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

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

THEREFORE, THE VALUE OF PROBABILITY IS 40.2%

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Short Answer

Step by step solution

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

Thus the step was

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

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