What is the probability (in %) of locating a particle in aone-dimensional box be

Question

What is the probability (in %) of locating a particle in aone-dimensional box between L/4 and 3L/4, where L is the length ofthe box? Assume the particle to be in the lowest level.

Explanation / Answer

(-hbar2/2m)*d2/dx2 =E*               V(x) = 0 for a particle in a box d2/dx2 = -(2mE/hbar2)*= -k2* = A*sin(kx) + B*cos(kx) For a particle in a box, = 0 at x = 0 and x = L (0) = A*sin(k*0) + B*cos(k*0) = A*0 + B*1 = 0 => B = 0 = A*sin(kx) at x = L, (L) = 0 = A*sin(k*L) => k*L =n* Thus = A*sin(n*x/L) Normalize the wavefunction integral [*]dx = 1 = A*A*integral [sin(nx/L)*sin(nx/L)]dx = A2* integral [1/2 - (1/2)*cos(2nx/L)]dx = A2* [x/2 -(L/4)sin(2nx/L)]       evaluate from 0 to L = A2*[(L/2 - (L/4n)*sin(2nL/L) - (0/2 -(L/4n)*sin(0/L)] =A2*L/2                 Recall sin(2n) = 0 Recall that normalize A2*L/2 = 1 =>  A = (2/L)1/2 for = A*sin(n*x/L) Now to find probability, n = 1, for lowest level. P = integral[*]dx            but limits of integration are from L/4 to 3L/4 P = A2*integral [sin2(nx/L)]dx =A2* (x/2 -(L/4)sin(2nx/L)          (recall this was found during normalization) P = (2/L)*[x/2 -(L/4n)sin(2nx/L)]           evaluate from L/4 to 3L/4 P = (2/L)*[(3L/8 - (L/4)*sin(3/2)) - (L/8 -(L/4)*sin(/2)] P = (2/L) *[(3L/8 - (L/4)*-1) - (L/8 - L/4)] P = (2/L)*(2L/8 + L/2) P = 1/2 + 1/

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