Question

The limiting line in balmer series will have frequency of ?

Answer 1

Well, it depends on the atomic number...

`nu = Z^2 cdot 8.22467 xx 10^14 "s"^(-1)`

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The Balmer series for the hydrogen-like atom is for transitions that end at `n = 2`.

The limiting line is basically when you consider shoving a new electron into the atom and forcing it to land in the destination energy level `n_f` for the particular series...

So, we have `n_f = 2` and `n_i = oo` in the Rydberg equation for hydrogen-like atoms (e.g. `"He"^(+)`, `"Li"^(2+)`, etc):

`E_n = -"13.6058 eV" cdot Z^2/n^2`

Thus, the change in energy for a relaxation is given by

`DeltaE_(oo->2) = -"13.6058 eV" cdot Z^2 (1/2^2 - cancel(1/oo^2)^(0))`

`= -Z^2cdot"3.4015 eV"`

The atom will then emit light... with photon energy

`|E| = DeltaE_(oo->2) = hnu`

`= Z^2cdot3.4015 cancel"eV" xx (1.60217662 xx 10^(-19) "J")/(cancel"1 eV")`

`= Z^2 cdot 5.4497 xx 10^(-19) "J"`.

where `h = 6.62607004 xx 10^(-34) "J"cdot"s"` is Planck's constant and `nu` is the frequency in `"s"^(-1)` (`"Hz"`).

Hence, the frequency is:

`color(blue)(nu) = [Z^2 cdot 5.4497 xx 10^(-19) cancel"J"]/(6.62607004 xx 10^(-34) cancel"J"cdot"s")`

`= color(blue)(Z^2 cdot 8.22467 xx 10^14 "s"^(-1))`