Question

The wavefront for a particle is defined by:
Ψ(x)= {Ncos(2πx/L) for -L/4≤x≤L/4
{0 otherwise
Determine:
i) the normalisation constant N
ii) the probability that the particle will be found between x=0 and x=L/8.

Answer 1

Answer on Question #84512 Physics / Atomic and Nuclear Physics

The wave front for a particle is defined by:

$$\Psi (x) = \left\{ \begin{array}{l l} N \cos \left(\frac {2 \pi x}{L}\right), & \text {for} - L / 4 \leq x \leq L / 4 \\ 0, & \text {otherwise} \end{array} \right.$$

Determine:

i) the normalization constant $N$

ii) the probability that the particle will be found between $x = 0$ and $x = L / 8$ .

Solution:

i) The normalization condition

$$\int_{-\infty}^{\infty} |\Psi(x)|^2 dx = 1$$

So

$$\begin{array}{l} 1 = N ^ {2} \int_{-\frac{L}{4}}^{\frac{L}{4}} \cos^2 \left(\frac {2 \pi x}{L}\right) dx \\ = \frac {N ^ {2}}{2} \int_{-\frac{L}{4}}^{\frac{L}{4}} \left(1 + \cos \left(\frac {4 \pi x}{L}\right)\right) dx = \frac {N ^ {2} L}{4} \\ \end{array}$$

Thus, the normalization constant

$$N = \sqrt{\frac{4}{L}}$$

ii) The probability that the particle will be found between $x = 0$ and $x = L / 8$

$$\begin{array}{l} P = \int_{0}^{\frac{L}{8}} |\Psi(x)|^2 dx = \frac{4}{L} \int_{0}^{\frac{L}{8}} \cos^2 \left(\frac {2 \pi x}{L}\right) dx = \frac{2}{L} \int_{0}^{\frac{L}{8}} \left(1 + \cos \left(\frac {4 \pi x}{L}\right)\right) dx \\ = \frac {2}{L} \left(\frac {L}{8} + \frac {L}{4 \pi} \sin \left(\frac {4 \pi x}{L}\right) \Bigg | _ {0} ^ {\frac {L}{8}}\right) = \frac {2}{L} \left(\frac {L}{8} + \frac {L}{4 \pi} \sin \left(\frac {\pi}{2}\right)\right) = \frac {1}{2} \left(\frac {1}{2} + \frac {1}{\pi}\right) = 0.41 \\ \end{array}$$

Answer: $N = \sqrt{\frac{4}{L}}$ , $P = 0.41$

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