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JEE Mains · Chemistry · STD 11 - 2. structure of atom

 \(2 s\) के लिए तरंग फलन \((\Psi)\) दिया है- \(\Psi_{2 s}=\frac{1}{2 \sqrt{2 \pi}}\left(\frac{1}{a_0}\right)^{1 / 2}\left(2-\frac{ r }{ a _0}\right) e ^{- r / 2 a _0}\) \(\mathrm{r}=\mathrm{r}_0\), पर त्रिज्या (रिडियल) नोड बनता है। अतः \(\mathrm{a}_0\) के पदो में, \(\mathrm{r}_0\) :

  1. A \(r_0=a_0\)
  2. B \(r_0=4 a_0\)
  3. C \(r _0=\frac{ a _0}{2}\)
  4. D \(r_0=2 a_0\)
Verified Solution

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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