Chemical Forums
Specialty Chemistry Forums => Materials and Nanochemistry forum => Topic started by: intrepid_nerd on December 27, 2008, 02:33:40 PM
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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: E1 :rarrow: 6 = qV (coulomb times potential).
then V = E/q and q = +2e = 2(1.6E-18).
then E = ? so far i'm quite sure i have to use 1/2 of Rydberg's equation
[ -R(1/n2 - 1/n1) where R = constant, n2 > n1 ].
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 = 2.18E-18J because i don't see the relation between the constant,
k = -((me4)/(8h2ε02)) 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!
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VICTORY! answered my own question -- in case anyone is curious, the constant "k" and the constant R can be used interchangeably. note, however, in the equation I posted above:
1/λ = -R(1/n2 - 1/n1) works ONLY with hydrogen.
1/λ = -RZ2(1/n2 - 1/n1) works with other atoms, z = proton #.
hc/λ = -kZ2(1/n2 - 1/n1) works for solving energy problems.
keep in mind friends, chemistry ignores the strong, the weak and the gravitational forces but is very cognizant of electromagnetic forces. math is wickedly important for understanding this at the quantum/molecular level!!