- #1

- 20

- 0

V

_{e}=√(2kq

_{1}/r)

The following bit I think is wrong, but fetches an interesting result.

As E=1/2mv

^{2}, the analogous equation for electrostatics (with m=q) is E=1/2qv

^{2}, so

E

_{e}=kq

_{1}q

_{2}/r

The charge on the electron and proton is e, (1.6x10

^{-19}).

E

_{e}=ke

^{2}/r

To convert this from joules to electronvolts

E

_{e}=(ke

^{2}/r)/e

so

E

_{e}=ke/r

So using k=8.99x10

^{9}, e=1.6x10

^{-19}, and the Bohr radius of r= 5.3x10

^{-11}

I

_{e}=27.14eVe

^{-1}

I

_{e}=27.14eVe

^{-1}

The experimental value of the ionisation of H is I

_{e}= 1312kJ.mol

^{-1},

I

_{e}= 1.312x10

^{7}J.mol

^{-1}

So convert to J.e

^{-1}(per electron, not the inverse of the charge of an electron)

I

_{e}= 1.312x10

^{7}J.mol

^{-1}*mol

I

_{e}= 1.312x10

^{7}J.mol

^{-1}/6.022x10

^{23}

I

_{e}= 2.18x10

^{-17}J.e

^{-1}

So to convert from J to eV

I

_{e}= 2.18x10

^{-17}J.e

^{-1}/e

I

_{e}= 2.18x10

^{-17}J.e

^{-1}/1.6x10

^{-19}

I

_{e}= 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 I

_{e}=ke/(2r)?

And as F=kq

_{1}q

_{2}/r

^{2}, does this then mean that

I

_{e}=(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. ^^