Medium
The orbital angular momentum of an electron in an orbital is given by $\sqrt{l(l + 1)} \cdot \frac{h}{2\pi}$. For an $s$-electron,this momentum is:
- A$\sqrt{2} \cdot \frac{h}{2\pi}$
- B$+\frac{1}{2} \cdot \frac{h}{2\pi}$
- C$0$
- D$\frac{h}{2\pi}$
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Similar Questions
The atomic number of the element (in ground state) having the maximum number of unpaired $3p$ electrons is:
Medium
View SolutionThe angular momentum of a $p$-orbital electron is given by:
Medium
View SolutionThe correct order of decreasing energy for the electrons whose quantum numbers $n$ and $l$ are given below, is
$A$. $n=5, l=2$
$B$. $n=5, l=0$
$C$. $n=4, l=3$
$D$. $n=4, l=1$
$A$. $n=5, l=2$
$B$. $n=5, l=0$
$C$. $n=4, l=3$
$D$. $n=4, l=1$
Identify the element having the electronic configuration $1s^2, 2s^2, 2p^2$.
Easy
View SolutionThe two electrons occupying the same orbital are distinguished by:
Easy
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