For a Hydrogen Atom these energies can be calculated by the following equation: Lan he -- Ar - R R-2.179 x 10g (Hydrogen atom) The Helium Ion, He*, has energy levels similar to those of the hydrogen atom. The helium ion, like the hydrogen atom, has only one electron. This eliminates electron-electron interactions, and allows us to calculate the energies of the electron in each of the energy levels (n - 1, 2, 3 .) of the helium ion: 8.7149 x 10 Ede R-8.7149 x 10 JHelian len) The energy released as a photon (Ephoton), when the electron in the helium ion (He") transitions from higher energy levels to lower energy levels in the lon, can be calculated by the equation: Egtut- he --AEd - R n Jolelam lon) = 8.7149 x 10ag. n (Helium lon) Calculate the electron energies (Eelectron) of the first 4 energy levels in the helium ion (He*): ISelest v x 10-18 ) - ISel v x 1018) v x 1018) - I Select v x 1018 ) Determine the electron transitions, (n n), that release photons with the following wavelengths (A): • wavelength (A) = 121.57 nm • wavelength (A) = 468.90 nm Calculate the Energy of the photon with a wavelength of: E12157 nm - ISelect v x 1018) Esaso m - Select) x 1018) Using the 4 Energy Levels of the Helium Ion, determine the electron transitions, (n → n), that release the photons with the wavelengths (A) of 121.57 nm and 468.90 nm. (A) = 121.57 nm: nitial = Select ial= Selec (A) = 468.90 nm: ISelect ISelec Ninitial = Nfinal =
For a Hydrogen Atom these energies can be calculated by the following equation: Lan he -- Ar - R R-2.179 x 10g (Hydrogen atom) The Helium Ion, He*, has energy levels similar to those of the hydrogen atom. The helium ion, like the hydrogen atom, has only one electron. This eliminates electron-electron interactions, and allows us to calculate the energies of the electron in each of the energy levels (n - 1, 2, 3 .) of the helium ion: 8.7149 x 10 Ede R-8.7149 x 10 JHelian len) The energy released as a photon (Ephoton), when the electron in the helium ion (He") transitions from higher energy levels to lower energy levels in the lon, can be calculated by the equation: Egtut- he --AEd - R n Jolelam lon) = 8.7149 x 10ag. n (Helium lon) Calculate the electron energies (Eelectron) of the first 4 energy levels in the helium ion (He*): ISelest v x 10-18 ) - ISel v x 1018) v x 1018) - I Select v x 1018 ) Determine the electron transitions, (n n), that release photons with the following wavelengths (A): • wavelength (A) = 121.57 nm • wavelength (A) = 468.90 nm Calculate the Energy of the photon with a wavelength of: E12157 nm - ISelect v x 1018) Esaso m - Select) x 1018) Using the 4 Energy Levels of the Helium Ion, determine the electron transitions, (n → n), that release the photons with the wavelengths (A) of 121.57 nm and 468.90 nm. (A) = 121.57 nm: nitial = Select ial= Selec (A) = 468.90 nm: ISelect ISelec Ninitial = Nfinal =
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter7: Quantum Theory Of The Atom
Section: Chapter Questions
Problem 7.24QP: Investigating Energy Levels Consider the hypothetical atom X that has one electron like the H atom...
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