Chapter 6: Q49P (page 1356)

Repeat Problem 40.48 for a particle in the first excited level.

Expert verified

The probability that the particle will be found in the interval x to x+ dxfor x= L/4 is $\frac{dx}{L}$
The probability that the particle will be found in the interval x to x+ dxfor x= L/2 is $\frac{2 dx}{L}$
The probability that the particle will be found in the interval x to x+ dxfor x= 3L/4 is $\frac{dx}{L}$

Step by step solution

The probability distribution between x and x+dx is given as,

$P = {|ψ|}^{2} dx$

Also, the first excited state (n = 2) wave function for particle in a box is given as,

$ψ_{1} \sqrt{\frac{2}{L}} \sin \frac{2 πx}{L}$

Thus, the wave function at x = L/4,

${|ψ|}^{2} dx = (\frac{2}{L}) \sin^{2} (\frac{2 π}{L} \frac{L}{4}) dx = (\frac{2}{L}) \sin^{2} (\frac{π}{2}) dx = \frac{2 dx}{L}$

Thus, P = $\frac{2 dx}{L}$

Similarly, repeat the above calculation for x = L/2, the wave function is,

${|ψ|}^{2} dx = (\frac{2}{L}) \sin^{2} (\frac{2 π}{L} \frac{L}{2}) dx = (\frac{2}{L}) \sin^{2} (π) dx = 0$

Thus, P = 0

Similarly, repeat the above calculation for x = 3L/4, the wave function is,

${|ψ|}^{2} dx = (\frac{2}{L}) \sin^{2} (\frac{2 π}{L} \frac{3 L}{4}) dx = (\frac{2}{L}) \sin^{2} (\frac{3 π}{2}) dx = \frac{2 dx}{L}$

Thus, P = $\frac{2 dx}{L}$

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