Calculating Electron Speed from Ionized Light

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The discussion focuses on calculating the speed of an electron ejected from a hydrogen atom by a photon of 70nm wavelength. The energy of the photon is calculated using E_p = hc/λ, and the kinetic energy of the electron is expressed as E_e = mv²/2. It is noted that the binding energy of the electron, specifically 13.6 eV for hydrogen, must be subtracted from the photon energy to find the actual kinetic energy. The final formula for electron speed incorporates this binding energy adjustment, resulting in v = √(2/m_e * (hc/λ - 13.6)). The conversation highlights the difference in energy considerations for hydrogen versus larger atoms, where multiple electrons might be involved.
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:
E_p = h\nu = \frac{hc}{\lambda}

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

Wish gives the speed:

v = \sqrt (\frac{2}{m_e} \frac{hc}{\lambda} )
 
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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.

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

v = \sqrt{\frac{2}{m_e} \left( \frac{hc}{\lambda}-13.6\right)}
 
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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