Question

There are two samples of $H$ and $H{e}^{+}$ atom. Both are in some excited state. In hydrogen atom, total number of lines observed in Balmer series is $4$ and in $H{e}^{+}$ 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 $H{e}^{+}$ sample. Then maximum excitation level of $H{e}^{+}$ 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 $H{e}^{+}$ ion, one line is observed in Paschen series. So electron is in n = $4(4\to 3)$.
$(H{)}_{6\to 2}=(He{)}_{n\to 4}$
$(H{)}_{6\to 2}=(H{e}^{+}{)}_{12\to 4}$
$\therefore$ electron in $H{e}^{+}$ will jump from $n=4$ to $n=12$.