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
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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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![For a Hydrogen Atom these energies can be calculated by the following equation:
1
Latulan he - - ATelacirun- R
n
Ru-2.179 x 10-18J
(Hydrogen atom)
The Helium Ion, Het, 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 10-18J
Eçloetron -
RH - 8.7149 x 1018J
n
(Helium Iou)
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 ion, can be calculated by the equation:
1
+ Ephoton - hc -- AEelectron - RHet -----
n? (Helitun Ion)
1
8.7149 x 1018J.
n
ni) (Helium Ion)
Calculate the electron energies (Eelectron) of the first 4 energy levels in the helium ion (He*):
n =
[ Select ]
x 10-18 J
n2 =
[ Select]
v x 10-18 J
n3 =
[ Select]
v x 1018 J
n4 =
[ Select)
v x 10-18 J
Determine the electron transitions, (n; > nf), 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:
E121,57 nm -
| Select]
x 10-18 J
E468.90 nm =
I Select]
х 10-18 ]
Using the 4 Energy Levels of the Helium Ion, determine the electron transitions, (nj > nf), that release the photons with the wavelengths (A) of 121.57 nm and 468.90 nm.
(A) = 121.57 nm:
| Select]
[ Select 1
Ninitial =
nfinal =
(A) = 468.90 nm:
ninitial =
| Select ]
nfinal
[ Select]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F66173c1a-f883-4b2a-9c4a-34112153fb2f%2F6315b7b6-e9c0-484c-9665-a4118922c8f6%2F5g7zrh_processed.png&w=3840&q=75)
Transcribed Image Text:For a Hydrogen Atom these energies can be calculated by the following equation:
1
Latulan he - - ATelacirun- R
n
Ru-2.179 x 10-18J
(Hydrogen atom)
The Helium Ion, Het, 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 10-18J
Eçloetron -
RH - 8.7149 x 1018J
n
(Helium Iou)
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 ion, can be calculated by the equation:
1
+ Ephoton - hc -- AEelectron - RHet -----
n? (Helitun Ion)
1
8.7149 x 1018J.
n
ni) (Helium Ion)
Calculate the electron energies (Eelectron) of the first 4 energy levels in the helium ion (He*):
n =
[ Select ]
x 10-18 J
n2 =
[ Select]
v x 10-18 J
n3 =
[ Select]
v x 1018 J
n4 =
[ Select)
v x 10-18 J
Determine the electron transitions, (n; > nf), 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:
E121,57 nm -
| Select]
x 10-18 J
E468.90 nm =
I Select]
х 10-18 ]
Using the 4 Energy Levels of the Helium Ion, determine the electron transitions, (nj > nf), that release the photons with the wavelengths (A) of 121.57 nm and 468.90 nm.
(A) = 121.57 nm:
| Select]
[ Select 1
Ninitial =
nfinal =
(A) = 468.90 nm:
ninitial =
| Select ]
nfinal
[ Select]
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