Atomic StructureHard
Question
The orbital angular momentum of an electron is $\sqrt{3}\frac{h}{\pi}$. Which of the following may be the permissible value of angular momentum of this electron revolving in unknown Bohr's orbit?
Options
A.$\frac{h}{\pi}$
B.$\frac{h}{2\pi}$
C.$\frac{3h}{2\pi}$
D.$\frac{2h}{\pi}$
Solution
$\sqrt{l(l + 1)}.\frac{h}{2\pi} = \sqrt{3}.\frac{h}{\pi} \Rightarrow l = 3 \Rightarrow n \geq 4$
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
An electron that has the quantum numbers n = 3 and m = 2...If ᐃ0 < P.E. the correct electronic configuration for d6 system will be :-...An electron revolves round Li2+ nucleus at a distance of 1.587 Å. The speed of electron should be...The orbital angular momentum of an electron in 2s orbital is :...The total number of atomic orbitals in fourth energy level of an atom is...