Fundamentals Of Physics 11th Edition Loose-leaf Print Companion Volume 2 With Wileyplus Card Set
11th Edition
ISBN: 9781119463252
Author: David Halliday
Publisher: John Wiley and Sons
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Chapter 38, Problem 61P
To determine
To show:
Eq.38-27 is a solution of Schrödinger’s equation, with
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Find the angular momentum and kinetic energy in the z axis for the
(cos(30))*e(iΦ)+(sin(30))*e(-iΦ) wave function.
*24 Figure 39-30 shows a two-dimen-
sional, infinite-potential well lying in an
xy plane that contains an electron. We
probe for the electron along a line that
bisects L, and find three points at which
the detection probability is maximum. Figure 39-30 Problem 24.
Those points are separated by 2.00 nm.
Then we probe along a line that bisects L, and find five points at
which the detection probability is maximum. Those points are sep-
arated by 3.00 nm. What is the energy of the electron?
16 For three experiments, Fig. 38-25
gives the transmission coefficient T
for electron tunneling through a po-
tential barrier, plotted versus barrier
thickness L. The de Broglie wave-
lengths of the electrons are identical
in the three experiments. The only
difference in the physical setups is
the barrier heights U. Rank the
three experiments according to U,
greatest first.
T:
Figure 38-25 Question 16.
Chapter 38 Solutions
Fundamentals Of Physics 11th Edition Loose-leaf Print Companion Volume 2 With Wileyplus Card Set
Ch. 38 - Prob. 1QCh. 38 - Prob. 2QCh. 38 - Prob. 3QCh. 38 - Prob. 4QCh. 38 - Prob. 5QCh. 38 - Prob. 6QCh. 38 - Prob. 7QCh. 38 - Prob. 8QCh. 38 - Prob. 9QCh. 38 - Prob. 10Q
Ch. 38 - Prob. 11QCh. 38 - Prob. 12QCh. 38 - Prob. 13QCh. 38 - Prob. 14QCh. 38 - Prob. 15QCh. 38 - Prob. 16QCh. 38 - Prob. 1PCh. 38 - Prob. 2PCh. 38 - Prob. 3PCh. 38 - Prob. 4PCh. 38 - Prob. 5PCh. 38 - Prob. 6PCh. 38 - Prob. 7PCh. 38 - Prob. 8PCh. 38 - Prob. 9PCh. 38 - Prob. 10PCh. 38 - Prob. 11PCh. 38 - Prob. 12PCh. 38 - Prob. 13PCh. 38 - Prob. 14PCh. 38 - Prob. 15PCh. 38 - Prob. 16PCh. 38 - Prob. 17PCh. 38 - Prob. 18PCh. 38 - Prob. 19PCh. 38 - Prob. 20PCh. 38 - Prob. 21PCh. 38 - Prob. 22PCh. 38 - Prob. 23PCh. 38 - Prob. 24PCh. 38 - Prob. 25PCh. 38 - Prob. 26PCh. 38 - Prob. 27PCh. 38 - Prob. 28PCh. 38 - Prob. 29PCh. 38 - Prob. 30PCh. 38 - Prob. 31PCh. 38 - Prob. 32PCh. 38 - Prob. 33PCh. 38 - Prob. 34PCh. 38 - Prob. 35PCh. 38 - Prob. 36PCh. 38 - Prob. 37PCh. 38 - Prob. 38PCh. 38 - Prob. 39PCh. 38 - Prob. 40PCh. 38 - Prob. 41PCh. 38 - Prob. 42PCh. 38 - Prob. 43PCh. 38 - Prob. 44PCh. 38 - Prob. 45PCh. 38 - Prob. 46PCh. 38 - Prob. 47PCh. 38 - Prob. 48PCh. 38 - Prob. 49PCh. 38 - Prob. 50PCh. 38 - Prob. 51PCh. 38 - Prob. 52PCh. 38 - Prob. 53PCh. 38 - Prob. 54PCh. 38 - Prob. 55PCh. 38 - Prob. 56PCh. 38 - Prob. 57PCh. 38 - Prob. 58PCh. 38 - Prob. 59PCh. 38 - Prob. 60PCh. 38 - Prob. 61PCh. 38 - Prob. 62PCh. 38 - Prob. 63PCh. 38 - Prob. 64PCh. 38 - Prob. 65PCh. 38 - Prob. 66PCh. 38 - Prob. 67PCh. 38 - Prob. 68PCh. 38 - Prob. 69PCh. 38 - Prob. 70PCh. 38 - Prob. 71PCh. 38 - Prob. 72PCh. 38 - Prob. 73PCh. 38 - Prob. 74PCh. 38 - Prob. 75PCh. 38 - Prob. 76PCh. 38 - Prob. 77PCh. 38 - Prob. 78PCh. 38 - Prob. 79PCh. 38 - Prob. 80PCh. 38 - Prob. 81PCh. 38 - Prob. 82PCh. 38 - Prob. 83PCh. 38 - Prob. 84PCh. 38 - Prob. 85PCh. 38 - Prob. 86PCh. 38 - Prob. 87PCh. 38 - Prob. 88PCh. 38 - Prob. 89PCh. 38 - Prob. 90P
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- Consider a potential energy barrier like that of Fig. 39-13a but whose height Enot is 5.9 eV and whose thickness L is 0.84 nm. What is the energy of an incident electron whose transmission probability is 0.0040? eV pot E Electron T (a) (8) Figure 39-13. Probability density 4 ( En ergyarrow_forwardCheck which of the wavefunctions below represents a physically possible solution to the Schrodinger equation for a free electron? 1:) ▼ (r, t) = e'(2x102z-wt) 2:) V (x, t) = e'(2x10%z-wt) 3:) V (x, t) = e'(2x1014-wt) %3D 4:) V (r, t) = e"(2×1015z–ut) 5:) V (x, t) = e'(2x100z-wt)arrow_forwardA particle with mass m is moving in three-dimensions under the potential energy U(r), where r is the radial distance from the origin. The state of the particle is given by the time-independent wavefunction, Y(r) = Ce-kr. Because it is in three dimensions, it is the solution of the following time-independent Schrodinger equation dıp r2 + U(r)µ(r). dr h2 d EÞ(r) = 2mr2 dr In addition, 00 1 = | 4ar?y? (r)dr, (A(r)) = | 4r²p²(r)A(r)dr. a. Using the fact that the particle has to be somewhere in space, determine C. Express your answer in terms of k. b. Remembering that E is a constant, and the fact that p(r) must satisfy the time-independent wave equation, what is the energy E of the particle and the potential energy U(r). (As usual, E and U(r) will be determined up to a constant.) Express your answer in terms of m, k, and ħ.arrow_forward
- Show that the wave function ψ = Ae i(i-ωt) is a solution to the Schrödinger equation (as shown), where k = 2π/λ and U = 0.arrow_forward.8 O An electron is trapped in a one-dimensional infinite well and is in its first excited state. Figure 39-27 indicates the five longest wavelengths of light that the electron could absorb in transitions from this initial state via a single photon absorption: A, = 80.78 nm, A, = 33.66 nm, A = 19.23 nm, A, = 12.62 nm, and A, = 8.98 nm. What is the width of the potential well? %3D %3! 2 (nm) Figure 39-27 Problem 8.arrow_forwardIn your research on new solid-state devices, you are studying a solid-state structure that can be modeled accurately as an electron in a one-dimensional infinite potential well (box) of width L. In one of your experiments, electromagnetic radiation is absorbed in transitions in which the initial state is the n = 1 ground state. You measure that light of frequency f = 9.0x 1014 Hz is absorbed and that the next higher absorbed frequency is 16.9 x 1014 Hz. (a) What is quantum number n for the final state in each of the transitions that leads to the absorption of photons of these frequencies? (b) What is the width L of the potential well? (c) What is the longest wavelength in air of light that can be absorbed by an electron if it is initially in the n = 1 state?arrow_forward
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