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I am probably talking out of the back of my head but I will have a go.The energy required to completely remove an electron is the ionization potential something like 4 ev in the case of Hydrogen knowing the mass of the electron this amount of energy can be converted into a velocity according E=MV^2/2Is this anywhere near the truth or just a confused ramble ?
Although the effect would not be very large at .004c would not the apparent increase of mass of the electron with velocity have to be taken into account.Jeffery HAs the energy of 4ev is rather a guesstimate is not carrying out the calculation to 30 decimal places rather overkill!
My guess was widely out apparently the ionization energy is 13.6ev
If we have an atom with a valency of 1, say hydrogen, and strip the electron from the nucleus, what speed would a free electron have to be travelling near the nucleus to prevent a fall into the Coulomb potential well?
Well let's have a go. If we take velocity as c
Quote from: jeffreyH on 09/01/2014 23:42:19Well let's have a go. If we take velocity as c Why?That's the speed that we know is wrong (because an electron has mass so it can't travel at c).
Lightarrow, both JeffreyH and myself make it ~2.2*10^3 Km/s , are two of us wrong or are you ?
The energy of the electron can be determined from the wavelength of the light it would emit falling from infinity down to the 1s orbital, ie 91.2nm.The energy released by falling from closer distances is given by the Lyman series, in ultraviolet wavelengths around 100nm.If you want to find the energy falling into the second shell, this is given by the Balmer series.The general answer is provided by the Rydberg formula; set n'=∞.See: http://en.wikipedia.org/wiki/Hydrogen_spectral_series#PhysicsEnergy=hc/λhttp://en.wikipedia.org/wiki/Photon#Physical_properties