The shortest wavelength of the Brackett series of a hydrogen like atom (atomic number = Z ) is the same as the shortest wavelength of the Balmar series of hydrogen atom. The value of Z is

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Shortest wavelength of Brackett series corresponds to the transition of electron between $n_{1} = 4$ and $n_{2} = \infty$ and the shortest wavelength of Balmer series corresponds to the transition of electron between $n_{1} = 2$ and $n_{2} = \infty$ , So,

$(z^{2}) (\frac{13.6}{16}) = (\frac{13.6}{4})$

$∴ Z^{2} = 4$

or $Z = 2$

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