don't fret - this is not uncertainty or any string theory or any of that nonsense!
problem: determine min potential that must be applied to an alpha-particle so that on interaction with a hydrogen atom, the ground state electron will excite to n=6.
solution: so far what i figure is that i'm using :delta: E
1 6 = qV (coulomb times potential).
then V = E/q and q = +2e = 2(1.6
E-18).
then E = ? so far i'm quite sure i have to use 1/2 of Rydberg's equation
[ -R(1/n
2 - 1/n
1) where R = constant, n
2 > n
1 ].
but R doesn't work here because R deals with wavelength and i need to find potential energy, which V = J/C
but what is my J?
i'm hesitant to use k = 2.18
E-18J because i don't see the relation between the constant,
k = -((me
4)/(8h
2ε
02)) and the Rydberg equation.
answer using k instead of R, 6.62V
i figure this is the solution to the problem but don't see the correlation. any help understanding this is greatly appreciated!