EP PHYSICS -MOD.MASTERING (18W)
5th Edition
ISBN: 9780136782490
Author: Walker
Publisher: PEARSON CO
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Chapter 31, Problem 82GP
(a)
To determine
To validate: The radii of the allowed orbits have the value
(b)
To determine
The speed of the particle in each allowed orbit.
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Chapter 31 Solutions
EP PHYSICS -MOD.MASTERING (18W)
Ch. 31.1 - Prob. 1EYUCh. 31.2 - Prob. 2EYUCh. 31.3 - Prob. 3EYUCh. 31.4 - Prob. 4EYUCh. 31.5 - Prob. 5EYUCh. 31.6 - Prob. 6EYUCh. 31.7 - Prob. 7EYUCh. 31 - Prob. 1CQCh. 31 - Prob. 2CQCh. 31 - Prob. 3CQ
Ch. 31 - Prob. 4CQCh. 31 - Prob. 5CQCh. 31 - Prob. 6CQCh. 31 - Prob. 7CQCh. 31 - Prob. 8CQCh. 31 - Prob. 9CQCh. 31 - Prob. 1PCECh. 31 - Prob. 2PCECh. 31 - Prob. 3PCECh. 31 - Prob. 4PCECh. 31 - Prob. 5PCECh. 31 - Prob. 6PCECh. 31 - Prob. 7PCECh. 31 - Prob. 8PCECh. 31 - Prob. 9PCECh. 31 - Prob. 10PCECh. 31 - Prob. 11PCECh. 31 - Prob. 12PCECh. 31 - Prob. 13PCECh. 31 - Prob. 14PCECh. 31 - Prob. 15PCECh. 31 - Prob. 16PCECh. 31 - Prob. 17PCECh. 31 - Prob. 18PCECh. 31 - Prob. 19PCECh. 31 - Prob. 20PCECh. 31 - Prob. 21PCECh. 31 - Prob. 22PCECh. 31 - Prob. 23PCECh. 31 - Prob. 24PCECh. 31 - Prob. 25PCECh. 31 - Prob. 26PCECh. 31 - Prob. 27PCECh. 31 - Prob. 28PCECh. 31 - Prob. 29PCECh. 31 - Prob. 30PCECh. 31 - Prob. 31PCECh. 31 - Prob. 32PCECh. 31 - Prob. 33PCECh. 31 - Prob. 34PCECh. 31 - Prob. 35PCECh. 31 - Prob. 36PCECh. 31 - Prob. 37PCECh. 31 - Prob. 38PCECh. 31 - Prob. 39PCECh. 31 - Prob. 40PCECh. 31 - Prob. 41PCECh. 31 - Prob. 42PCECh. 31 - Prob. 43PCECh. 31 - Prob. 44PCECh. 31 - Prob. 45PCECh. 31 - Prob. 46PCECh. 31 - Prob. 47PCECh. 31 - Prob. 48PCECh. 31 - Prob. 49PCECh. 31 - Prob. 50PCECh. 31 - Prob. 51PCECh. 31 - Prob. 52PCECh. 31 - Give the electronic configuration for the ground...Ch. 31 - Prob. 54PCECh. 31 - Prob. 55PCECh. 31 - Prob. 56PCECh. 31 - The configuration of the outer electrons in Ni is...Ch. 31 - Prob. 58PCECh. 31 - Prob. 59PCECh. 31 - Prob. 60PCECh. 31 - Prob. 61PCECh. 31 - Prob. 62PCECh. 31 - Prob. 63PCECh. 31 - Prob. 64PCECh. 31 - Prob. 65PCECh. 31 - Prob. 66PCECh. 31 - Prob. 67PCECh. 31 - Prob. 68GPCh. 31 - Prob. 69GPCh. 31 - Prob. 70GPCh. 31 - Prob. 71GPCh. 31 - Prob. 72GPCh. 31 - Prob. 73GPCh. 31 - Prob. 74GPCh. 31 - Prob. 75GPCh. 31 - Prob. 76GPCh. 31 - Prob. 77GPCh. 31 - Prob. 78GPCh. 31 - Prob. 79GPCh. 31 - Prob. 80GPCh. 31 - Prob. 81GPCh. 31 - Prob. 82GPCh. 31 - Prob. 83GPCh. 31 - Prob. 84PPCh. 31 - Prob. 85PPCh. 31 - Prob. 86PPCh. 31 - Prob. 87PPCh. 31 - Prob. 88PPCh. 31 - Prob. 89PP
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- (a) Using the Pauli exclusion principle and the rules relating the allowed values of the quantum numbers (n,l,ml,ms), prove that the maximum number of electrons in a subshell is 2n2. (b) In a similar manner, prove that the maximum number of electrons in a shell is 2n2.arrow_forwardSeven possible transitions are identified on the energy-level diagram as shown. For each, is this an allowed transition? If allowed, is it an emission or an absorption transition, and is the photon infrared, visible, or ultraviolet? If not allowed, why not?arrow_forwardA physicist is watching a 15-kg orangutan at a zoo swing lazily in a tire at the end of a rope. He (the physicist) notices that each oscillation takes 3.80 s and hypothesizes that the energy is quantized. (a) What is the difference in energy in joules between allowed oscillator states? (b) What is the value of n for a state where the energy is 6.10 J?arrow_forward
- The K series of the discrete x-ray spectrum of tungsten contains wavelengths of 0.018 5 nm, 0.020 9 nm, and 0.021 5 nm. The K-shell ionization energy is 69.5 keV. (a) Determine the ionization energies of the L, M, and N shells. L shell _keV M shell keV N shell _keVarrow_forwardThe light observed that is emitted by a hydrogen atom is explained by a simple model of its structure with one proton in its nucleus and an electron bound to it, but only with internal energies of the atom satisfying EH=−RH/n2EH=−RH/n2 where RHRH is the Rydberg constant and nn is an integer such as 1, 2, 3 ... and so on. When a hydrogen atom in an excited state emits light, the photon carries away energy and the atom goes into a lower energy state. Be careful about units. The Rydberg constant in eV is 13.605693009 eV That would be multiplied by the charge on the electron 1.602× 10-19 C to give 2.18× 10-18 J A photon with this energy would have a frequency f such that E=hf. Its wavelength would be λ = c/f = hc/E. Sometimes it is handy to measure the Rydberg constant in units of 1/length for this reason. You may see it given as 109737 cm-1 if you search the web, so be aware that's not joules. The following questions are intended to help you understand the connection between…arrow_forwardShow that the following 4 lines in the Lyman series can be predicted: 91.127, 97.202, 102.52, and 121.57 nm.arrow_forward
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