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Evaluate the amplitude constant A in Eq. 38-10 for an infinite potential well extending from x=0" to "x=L. |
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Answer» Solution :KEY IDEAS The wave function of Eq. 38-10 must satisfy the normalisation requirement of Eq. 38-14, which states that the probability that the electron can be detected somewhere along the x axis is 1. Calculations : Substituting Eq. 38 - 10 into Eq. 38 - 14 and taking the constant A outside the integral yield `A^(2)int_(0)^(L)sin^(2)((npi)/(L)x)dx=1`. We have changed the limits of the integral from `-oo and +oo`to 0 and Lbecause the ..outside.. wave function is ZERO. We can simplify the indicated integration by changing the VARIABLE from the dimensionless variable y, where `y=(npi)/(L)x.` hence`""dx=(L)/(npi)dy.` When we change the varibale, we must also change the integration limits (again). Equation 38-16 tells us that y = 0 when x = 0 and that `y=npi` when `x=L`, thus 0 and `npi` are our new limits. With all these subtitutions, Eq. 38-15 BECOMES `A^(2)(L)/(npi)int_(0)^(npi)(sin^(2)y)dy=1.` We can use integral 11 in appendix E to value with the integral obtaining the equation `(A^(2)L)/(npi)[(y)/(2)-(sin2y)/(4)]_(0)^(npi)=1`. EVALUATING at the limits yields `(A^(2)L)/(npi)(npi)/(2)=1`, thus `A=sqrt((2)/(L))`. This result tells us that the dimension for `A^(2)`, and thus for `Psi_(n)^(2)(x)`, is an inverse length. This is appropriate because the probability density of Eq. 38-12 is a the probability per unit length. |
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