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A ground state electron is trapped in the one dimensional infinite potential well of fig.38-2, with width L = 100 pm. (a) What is the probability that the electron can be detected in the left one third of well (x_(1)=0" to "x_(2)=L//3) What is the probability that the electron can be detected in the middle one third of the well? |
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Answer» Solution :KEY IDEAS If we prove the left one third of the welcome there is no guarantee that we will detect the electron. HOWEVER, we can calculate the probability of detecting it with the integral of EQ. (2) The probability very much depends on which state of electron is in that is, the value of quantum number n. Calculations: Because there the electron is in the ground state, we set n =1 in Eq. 38-13 . we also set the limits of integration as the POSITIONS `x_(1)=0 and x_(2)=L//3`set the amplitude constant a s so that the wave function is normalised we then see that `({:("Probability of detection"),("in left one - third"):})=int_(0)^(L//3)(2)/(L)sin^(2)((1pi)/(L)x)dx.` We could find this probability by substituting `100xx10^(-12)m` for L and then using a graphing calculator for computer maths package to EVALUATE the integral. Here however we should evaluate the integral ..by hand... First we switch to and new integration variable y: `u=(pi)/(L)x and dx=(L)/(pi)dy.` From the first of these equation, we find the new limits of integration to be `y_(1)=0` and `y_(2)=pi//3` for `x_(2)=L//3`.We then must evaluate `"Probability"=((2)/(L))=((L)/(pi))int_(0)^(pi//3)(sin^(2)y)dy.` Using integral 11 in Appendix E, we then find `"Probability"=(2)/(pi)((y)/(2)-(sin2y)/(4))_(0)^(pi//3)=0.20`. Thus, we have `({:("Probability of detection"),("in left one - third"):})=0.20` That is if we repeatedly Pro the left one third of the well then on average we can detect the electron with `20%` of the probs . Reasoning : We now KNOW that the probability of the detection in the left one third of the well is 0.20 by symmetry the probability of the detection in the right one third of the well is also 0.20. because the electron is certainly in the welcome of the probability of detection in the entire well is one. Thus theprobability of detection in the middle one third of the well is `({:("Probability of detection"),("in middle one - third"):})=1-0.20-0.20` `=0.60` |
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