There are two samples of H and ${He}^{+}$ atom. Both are in some excited state. In hydrogen atom, total number of lines observed in Balmer series is 4 and in ${He}^{+}$ atom total number of lines observed in Paschen series is 1. Electron in hydrogen sample make transitions to lower states from its excited state, then the photon corresponding to the line of maximum energy line of Balmer series of H sample is used to further excite the already excited ${He}^{+}$ sample. Then maximum excitation level of ${He}^{+}$ sample will be

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n = 6

n = 8

n = 12

n = 9

answer is C.

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Detailed Solution

In H-atom, 4 lines are observed in Balmer series. So, electron is in n = 6 $(6 \to 2, 5 \to 2, 4 \to 2, 3 \to 2) .$ In ${He}^{+}$ ion, one line is observed in Paschen series. So electron is in n = 4 $(4 \to 3) .$

$\begin{array}{l} {(H)}_{6 \to 2} = {(He)}_{n \to 4} \\ {(H)}_{6 \to 2} = {({He}^{+})}_{12 \to 4} \\ ∴ electron in {He}^{+} will jump from n = 4 to n = 12 . \end{array}$