The continuity of wave function and their derivatives lead to the equations A sin ( ε ) = B e − μ − ε and ε A cos ( ε ) = − μ − ε B e − μ − ε

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Answer 1

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Chapter 35, Problem 59P

(a)

To determine

To Show: The continuity of wave function and their derivatives lead to the equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε

(b)

To determine

To Show: The equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε lead to single equation tan(ε)=−εμ−ε .

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A particle is in a three-dimensional box. The y length of the box is twice the x length, and the z length is one-third of the y length. (a) What is the energy difference between the first excited level and the ground level? (b) Is the first excited level degenerate? (c) In terms of the x length, where is the probability distribution the greatest in the lowest-energy level?

∆E ∆t ≥ ħTime is a parameter, not an observable. ∆t is some timescale over which the expectation value of an operator changes. For example, an electron's angular momentum in a hydrogen atom decays from 2p to 1s. These decays are relativistic, however the uncertainty principle is still valid, and we can use it to estimate uncertainties. The lifetime of hydrogen in the 2p state to decay to the Is ground state is 1.6 x 10-9 s. Estimate the uncertainty ∆E in energy of this excited state. What is the corresponding linewidth in angstroms?

∆E ∆t ≥ ħTime is a parameter, not an observable. ∆t is some timescale over which the expectation value of an operator changes. For example, an electron's angular momentum in a hydrogen atom decays from 2p to 1s. These decays are relativistic, however the uncertainty principle is still valid, and we can use it to estimate uncertainties. ∆E doesn't change in time, so when an excited state decays to the ground state (infinite lifetime, so no energy uncertainty), the energy uncertainty has to go somewhere. Usually, it’s in the frequency of a photon giving a width (through E = hν) to the transition line in an spectroscopy experiment. The linewidth of the 2p state in 9Be+ is 19.4 MHz. What is its lifetime? (Note: in the relativistic atom–photon system, the Hamiltonian is independent of time and both energy and its uncertainty are conserved.)

Answer 2

Question

Chapter 35, Problem 59P

(a)

To determine

To Show: The continuity of wave function and their derivatives lead to the equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε

(b)

To determine

To Show: The equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε lead to single equation tan(ε)=−εμ−ε .

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Answer 3

Question

Chapter 35, Problem 59P

(a)

To determine

To Show: The continuity of wave function and their derivatives lead to the equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε

(b)

To determine

To Show: The equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε lead to single equation tan(ε)=−εμ−ε .

Expert Solution & Answer

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Answer 4

Question

Chapter 35, Problem 59P

(a)

To determine

To Show: The continuity of wave function and their derivatives lead to the equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε

(b)

To determine

To Show: The equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε lead to single equation tan(ε)=−εμ−ε .

Expert Solution & Answer

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Answer 5

Chapter 35, Problem 59P

(a)

To determine

To Show: The continuity of wave function and their derivatives lead to the equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε

(b)

To determine

To Show: The equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε lead to single equation tan(ε)=−εμ−ε .

Answer 6

Chapter 35, Problem 59P

(a)

To determine

To Show: The continuity of wave function and their derivatives lead to the equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε

Answer 7

Chapter 35, Problem 59P

(a)

To determine

To Show: The continuity of wave function and their derivatives lead to the equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε

Answer 8

(b)

To determine

To Show: The equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε lead to single equation tan(ε)=−εμ−ε .

Answer 9

(b)

To determine

To Show: The equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε lead to single equation tan(ε)=−εμ−ε .

Answer 10

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Answer 11

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Answer 12

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Answer 13

Ch. 35.1 - Prob. 35.1GI Ch. 35.2 - Prob. 35.2GI Ch. 35.3 - Prob. 35.3GI Ch. 35.3 - Prob. 35.4GI Ch. 35.3 - Prob. 35.5GI Ch. 35.4 - Prob. 35.6GI Ch. 35 - Prob. 1FTD Ch. 35 - Prob. 2FTD Ch. 35 - Prob. 3FTD Ch. 35 - Prob. 4FTD

The continuity of wave function and their derivatives lead to the equations A sin ( ε ) = B e − μ − ε and ε A cos ( ε ) = − μ − ε B e − μ − ε

Solution Summary: The author explains the continuity of wave function and their derivatives lead to the equations Amathrmsinleft and

To Show: The continuity of wave function and their derivatives lead to the equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε

To Show: The equations Asin(ε)=Be−μ−ε and εAcos(ε)=−μ−εBe−μ−ε lead to single equation tan(ε)=−εμ−ε .

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Chapter 35 Solutions