction expression into the Hamiltonian for the particle to find an expression for the allowed energies of the particle; and graph the first two lowest energy solutions and comment on the changes if any that the additional walls make to the wavefunctions of the particle in the original box of length L.Explanation / Answer
The wave function n(x) for a particle in the nth energy state in an infinite square box with walls at x = 0 and x = L is n(x) = [(2/L)^1/2 * (sin nx/L)]
at :-
(i) x=o
the wave function will be
= [(2/L)^1/2 * (sin n*0/L)]
=0 ( because the sine function will become 0 as sine 0 is 0)
(ii) at x= L
[(2/L)^1/2 * (sin n* L/L)]
= [(2/L)^1/2 * (sin n)]
(iii) at x= L/3
= [(2/L)^1/2 * (sin n *L/3 /L)]
= [(2/L)^1/2 * (sin n/3)]
(iv) at x= 2L/3
=[(2/L)^1/2 * (sin n*2L/3/L)]
=[(2/L)^1/2 * (sin n2/3)]