JEE Advanced · Chemistry · 2. Atomic Structure
Paragraph:
The hydrogen-like species \(\mathrm{Li}^{2+}\) is in a spherically symmetric state \(S_1\) with one radial node. Upon absorbing light the ion undergoes transition to a state \(S_2\). The state \(S_2\) has one radial node and its energy is equal to the ground state energy of the hydrogen atom.
Question:
The state \(S_1\) is
- A \(1 s\)
- B \(2 \mathrm{~s}\)
- C \(2 p\)
- D \(3 \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(2 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
Given : Hydrogen like species, \(\mathrm{Li}^{2+}\)
Spherically symmetric state, \(S_1 \Rightarrow s\)-orbital
\(S_1\) has radial node \(1 \Rightarrow 2 s \quad\) (No. of radial node \(=1=n-l-1\), so \(n=2\) ) Species undergoes state \(S_2\)
\(S_2\) has energy equal to the ground state energy of \(\mathrm{H}\)-atom \(=-13.6 \mathrm{eV}\)
So, \(\quad-13.6=-13.6\left(\frac{3^2}{n^2}\right)\)
\(n=3\)
Again radial node \(=1=n-1-1\)
or \(1=3-l-1 \quad\) so \(\quad l=1\)
It suggests \(S_2\) is \(3 p\)
Spherically symmetric state, \(S_1 \Rightarrow s\)-orbital
\(S_1\) has radial node \(1 \Rightarrow 2 s \quad\) (No. of radial node \(=1=n-l-1\), so \(n=2\) ) Species undergoes state \(S_2\)
\(S_2\) has energy equal to the ground state energy of \(\mathrm{H}\)-atom \(=-13.6 \mathrm{eV}\)
So, \(\quad-13.6=-13.6\left(\frac{3^2}{n^2}\right)\)
\(n=3\)
Again radial node \(=1=n-1-1\)
or \(1=3-l-1 \quad\) so \(\quad l=1\)
It suggests \(S_2\) is \(3 p\)
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