Hwksolns12TA

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

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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 xx δ || 22 2 tot 2 11 tot (in at ) (in at ) N Ax x N N x N = We are given 1 6000, N = 1 010±mm , x =. 1 () 2 0 0 Vm , = / 2 3000, N = and 2 0 20 mm. x We are not given N tot but it cancels anyway. Solve: Solve the above equation for 2 . A x 12 2 21 1 (in at ) (0 10 mm)(3000) ( ) ( ) (200 V m) 100 V m (in at ) (0 20 mm)(6000) xN x x x x δδ . == / = / . 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 Px 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 = At x = 0 mm, the number of electrons landing is calculated as follows: 2 16 1 3 6 0 mm mm 0.010 mm 1.0 10 3333 1.0 10 N xN ψδ =⇒ = × = × (b) Likewise, the number of electrons landing at x = 2.0 mm is ( ) ( ) 2 total 2.0 mm 0.111 mm 0.010 mm 1.0 10 1111 Nx N == × =
40.10. Solve: () 2 x x ψ δ is a probability, which is dimensionless. The units of x are m , so the units of 2 x are m 1 and thus the units of are m 1/2 .

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40.14. Model: The probability of finding a particle is determined by the probability density ( ) ( ) 2 Px x ψ = . Solve: (a) The normalization condition for a wave function: () 2 x dx −∞ = area under the curve = 1. In the present case, the area under the 2 x -versus- x graph is 2 nm . a Hence, 1 1 2 nm . a = (b) Each point on the ( x ) graph is the square root of the corresponding point on the | ( x )| 2 graph. Where the | ( x )| 2
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## This note was uploaded on 12/26/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.

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

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