This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 35.1 QUANTUM MECHANICS EXERCISES Section 35.2 The Schrödinger Equation 10. The onedimensional wave function is related to the probability by Equation 35.2, 2 ( ) . dP x dx ψ = Since probability (a pure number) is dimensionless, the units of ψ must be the square root of the inverse of the units of x (a length), or 1/ 2 (meters) . − 11. INTERPRET We are given the wave function of a particle and asked about its probability distribution. DEVELOP Since the quantity 2 ( ) x ψ represents the probability density of finding the particle, the particle is most likely to be found at the position where the probability density 2 ( ) x ψ is a maximum. EVALUATE (a) The maximum of 2 2 2 2 2 / ( ) x a x A e ψ − = is at x = (calculate 2 ( )/ d dx ψ = for corroboration). (b) The probability density 2 ( ) x ψ falls to half its maximum value 2 2 ( (0) ) A ψ = when 2 2 2 / 1/2 , x a e − = or 1 2 ln2 0.589 . x a a = ± = ± ASSESS The probability distribution is shown below. Note that 2 2 2 2 2 / ( )/ x a x A e ψ − = peaks at x = and is halved at / 0.589. x a = ± 12. INTERPRET We normalize the wave function that is given graphically in Figure 35.19. The particle must be somewhere , so the normalization constant is chosen such that the integral of the wave function over all space is one. DEVELOP The probability of finding the particle within dx of x is 2 ( ) ( ). P x x ψ = We will write an equation for the wave function, square it, and integrate. Then we’ll set the value of A such that the integral over the entire region is one. EVALUATE For 2 , L x ≤ ≤ 2 ( ) . A L x x ψ = Since the wave function is symmetric, we’ll just integrate from x = to 2 L x = and multiply by 2 to obtain the integral over all space. 3 2 2 2 2 / 2 2 2 4 3 1 2 3 3 L A x A L A L dx A L L L = = = → = ∫ ASSESS The actual wave function for the region 2 L x ≤ ≤ is then 3/ 2 2 3 ( ) . L x x ψ = 13. INTERPRET We use the normalization constant and the wave function from Exercise 12 to find the probability of finding the particle in the region 4 . L x ≤ ≤ The probability of finding a particle in a region dx is 2 . dx Ψ DEVELOP We integrate the square of the normalized wave function 3/ 2 2 3 ( ) L x x ψ = from x = to 4 L x = to find the probability of finding the particle in the region 4 . L x ≤ ≤ 35 35.2 Chapter 35 EVALUATE 3 / 4 2 3 3 3 12 12 1 1 3 16 4 L L P x dx L L = = = ∫ ASSESS There is a one in sixteen chance of finding the particle in the region described. The region is a quarter of the entire region where the particle could be, but the wave function is smaller in this region than elsewhere so the probability is less than one in four....
View
Full
Document
This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.
 Fall '10
 Pheong

Click to edit the document details