Hwksolns12TA

# Hwksolns12TA - 4 0.2 The rel at ionshi p bet ween pr obabil...

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40.2. The relationship between probability and probability density is similar to the relationship between mass m and mass density ρ . Regions of higher mass density tell us where mass is concentrated. The mass itself is a more tangible quantity that depends both on the mass density and on the size of a specific piece of material. Similarly, probability density tells us regions in which a particle is more likely, or less likely, to be found. The probability is a definite number between 0 and 1. Probability depends both on the probability density and on the size of the specific region we are considering. 40.4. (a) The probability density is maximum at x ± 2 mm. (b) We cannot tell where the wave function is most positive; it could be at either x = 2 mm or x = −2 mm. It will be positive at one and negative at the other. 40.6. Particle 1 because it has a less definite Δx and therefore a more definite Δp = Δmv. Problems and Exercises 40.4. Model: The probability that the outcome will be A or B is the sum of P A and P B. The expected value is your best possible prediction of the outcome of an experiment. Solve: For each deck, there are 12 picture cards (4 Jacks, 4 Queens, and 4 Kings). Because the probability of drawing one card out of 52 cards is 1/52, the probability of drawing a card that is a picture card is 12/52 = 23.1%. The number of picture cards that will be drawn is 0.231 × 1000 = 231.

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40.7. Visualize: Combine Equations 40.10 and 40.11 to show that N is proportional to 2 ( ) . A x x δ | | 2 2 2 2 2 tot 2 1 1 1 1 tot (in at ) ( ) (in at ) ( ) N x x A x x N N x x A x x N δ δ δ δ = We are given 1 6000, N = 1 0 10 mm, x δ = . 1 ( ) 200 V m, A x = / 2 3000, N = and 2 0 20 mm. x δ = . We are not given N tot but it cancels anyway. Solve: Solve the above equation for 2 ( ) . A x 1 2 2 2 1 2 1 1 (in at ) (0 10 mm)(3000) ( ) ( ) (200 V m) 100 V m (in at ) (0 20 mm)(6000) x N x x A x A x x N x x δ δ δ δ . = = / = / . Assess: The answer is half of the wave amplitude at the other strip, which seems reasonable.
40.8. Solve: The probability that a photon arrives at this 0.10-mm-wide strip is Prob(in 0.10 mm at x ) ( ) 10 1.0 10 N P x x δ = = × where N is the number of photons detected in the strip and the total number of photons is 1.0 × 10 10 . We have ( )( )( ) 10 1 3 1.0 10 20 m 0.10 10 m N = × × N = 2.0 × 10 7

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40.11. Model: The probability of finding a particle at position x is determined by ( ) 2 . x ψ Solve: (a) The probability of detecting an electron is Prob(in δ x at x ) ( ) 2 . x x ψ δ = At x = 0 mm, the number of electrons landing is calculated as follows: ( ) ( ) ( ) ( ) 2 1 6 1 3 6 0 mm mm 0.010 mm 1.0 10 3333 1.0 10 N x N ψ δ = = × = × (b) Likewise, the number of electrons landing at x = 2.0 mm is ( ) ( ) ( ) ( ) 2 1 6 total 2.0 mm 0.111 mm 0.010 mm 1.0 10 1111 N xN ψ δ = = × =
40.10. Solve: ( ) 2 x x ψ δ is a probability, which is dimensionless. The units of x δ are m , so the units of ( ) 2 x ψ are m 1 and thus the units of ψ are m 1/2 .

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40.14. Model: The probability of finding a particle is determined by the probability density ( ) ( ) 2 P x x ψ = .
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