Bohr Model: Calculating Wavelengths of Emitted Photons

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SUMMARY

The discussion focuses on calculating the wavelengths of emitted photons from a singly ionized helium atom (He+) transitioning from the n = 3 excited state to the ground state (n = 1). The process involves two photon emissions: first from n = 3 to n = 2, and then from n = 2 to n = 1. The wavelengths of these photons can be calculated using the formula for hydrogen-like ions, incorporating the Rydberg constant for He+ (109722.263 1/cm) and the Z^2 factor, which results in a wavelength of 303.777 Angstroms for the transition from n = 2 to n = 1.

PREREQUISITES
  • Understanding of the Bohr model of the atom
  • Familiarity with quantum mechanics concepts, specifically energy levels
  • Knowledge of the Rydberg formula for hydrogen-like ions
  • Basic proficiency in calculating wavelengths using the formula wavelength = E/hc
NEXT STEPS
  • Study the Rydberg formula and its applications to various hydrogen-like ions
  • Learn about the significance of the Z^2 factor in spectral calculations
  • Explore the concept of photon energy and its relationship to wavelength
  • Investigate the differences in spectral lines between hydrogen and helium-like atoms
USEFUL FOR

Students and professionals in physics, particularly those focusing on atomic structure, quantum mechanics, and spectroscopy. This discussion is beneficial for anyone looking to deepen their understanding of photon emissions in hydrogen-like atoms.

tigerguy
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Hi,

I'm having trouble on the following question:
A singly ionized helium atom is in the ground state. It absorbs energy and makes a transition to the n = 3 excited state. The ion returns to the ground state by emitting two photons. What are the wavelengths of the lowest energy photon and highest energy photon?

I think I have to use the conservation of energy approach where nf=1 and ni=3, and also account for the energy lost to the photon. But, I'm lost of how to actually do this, and how to differentiate between the lowest energy photon and highest energy photon (I figure it has something to do with the length of the wavelength - longer wavelength leading to less energy, and vice-versa).

Thanks for your help!
 
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I think that problem suggests that the electron returns to the n = 1
state via the n = 2 state. So when the electron goes from the n = 3
state to the n = 2 state, it emits 1 photon then emits another photon
going from the n = 2 state to the n = 1 state.
 
what is the energy difference between level 3 and level 2? level 2 and level 1? wavelength=E/hc
 
The wavelength will be very close to 1/4 of that of the corresponding
transitions for hydrogen due to the Z^2 factor. For hydrogen the wavelengths in Angstroms are: N2 to N1 - 1215.66
N3 to N2 - 6562.79

The wavelength for He (He+) N2 to N1 - 303.777 Angstroms

The general formula for hydrogen-like ions is wavenumber (1/wavelength in centimeters) =

RZ^2 * (1/n2 ^2 - 1/n1^2) where n2 is the principal quantum
number of the lower state and n1 the principal quantum
number of the upper state.
R is the Rydberg constant and for He+ = 109722.263 (1/cm)
for H = 109677.581
the difference being in the reduced mass of the nucleus-electron
system.
 

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