Calculating Electron Speed from Ionized Light

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Discussion Overview

The discussion revolves around calculating the speed of an electron that has been ionized by a photon of light with a specific wavelength, focusing on the energy conservation during the ionization process. Participants explore the relationship between photon energy and electron speed, considering the effects of binding energy in hydrogen and other atoms.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes calculating the energy of a photon using the formula E_p = hc/λ and equating it to the kinetic energy of the electron after ionization.
  • Another participant points out the need to consider the potential energy associated with the electron's binding in the atom, suggesting that this energy must be accounted for in the calculations.
  • A subsequent reply acknowledges the binding energy and suggests using the energy of the first energy level of hydrogen to adjust the kinetic energy calculation.
  • Further discussion highlights that for hydrogen, only one electron is involved in the ionization process, while for larger atoms, multiple electrons may be ejected, leading to different binding energies that must be considered.

Areas of Agreement / Disagreement

Participants generally agree on the need to account for binding energy in the calculations, but there is no consensus on how to approach the energy adjustments for different atomic structures.

Contextual Notes

Participants have not resolved the implications of varying binding energies for different atoms, nor have they fully clarified the assumptions regarding the energy levels involved in the calculations.

liquidFuzz
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If I have, let's say, light of 70nm wavelength. I should be able to calculate the energy like this:
[itex]E_p = h\nu = \frac{hc}{\lambda}[/itex]

Now, if the photon ionize a hydrogen atom. Can can I calculate the electron speed using. [itex]E_e = \frac{mv^2}{2}[/itex] And say that the energy is preserved in the process?
[itex]E_e = E_p[/itex]
[itex]\frac{m_ev^2}{2} = \frac{hc}{\lambda}[/itex]

Wish gives the speed:

[itex]v = \sqrt (\frac{2}{m_e} \frac{hc}{\lambda} )[/itex]
 
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You forgot about the potential energy of binding the electron in the atom.
 
Ahh... You're right. The electron has the be, sort of, pulled away. I take it I can discard all but the first energy level, or? Giving me something like this.

[itex]\frac{13.6*num protons}{1^2} = 13.6 (eV)[/itex] ; number of protons = 1

[itex]v = \sqrt{\frac{2}{m_e} \left( \frac{hc}{\lambda}-13.6\right)}[/itex]
 
For Hydrogen you have only one electron to kick out, which require 13.6eV. For bigger atoms your photon may kick out any of electrons (if ots energy is sufficient) - not only 1s one, so you'll have choice of energies to subtract.
 

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