The limiting line in balmer series will have frequency of ?

Well, it depends on the atomic number...

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

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))#