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\( psi(x) = 0, x < = -L/2 \)

= C(2x/L+1), -L/2 < x < 0

= C(-2x/L+1), 0 < x < + L/2

= 0, x > +L/2

(a) Use the normalization condition to find C.

(b) Evaluate the probability to find the particle in an interval of width 0.010L at x = L/4 (That is, between x = 0.245L and x = 0.255L. No integral calculation is necessary for this calculation.)

(c) Evaluate the probability to find the particle between x = 0 and x = +L/4

(d) Find the average value of x and rms of x:

\(x = \sqrt{(x^2)_{avg}}\)

Edit
Added Wed, 24 Feb '16
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part (a)

∫ψ2 dx = 1

∫C^2(2x/L + 1)^2 dx + ∫C^2(-2x/L + 1)^2 = 1

∫(2x/L + 1)^2 dx + ∫(-2x/L + 1)^2 = 1/C^2

-----------------------------------------------------------

Finding ∫(2x/L + 1)^2 dx :

∫(2x/L + 1)^2 dx = ∫(2x/L)^2 + 4x/L + 1 dx

= 4x^3/3L^2 + 2x^2/L + x

= -4(-L/2)^3/3L^2 - 2(-L/2)^2/L - (-L/2)

= 4(L/2)^3/3L^2 - 2(L/2)^2/L + (L/2)

= (L)/6 - (L/2) + (L/2)

= L/6

-------------------------------------------------------

Findin ∫(-2x/L + 1)^2 :

∫(-2x/L + 1)^2 = ∫(2x/L)^2 - 4x/L + 1 dx

= 4x^3/3L^2 - 2x^2/L + x

= 4(L/2)^3/3L^2 - 2(L/2)^2/L + (L/2)

= 4(L/2)^3/3L^2 - 2(L/2)^2/L + (L/2)

= (L)/6 - (L/2) + (L/2)

= L/6

------------------------------------------------------

we had ∫(2x/L + 1)^2 dx + ∫(-2x/L + 1)^2 = 1/C^2

therefore

==> L/6 + L/6 = 1/C^2

==> C = √(3/L)

part (b):

P = ψ^2 Δx

P = C^2 (-2x/L + 1)^2 Δx

P = (√(3/L))^2 (-2(L/4)/L + 1)^2 (0.010 L)

P = (3/L) (-1/2 + 1)^2 (0.010 L)

P = (3) (0.5)^2 (0.010)

P = 0.0075

part (c)

P = ∫C^2 (2x/L + 1)^2 dx

integration from 0 to L/4

P = (3/L) (4x^3/3L^2 - 2x^2/L + x)

P = (3/L) (4(L/4)^3/3L^2 - 2(L/4)^2/L + (L/4))

P = (3/L) (L/48 - L/8 + L/4)

P = (3) (1/48 - 1/8 + 1/4)

P = 0.4375

part (d)

<x> = ∫ψ* x ψ dx

= ∫ψ^2 x dx

= ∫C^2(2x/L + 1)^2 x dx + ∫C^2 (-2x/L + 1)^2 x dx

-----------------------------------------------------------

Finding ∫C^2(2x/L + 1)^2 x dx :

∫(3/L)(2x/L + 1)^2 x dx = ∫(3/L) (2x/L)^2 * x + 4x^2/L + x dx

= (3/L) (4x^4/(4L^2) + 4x^3/(3L) + x^2/2)

= (-3/L) (4(-L/2)^4/(4L^2) + 4(-L/2)^3/(3L) + (-L/2)^2/2)

= -L/16

-------------------------------------------------------

Finding ∫C^2 (-2x/L + 1)^2 x dx :

∫(3/L)(-2x/L + 1)^2 x dx = ∫(3/L) (2x/L)^2 * x - 4x^2/L + x dx

= (3/L) (4x^4/4L^2 - 4x^3/3L + x^2/2)

= (3/L) (4(L/2)^4/4L^2 - 4(L/2)^3/3L + (L/2)^2/2)

= L/16

------------------------------------------------------

Therefore:

<x> = ∫ψ* x ψ dx

<x> = ∫ψ^2 x dx

<x> = ∫C^2(2x/L + 1)^2 x dx + ∫C^2 (-2x/L + 1)^2 x dx

<x> = (-L/16) + L/16

<x> = 0

--------------------------------------------------------

---------------------------------------------------------

<x2> = ∫ψ* x ψ dx

= ∫ψ^2 x dx

= ∫C^2(2x/L + 1)^2 x2 dx + ∫C^2 (-2x/L + 1)^2 x2 dx

-----------------------------------------------------------

Finding ∫C^2(2x/L + 1)^2 x2 dx :

∫(3/L)(2x/L + 1)^2 x2 dx = ∫(3/L) ((2x/L)^2 * x^2 + 4x^3/L + x^2) dx

= (3/L) (4x^5/(5L^2) + 4x^4/(4L) + x^3/3)

= (-3/L) (4(-L/2)^5/(5L^2) + 4(-L/2)^4/(4L) + (-L/2)^3/3)

= L^2/80

-------------------------------------------------------

Finding ∫C^2 (-2x/L + 1)^2 x dx :

∫(3/L)(-2x/L + 1)^2 x2 dx = ∫(3/L) ((2x/L)^2 * x^2 - 4x^3/L + x^2) dx

= (3/L) (4x^5/(5L^2) - 4x^4/(4L) + x^3/3)

= (3/L) (4(L/2)^5/(5L^2) - 4(L/2)^4/(4L) + (L/2)^3/3)

= L^2/80

------------------------------------------------------

Therefore:

<x2> = ∫ψ* x^2 ψ dx

<x2> = ∫ψ^2 x^2 dx

<x2> = ∫C^2(2x/L + 1)^2 x2 dx + ∫C^2 (-2x/L + 1)^2 x2 dx

<x2> = L^2/80 + L^2/80

<x2> = L2/40

√(<x2>) = L/√40

Edit
Added Wed, 24 Feb '16
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