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MArishka [77]
3 years ago
9

For a solution to be a physical solution, it must satisfy several criteria. First, it must be continuous everywhere. Second, it

must have a continuous derivative everywhere, except for points where the potential energy becomes infinite (as it does at the walls of the box). Finally, it must be normalizable. In this case, you can check the first criterion by noting that the two functions Csin(nT/L) and 0 are continuous and that they have the same value at x = 0 and z = L, where their domains meet. To check the second criterion, simply take the first derivative of Csin(nTr/L) and 0. Both derivatives are continuous functions in their domains. The points where the domains meet are exactly the points where the potential energy becomes infinite, so you don't have to check for continuity there. The third criterion requires that there exist some value of C such that Since (z) is zero outside of the interval 0 < ¢ < L, this equation reduces to Use this equation to find the unique positive value of C. Express your answer in terms of L.

Physics
1 answer:
Zigmanuir [339]3 years ago
5 0

Answer:

Explanation:

solution

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