MCQ
If there is $2$ nodal surfaces in third excited state. Find the orbital angular momentum
- A$\sqrt 3 \,\hbar $
- ✓$\sqrt 2 \,\hbar $
- C$4\,\hbar $
- D$\frac{1}{{\sqrt 2\, \hbar }}\,$
✓
Answer
Correct option: B.
$\sqrt 2 \,\hbar $
b
Third excited state $n=4$
Third excited state $n=4$
$n-\ell-1=2$
$4-\ell-1=2$
$C=1$
$=\sqrt{\ell(\ell+1)} \hbar$
$=\sqrt{1(1+1)} h=\sqrt{2} \hbar$
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