- #1
Kael42
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So, as a matter of interest, I tried predicting the ionisation energy of a Hydrogen atom according to classical electrostatics. I started from the point of escape velocity, with
Ve=√(2kq1/r)
The following bit I think is wrong, but fetches an interesting result.
As E=1/2mv2, the analogous equation for electrostatics (with m=q) is E=1/2qv2, so
Ee=kq1q2/r
The charge on the electron and proton is e, (1.6x10-19).
Ee=ke2/r
To convert this from joules to electronvolts
Ee=(ke2/r)/e
so
Ee=ke/r
So using k=8.99x109, e=1.6x10-19, and the Bohr radius of r= 5.3x10-11
Ie=27.14eVe-1
Ie=27.14eVe-1
The experimental value of the ionisation of H is Ie = 1312kJ.mol-1,
Ie = 1.312x107J.mol-1
So convert to J.e-1 (per electron, not the inverse of the charge of an electron)
Ie = 1.312x107J.mol-1*mol
Ie = 1.312x107J.mol-1/6.022x1023
Ie = 2.18x10-17J.e-1
So to convert from J to eV
Ie = 2.18x10-17J.e-1/e
Ie = 2.18x10-17J.e-1/1.6x10-19
Ie = 136.17eVe-1
This experimental value is ridiculous as an ionisation value, so I imagine I have forgotten something. Interestingly though, the value I have found from various sources is 13.6eVe-1, exactly 1 decimal place shifted. I will assume then that the experimental value is 13.6, and I have messed up somewhere.
Interestingly, the correct experimental of 13.6 is nearly EXACTLY half of 27.14. Is this a coincidence, or does it imply that Ie=ke/(2r)?
And as F=kq1q2/r2, does this then mean that
Ie=(1/2)Fr?
Perhaps this is an equation I do not know of?
Please feel free to pick apart my logic and findings - it seems a bit simple. ^^
Ve=√(2kq1/r)
The following bit I think is wrong, but fetches an interesting result.
As E=1/2mv2, the analogous equation for electrostatics (with m=q) is E=1/2qv2, so
Ee=kq1q2/r
The charge on the electron and proton is e, (1.6x10-19).
Ee=ke2/r
To convert this from joules to electronvolts
Ee=(ke2/r)/e
so
Ee=ke/r
So using k=8.99x109, e=1.6x10-19, and the Bohr radius of r= 5.3x10-11
Ie=27.14eVe-1
Ie=27.14eVe-1
The experimental value of the ionisation of H is Ie = 1312kJ.mol-1,
Ie = 1.312x107J.mol-1
So convert to J.e-1 (per electron, not the inverse of the charge of an electron)
Ie = 1.312x107J.mol-1*mol
Ie = 1.312x107J.mol-1/6.022x1023
Ie = 2.18x10-17J.e-1
So to convert from J to eV
Ie = 2.18x10-17J.e-1/e
Ie = 2.18x10-17J.e-1/1.6x10-19
Ie = 136.17eVe-1
This experimental value is ridiculous as an ionisation value, so I imagine I have forgotten something. Interestingly though, the value I have found from various sources is 13.6eVe-1, exactly 1 decimal place shifted. I will assume then that the experimental value is 13.6, and I have messed up somewhere.
Interestingly, the correct experimental of 13.6 is nearly EXACTLY half of 27.14. Is this a coincidence, or does it imply that Ie=ke/(2r)?
And as F=kq1q2/r2, does this then mean that
Ie=(1/2)Fr?
Perhaps this is an equation I do not know of?
Please feel free to pick apart my logic and findings - it seems a bit simple. ^^