Chapter 39: Q.34 (page 1138)

The probability density for finding a particle at position x is P1x2 = • a 11 - x2 -1 mm … x 6 0 mm b11 - x2 0 mm … x … 1 mm and zero elsewhere. a. You will learn in Chapter 40 that the wave function must be a continuous function. Assuming that to be the case, what can you conclude about the relationship between a and b? b. Determine values for a and b. c. Draw a graph of the probability density over the interval -2 mm … x … 2 mm. d. What is the probability that the particle will be found to the left of the origin?

Short Answer

Expert verified

The wave function of a particle confined between $x = - 4 m m a n d x = 4 m m$

Step by step solution

(a)The wave function of a particle confirmed between x=-4mm and x=4mm is shown in the figure below

Using the concept of similar triangles, we have

$\frac{c}{ψ (x)} = \frac{4}{x} ψ (x) = \frac{c}{4} x \int_{- \infty}^{+ \infty} {|ψ (x)|}^{2} d x = 1 \int_{- 4}^{+ 4} {|ψ (x)|}^{2} d x = 1 \int_{- 4}^{0} {|ψ (x)|}^{2} d x + \int_{0}^{4} {|ψ (x)|}^{2} d x = 1$

The statement particle has to land somewhere on the x-axis is expressed mathematically as

Now in the case

$\int_{- 4}^{0} {(\frac{C}{4} x)}^{2} d x + \int_{0}^{4} {(\frac{C}{4} x)}^{2} d x = 1 \frac{C^{2}}{16} \int_{- 4}^{0} x^{2} d x + \int_{0}^{4} \frac{C^{2}}{16} x^{2} d x = 1 \frac{C^{2}}{16} {[\frac{x^{3}}{3}]}_{- 4}^{0} + \frac{C^{2}}{16} {[\frac{x^{3}}{3}]}_{0}^{4} = 1 \frac{C^{2}}{16} (\frac{64}{3}) + \frac{C^{2}}{16} (\frac{64}{4 = 1}) = 1 \frac{4 C^{2}}{3} + \frac{4 C^{2}}{3} = 1 8 C^{2}$

(b)The probability density curve ψx2 of the particle is shown in figure below

(c)

(d)The probability of finding the particle in the interval -2.0mm ≤x≤2.0mm is

$P (- 2.0 m m \leq x \leq 2.0 m m) = \int_{- 2}^{2} {|ψ (x)|}^{2} d x = \int_{- 2}^{2} {(\frac{C}{4} x)}^{2} d x = \frac{C^{2}}{16} \int_{- 2}^{2} x^{2} d x = \frac{3}{128} {[\frac{x^{3}}{3}]}_{- 2}^{2} = (\frac{3}{128}) (\frac{16}{3}) = \frac{1}{8} = 0.125$