Medium
For an electron in the $d$-orbital,the orbital angular momentum is
- A$\frac{h}{2 \pi}$
- B$\frac{\sqrt{2} h}{2 \pi}$
- C$\frac{\sqrt{6} h}{2 \pi}$
- D$\frac{\sqrt{6} h}{2 \pi}$
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Similar Questions
The azimuthal quantum number for the valence electrons of $Ga^{+}$ ion is $.....$
(Atomic number of $Ga = 31$)
(Atomic number of $Ga = 31$)
The electrons identified by quantum numbers $n$ and $l$ :
$A. n=4, l=1$ $B. n=4, l=0$
$C. n=3, l=2$ $D. n=3, l=1$
can be placed in order of increasing energy as :
$A. n=4, l=1$ $B. n=4, l=0$
$C. n=3, l=2$ $D. n=3, l=1$
can be placed in order of increasing energy as :
MediumAIEEE 2012
View SolutionNickel atom can lose two electrons to form $Ni^{2+}$ ion. The atomic number of nickel is $28$. From which orbital will nickel lose two electrons?
Easy
View SolutionThe orbital angular momentum of an electron in $3s$ orbital is $\frac{xh}{2\pi}$. The value of $x$ is . . . . . . .
Given below are the quantum numbers for $4$ electrons.
$A. \; n = 3, l = 2, m_l = 1, m_s = +1/2$
$B. \; n = 4, l = 1, m_l = 0, m_s = +1/2$
$C. \; n = 4, l = 2, m_l = -2, m_s = -1/2$
$D. \; n = 3, l = 1, m_l = -1, m_s = +1/2$
The correct order of increasing energy is
$A. \; n = 3, l = 2, m_l = 1, m_s = +1/2$
$B. \; n = 4, l = 1, m_l = 0, m_s = +1/2$
$C. \; n = 4, l = 2, m_l = -2, m_s = -1/2$
$D. \; n = 3, l = 1, m_l = -1, m_s = +1/2$
The correct order of increasing energy is
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