Calculate the wavelength when electron transits.

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SUMMARY

The discussion centers on calculating the wavelength of a photon emitted during an electron transition from an energy state of -6 eV to -8 eV. Using the equation E=hc/λ, participants deduced that the energy difference between these states is -2 eV. Consequently, since energy is inversely proportional to wavelength, the wavelength for this transition is determined to be 600 nm, effectively doubling from the initial 300 nm photon emitted during a transition from -4 eV to -8 eV.

PREREQUISITES
  • Understanding of photon emission and electron transitions in quantum mechanics.
  • Familiarity with the energy equation E=hc/λ.
  • Knowledge of electron energy states in atoms.
  • Basic principles of wavelength and energy relationship.
NEXT STEPS
  • Study the concept of energy level transitions in quantum mechanics.
  • Learn about the Planck constant (h) and its role in photon energy calculations.
  • Explore the implications of wavelength changes on energy transitions.
  • Investigate other examples of photon emissions and their corresponding wavelengths.
USEFUL FOR

Students studying quantum mechanics, physics educators, and anyone interested in understanding photon emissions and electron transitions in atomic structures.

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Homework Statement


A photon of wavelength 300nm is emitted from an atom when an electron makes a transition from an energy state of -4 eV to a state of -8 eV. IF the initial state had been at enerygy of -6 eV, the wavelength of the photon emitted in a transition to -8eV is..??

Homework Equations


E=hc/λ

The Attempt at a Solution


I don’t know the basis of this question, but here is my attempt. Since the change in energy from -4eV to -8eV is -4 eV and from -6ev to -8eV is -2, so the energy is halved. If E is halved, then I think λ doubles and the anwer would be 600nm (not sure)
 
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Apparently you do know the basis of the question since your reasoning is correct.

You can use the relevant equation to guide your thinking about how energy and wavelength are related. If you doubled the wavelength ##\lambda##, what would happen to E?
 
vela said:
Apparently you do know the basis of the question since your reasoning is correct.
Oh, great! I wasn’t sure I know that topic well. :oldsmile:
 

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