JEE Mains · Chemistry · STD 11 - 2. structure of atom
The wave function \((\Psi)\) of \(2 s\) is given by \(\Psi_{2 s}=\frac{1}{2 \sqrt{2 \pi}}\left(\frac{1}{a_0}\right)^{1 / 2}\left(2-\frac{ r }{ a _0}\right) e ^{- r / 2 a _0}\) At \(r=r_0\), radial node is formed. Thus, \(r_0\) in terms of \(a_0\)
- A \(r_0=a_0\)
- B \(r_0=4 a_0\)
- C \(r _0=\frac{ a _0}{2}\)
- D \(r_0=2 a_0\)
Answer & Solution
Correct Answer
(D) \(r_0=2 a_0\)
Step-by-step Solution
Detailed explanation
At node \(\Psi_{2 s }=0\) \(\therefore 2-\frac{ r _0}{ a _0}=0\) \(\therefore r _0=2 a _0\)
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