Note this is a Physics/Calculus question. A particle is described by the wave function: `psi(x)={(ce^(x/L) if x<=0mm),(ce^(-x/L) if x>=0mm):}` where L=2mm a) Sketch graphs of both the wave function and the probability density as functions of x (meaning x on the x-axis) b) Determine the normalization constant c (should = .707 by c=1/sqrt(L) c) Calculate the probability of finding the particle within 1mm of the origin (should = 63.2%) d) Interpret the answer from part b) by shading the region of probability from part c) on the graph in part a) (poorly written question but still clear)

Expert Answers

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The probability density is the square of the absolute value of a wave function, in our case it is `P_d(x)=|Psi(x)|^2=c^2 e^(-2x/L)` for positive x's and `c^2 e^(2x/L)` for negative x's.

The integral over the real axis of a probability density must be `1` (the total probability). From this statement we can find `c:`

`int_(RR) p_d(x) dx = 2 int_0^(+oo) c^2 e^(-2x/L) dx = -2c^2 L/2e^(-2x/L) |_(x=0)^(+oo) = c^2 L=1.`

So really `c = 1/sqrt(L) approx 0.707.` (b)

(c) the probability of finding the particle within 1 mm of the origin is the integral of the probability density from `-1` to `1.` It is evidently

`2 int_0^1 1/L e^(-2x/L) dx = -2/L L/2e^(-2x/L) |_(x=0)^1 = 1-e^(-2/L) = 1-e^(-1) approx 0.632,`

which is really 63.2%.

(a), (d) look at the graph at

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