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

I got #"91.1 nm"#.

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")#