MCQ
The orbital angular momentum of $3p$ electron is
  • A
    $\sqrt 3\ h$
  • B
    $\sqrt 6\ h$
  • C
    zero
  • $\sqrt 2 \frac{h}{{2\pi }}$

Answer

Correct option: D.
$\sqrt 2 \frac{h}{{2\pi }}$
d
Orbital angular momentum $ = \sqrt {l(l + 1)} \frac{h}{{2\pi }}$ $l=1$ for $\mathrm{p}-$ orbital.

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Similar questions

Consider the following reactions in which all the reactants and products are in gaseous state

${\text{2PQ}} \rightleftharpoons {{\text{P}}_2}{\text{  +  }}{{\text{Q}}_2}\,;\,\,\,\,{{\text{K}}_1}{\text{ =  2}}{\text{.5 }} \times \,{\text{1}}{{\text{0}}^5}\,$

${\text{PQ + }}\frac{1}{2}{R_2} \rightleftharpoons PQR\,;\,\,\,\,{{\text{K}}_2}{\text{ =  5 }} \times \,{\text{1}}{{\text{0}}^{ - 3}}\,$

The value of equilibrium constant for reaction

$\frac{1}{2}{{\text{P}}_2}{\text{ + }}\frac{1}{2}{{\text{Q}}_2}{\text{ + }}\frac{1}{2}{R_2} \rightleftharpoons PQR\,$ is

The $IUPAC$ name of compound

$\begin{array}{*{20}{c}}
  {HO - C = O\,\,\,\,\,\,\,C{H_3}} \\ 
  {\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {C{H_3} - C = C-C - H} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,N{H_2}\,\,\,\,\,Cl} 
\end{array}$

is :

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