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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.

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  • $\begingroup$ Could you elaborate? $\endgroup$
    – Mithoron
    Commented Jul 31, 2017 at 14:54

1 Answer 1

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$$E_n = Z^2\mathcal{R}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

You can easily verify that if you replace $Z$ with $2Z$, then the energy of the corresponding transition will be four times the original energy. Now if you replace both $n_1$ and $n_2$ with $2n_1$ and $2n_2$, then you can take out a factor of $1/4$; so the transition $2n_1 \leftarrow 2n_2$ in an ion with nuclear charge $2Z$ occurs at the same energy/wavelength/wavenumber as the transition $n_1 \leftarrow n_2$ in an ion with nuclear charge $Z$.

Same applies for any other factor $k$: the $1 \leftarrow 2$ transition in hydrogen occurs at the same wavelength as the $4 \leftarrow 8$ transition in $\ce{Be^3+}$. Obviously this discussion ignores the variation in the Rydberg constants for different nuclei.

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