The hydrogen like species Li^2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy of the hydrogen atom. The orbital angular momentum quantum number of the state S2 is

Structure of Atom PYQ — JEE Advanced 2010 · Shift 2 · Q5 with Solution

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Correct answer: B

For a hydrogen-like species, the number of radial nodes is given by

\text{radial nodes} = n-l-1

Since $S_1$ is spherically symmetric, it must be an $s$ -state, so

l=0

Given that it has one radial node:

n-0-1=1 \Rightarrow n=2

So,

S_1 = 2s

For $Li^{2+}$ , $Z=3$ .

Energy of a hydrogen-like ion is

E_n=-\frac{13.6 Z^2}{n^2}\,\text{eV}

For $Li^{2+}$ ,

E_n=-\frac{13.6\times 3^2}{n^2}=-\frac{122.4}{n^2}\,\text{eV}

The question says the energy of $S_2$ equals the ground state energy of hydrogen atom:

E=-13.6\,\text{eV}

Thus,

-\frac{122.4}{n^2}=-13.6

\frac{122.4}{n^2}=13.6

n^2=9 \Rightarrow n=3

So the final state belongs to the $n=3$ shell.

$S_2$ also has one radial node, so

n-l-1=1

With $n=3$ ,

3-l-1=1

2-l=1

l=1

Hence the orbital angular momentum quantum number of $S_2$ is

\boxed{1}

Therefore, the correct option is

\boxed{\text{B}}