Consider a particle trapped in a 1D box with zero potential energy with walls at x = oand x = L. The general wavefunction solutions for this problem with quantum number,n, are:V,6) = sin )4n(x) =The corresponding energy (level) for each wavefunction solution is:n²h?En8mL?a) What is the probability of finding the particle between x = L/4 and x = 3L/4 whenthe particle is in quantum state n = 1, 2 and 3.You can use calculator or a numerical program to do the integral. For people who wantto try doing the integral by hand, the following identity will be helpful: sin²(x) = (1 – cos(2x))/2.

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Consider a particle trapped in a 1D box with zero potential energy with walls at x = 0 and x = L. The general wavefunction solutions for this problem with quantum number, n, are: Vn(x) = sin ( The corresponding energy (level) for each wavefunction solution is: n²h? En 8mL² a) What is the probability of finding the particle between x = L/4 and x = 3L/4 when the particle is in quantum state n = 1, 2 and 3. You can use calculator or a numerical program to do the integral. For people who want to try doing the integral by hand, the following identity will be helpful: sin²(x) = (1 – cos (2x))/2.

Consider a particle trapped in a 1D box with zero potential energy with walls at x = 0 and x = L. The general wavefunction solutions for this problem with quantum number, n, are: Vn(x) = sin ( The corresponding energy (level) for each wavefunction solution is: n²h? En 8mL² a) What is the probability of finding the particle between x = L/4 and x = 3L/4 when the particle is in quantum state n = 1, 2 and 3. You can use calculator or a numerical program to do the integral. For people who want to try doing the integral by hand, the following identity will be helpful: sin²(x) = (1 – cos (2x))/2.