Question

What wavelength in `"nm"` corresponds to the limiting line of the Lyman series for hydrogen atom?

Answer 1

I got `"91.1 nm"`.

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Well, you remember limits? You can approach `oo`, and so, the limiting line is simply the change in `n` (the principal quantum number) going towards `n = oo`.

The Lyman series is in reference to `n = 1`. That is, it always stops on `n = 1`.

Any emission line in hydrogen atom will involve a change in energy from the Rydberg equation:

`DeltaE = -"13.61 eV" (1/n_f^2 - 1/n_i^2)`

where `n` is as defined before. `f` indicates final and `i` indicates initial states.

The limiting line is simply when `n_i -> oo`, i.e. the change in `n` is infinitely large so that the electron can escape the atom (i.e. so that the atom is ionized).

Thus, we expect this to be the ionization energy of hydrogen atom.

`color(green)(|DeltaE|) = lim_(n_i->oo) [|-"13.61 eV"| cdot (1/1^2 - 1/n_i^2)]`

`=` `color(green)("13.61 eV")`

and this is indeed the ionization energy of hydrogen atom.

As a result, we can now find the wavelength needed to ionize hydrogen atom.

`|DeltaE| = E_"photon" = hnu = (hc)/lambda`

where `c = 2.998 xx 10^(8) "m/s"` is the speed of light and `h = 6.626 xx 10^(-34) "J"cdot"s"` is Planck's constant.

Therefore,

`color(blue)(lambda) = (hc)/(|DeltaE|)`

`= (6.626 xx 10^(-34) cancel"J"cdotcancel"s" cdot 2.998 xx 10^(8) "m" cdot cancel("s"^(-1)))/(13.61 cancel"eV" xx (1.602 xx 10^(-19) cancel"J")/(cancel"1 eV"))`

`= 9.11 xx 10^(-8) "m"`

`=` `color(blue)ul("91.1 nm")`