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The hydrogen-like species $Li^{2+}$ is in a spherically symmetric state $S_1$ with one radial node. Upon absorbing light the ion undergoes transition to a state $S_2$ has one radial node and its energy is equal to the ground state energy of hydrogen atom. The state $S_1$ is

$(a)\;1s\qquad(b)\;2s\qquad(c)\;2p\qquad(d)\;3s$

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For $S_1$ (spherically symmetrical) state, number of nodes = 1
1 = n-1
$\therefore$ n = 2
i.e it is 2s
For $S_2$ radial node = 1
$E_{s_2} = \large\frac{-13.6\times z^2}{n^2} = E_H$ in ground state
= -13.6
$E = \large\frac{-13.6\times9}{n^2}$
$\Rightarrow n = 3$
So state is $S_1$ is 2s and $S_2$ is 3p
Hence answer is (b)
answered Feb 10, 2014 by sharmaaparna1
 

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