The longest wavelength of the Lyman series to Hydrogen atom is the same as the wavelength of certain line in the spectrum of He+ when the electron makes a transition from n→2 The value of n is:

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The longest wavelength of the Lyman series to Hydrogen atom is the same as the wavelength of certain line in the spectrum of ${He}^{+}$ when the electron makes a transition from $n \to 2$ The value of n is:

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answer is B.

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Detailed Solution

According to the question,

$λ_{{He}^{+}}$ =

$= R_{H} Z_{{He}^{+}}^{2} [\frac{1}{2^{2}} - \frac{1}{4^{2}}]$

Z = Atomic number of He

$[\frac{1}{1^{2}} - \frac{1}{2^{2}}] = 2^{2} [\frac{1}{2^{2}} - \frac{1}{n^{2}}] \Rightarrow n = 4$

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