# 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) Hello!

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 https://www.desmos.com/calculator/mmija615t1

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