InterviewSolution
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InterviewSolution
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A particle in the ground state is located in a unidimesional square potential well of length `l` with absolutely impenetrable walls `(0 lt x lt l)`. Find the probability of the particle staying within a region `(1)/(3)l le x le (2)/(3)l`. |
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Answer» We look for the solution of Schrodinger eqn. with `-(ħ^(2))/(2m)(d^(2)Psi)/(dx^(2))=E Psi,0lexlel`….(1) The boundary condition of impenetrable walls means `Psi(x)=0` for `x=0` and `x=l` (as `Psi(x)=0` for `xlt0` and `xgtl`,) Then solution of (1) is `Psi(x)=A "sin"sqrt(2mE)/(ħ)x+B "cos"sqrt(2mE)/(ħ)x` Then `Psi(0)=0implies B=0` `Psi(l)=0implies A "sin"sqrt(mE)/(ħ)l-0` `A~~ 0` so `sqrt(2mE)/(ħ)l=n pii` Hence `E_(n)=(n^(2)pi^(2)ħ^(2))/(2ml^(2)),n= 1,2,3`.... Thus the ground state wave fucnction is `Psi(x)=A "sin"(pi x)/(l)` We evaluate `A` by nomalization `1=A^(2)int_(0)^(i)"sin"^(2)(pi x)/(l)dx=A^(2)(l)/(pi)int_(0)^(x)sin^(2) theta d theta=A^(2)(l)/(pi).(pi)/(2)` Thus `A+(sqrt(2)/(l))` Finally, the probability `P` for the particle to lie in `(l)/(2) le (2l)/(3)` is `P=P((l)/(3)lexle(2l)/(3))=(2)/(l)int_(l/3)"sin"^(2)(pix)/(l)dx` `=(2)/(pi)int_(pi//3) sin^(2) theta d theta=(1)/(pi)int_(pi//3)^(2pi//3)(1- cos 2 theta)d theta` `(1)/(pi)(theta-(1)/(2) sin 2 theta)^(2pi//3)=(1)/(pi)((2pi)/(3)-(pi)/(3)-(1)/(2)"sin"(4pi)/(3)+(1)/(2)`sin`(2pi)/(3))` `=(1)/(pi)((pi)/(3)+(1)/(2)(sqrt(3))/(2)+(1)/(2)(sqrt(3))/(2))=(1)/(3)+(sqrt(3))/(2pi)= 0.609` |
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