Question

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:

• A. $n = 6$
• B. $n = 8$
• C. $n = 12$
• D. $n = 9$

Answer 1

Answer: (C)

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 = $4 (4 \to 3)$.
$(H) _{6\to2} =(He)_{n\to4}$
$(H)_{6\to2} =(He^{+})_{12\to4}$
$\therefore $ electron in $He^+$ will jump from $n = 4$ to $n = 12$.