Can we use the same spectral lines for a hydrogenoid like $\rm He^{+1}$ I need to determine the transition the resulted in a wave with wavelength $4689\rm{\mathring{A}}$, from a problem I tried to solve. I know that the concerned hydrogenoid is a helium hydrogenoid: $\rm He^{+1}$, so this is a wave length from its emission spectrum.
I've computed the Rydberg constant for this hydrogenoid: $R_{\rm He^{+1}}=4.4\cdot10^7\rm m^{-1}$. My understanding is that I can determine the transition type (starting orbit and final orbit number) using Ritz's law,
$$\frac{1}{\lambda}=R_{\rm H}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right).$$
I have that $n_f=2$  because it's a Balmer series since this is a visible wave so I have to determine the $n_i$.
However this reasoning is true for hydrogen atom and hydrogen's emission spectrum, I don't know if this apply to other hydrogenoid like $\rm He^{+1}$? Is the Balmer series the same for all hydrogenoids?
 A: Eranreches already gave a good answer, but some additional comments might be helpful. For one, giving transition wavelengths to four significant figures is rather too precise for the purely electrostatic dynamics your set-piece is asking you to consider, and in general there will be fine-structure corrections that will mess with the details of the transition frequencies at roughly that level.
For a good grounding in reality, an excellent resource is the NIST Atomic Spectra Database, which contains a huge range of experimentally-measured transition frequencies. For singly ionized helium (which you input as He II; neutral helium is He I, the "first spectrum" of helium), the database indicates that your fourth significant figure is off - there's simply no transition at $4689\:\rm\mathring A$ in $\rm He^+$, but you do get a transition at $4685\:\rm\mathring A$ instead. (This is then split at the fifth significant figure by fine-structure contributions into transitions between different subshells with different angular momentum.)

And, on another vein, the deduction that you must have $n_f=2$ because the wavelength is in the visible range is flawed. The names Balmer, Lyman, Paschen, &c, are generally only used for the hydrogen spectral series, and not for the rest of its isoelectronic sequence. At the electrostatic level, as eranreches points out in their answer, the transition wavelength depends on both the initial and final shells as well as on the nuclear charge, via
$$\frac{1}{\lambda_{n_{1},n_{2}}}=R_{\infty}Z^2\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right).$$
That means that if a given $n_1,n_2$ pair gives a line in the visible range for hydrogen, then it will go down in wavelength with $1/Z^2$ as the nuclear charge increases. For helium that already gives a factor of $1/4$: if a given line is in the visible for hydrogen, it definitely won't be there for $\rm He^+$, and it only gets worse as the charge increases.
More practically, in your particular example, the deduction that $n_f=2$ is simply wrong; you can get the right answer by (i) checking the ASD link above, or (ii) playing around and seeing what $n$s satisfy the correct relationship via trial and error, or (iii) by multiplying the wavelength by $4=Z^2$ and seeing what hydrogen transition it corresponds to.
A: Indeed, for Hydrogen-like atoms with atomic number $Z$ you can use the following formula for the wavelength of a transition
$$\frac{1}{\lambda_{n_{1},n_{2}}}=R_{\infty}Z^2\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$$
where $R_{\infty}$ is the usual Rydberg's constant. 
