If electron makes transition from, in $H$ atom, $6\rightarrow 2$ (to Balmer), then $\bar{\nu}_{{H}_{6\rightarrow 2}}=x$ (let).
Also if electron makes transition from, in $He^{+}$ ion, so that $\bar{\nu}_{{He^{+}}_{a\rightarrow b}}=x$, gives me $a=12$ to $b=4$, by numerical caulations.
Now, what is observed is if I multiply $Z_{He^{+}}=2$ to transition of $H$ i.e. by $2\times\big({6\rightarrow 2}\big)=12\rightarrow 4$, gives me same result without any nasty calculation, why this works I don't know can you help me with this.
Thanks.