$(a)\;2^{nd}\;orbit\qquad(b)\;1^{st}\;orbit\qquad(c)\;3^{rd}\;orbit\qquad(d)\;5^{th}\;orbit$

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Given :

$\lambda_2=30.4\times 10^{-7}cm$

$\lambda_1=108.5\times 10^{-7}cm$

Let excited state of $He^+$ be $n_2$.It comes from $n_2$ to $n_1$ and then $n_1$ to 1 to emit two successive photon.

$\large\frac{1}{\lambda_2}=$$R_H.z^2\big[\large\frac{1}{1^2}-\frac{1}{n_1^2}\big]$

$\large\frac{1}{30.4\times 10^{-7}}=$$109678\times 4\big[\large\frac{1}{1^2}-\frac{1}{n_1^2}\big]$

$\therefore n_1=2$

Now for $\lambda_1:n_1=2$

$\large\frac{1}{\lambda_1}=$$R_H.z^2\big[\large\frac{1}{2^2}-\frac{1}{n_2^2}\big]$

$\large\frac{1}{108.5\times 10^{-7}}=$$109678\times 4\big[\large\frac{1}{2^2}-\frac{1}{n_2^2}\big]$

$n_2=5$

$\therefore$ excited state of He is $5^{th}$ orbit.

Hence (d) is the correct answer.

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