Chapter 39: Q.37 (page 1139)

Consider the electron wave function
$ψ (X) = \{\begin{cases} c \sin (\frac{2 πx}{L}) 0 \leq x \leq L \\ 0 x < 0 o r x > L \end{cases}$
a. Determine the normalization constant c. Your answer will be in terms of L.
b. Draw a graph of $ψ (x)$ over the interval -L $\leq$ x $\leq$ 2L.
c. Draw a graph of ${|ψ (x)|}^{2}$ over the interval -L $\leq$ x $\leq$ 2L. d. What is the probability that an electron is in the interval 0 $\leq$ x $\leq$ L/3?

Short Answer

Expert verified

The Value of Probability is $40.2 %$ .

Step by step solution

Use the equations of normalization to determine the value of constant and the formula of probability to determine its value

Sub part (a) step1:

For the probability interpretation of $ψ (x)$ to make sense the wave function must satisfy the following equation

$\int_{- \infty}^{+ \infty} {|ψ (x)|}^{2} d x = 1$

The above integrals is expanded as follows $\int_{- \infty}^{0} {|ψ (x)|}^{2} d x + \int_{0}^{1} {|ψ (x)|}^{2} d x + \int_{1}^{+ \infty} {|ψ (x)|}^{2} d x = 1$

The wave function is defined only in the region $0 \leq x \leq L$ therefore substitute csin $(\frac{2 πr}{L})$ for thye second integral and zero for the rest of regions

$0 + \int_{0}^{L} {[c \sin (\frac{2 πx}{L})]}^{2} d x + 0 = 1 c^{2} \int_{0}^{L} {[c \sin (\frac{2 πx}{L})]}^{2} d x = 1$

consider the following trigonometric relation

$\sin^{2} θ = \frac{1 - \cos θ}{2}$

Hence substitute $\frac{1 - \cos [2 (\frac{2 πx}{L})]}{2}$ for sin² $(\frac{2 πx}{L})$ AND SOLVE FOR C

$C^{2} \int_{0}^{L} \frac{1 - \cos [2 (\frac{2 πx}{L})]}{2} d x = 1 C^{2} \int_{0}^{L} 1 - \cos (\frac{4 πx}{L}) d x = 1$

Thus the step was

${[x]}_{0}^{L} - {[\frac{\sin (\frac{4 πx}{L})}{\frac{4 π}{L}}]}_{0}^{L} = \frac{2}{C^{2}} L - \frac{L}{4 π} [\sin (\frac{4 πx}{L}) - \sin 0] = \frac{2}{C^{2}} L - 0 = \frac{2}{C^{2}} C = \sqrt{\frac{2}{L}}$

Sub part (b) step2:The graph of ψx over the interval-L≤X≤2L is represented as follows

Subpart (c) step 3:The graph of ψx2 over the interval -L≤x≤2L is represented as follows

$P = \int {|ψ (x)|}^{2} d x P = \int_{0}^{\frac{L}{3}} {[\sqrt{\frac{2}{L} \sin (\frac{2 πx}{l})}]}^{2} d x = \frac{1}{L} [\frac{L}{3} - \frac{L}{4 π} (- 0.866)] = 0.402 = 40.2 %$