Question

The angular momentum of \[3\]p-orbitals in terms of \[{{\text{h}}^{\text{*}}}{\text{(}}{{\text{h}}^{\text{*}}}{\text{ = }}\dfrac{{\text{h}}}{{{{2\pi }}}}{\text{)}}\,{\text{is}}\]
\[
  {\text{A}}{\text{.}}\sqrt {\text{2}} {{\text{h}}^{\text{*}}} \\
  {\text{B}}{\text{.2}}{{\text{h}}^{\text{*}}} \\
  {\text{C}}{\text{.}}\dfrac{{{{\text{h}}^{\text{*}}}}}{{\sqrt {\text{2}} }} \\
  {\text{D}}{\text{.}}{{\text{h}}^{\text{*}}} \\
 \]

Answer 1

Hint : The amount of motion that occurs in something that is moving, or the force that propels something forward to keep it moving, is known as momentum.

Complete Step By Step Answer:
A.\[\sqrt 2 h^*\] is the correct option because as we all know that,
Angular momentum in an orbital \[ = \sqrt {1(1 + 1)} .\dfrac{h}{{2\pi }}\]
\[1 = 1\]for \[3p\]orbital,
Angular momentum in \[3p\]is
\[ = \sqrt {1(1 + 1)} .\dfrac{h}{{2\pi }} = \sqrt 2 {h^*}\]
B. \[2{h^*}\] is not the correct answer. Because, while substituting the values in the corresponding equation of Angular Momentum in an orbital\[ = \sqrt {1(1 + 1)} .\dfrac{h}{{2\pi }}\], we will not get the angular momentum of \[3\]p-orbitals in terms of \[{{\text{h}}^{\text{*}}}{\text{(}}{{\text{h}}^{\text{*}}}{\text{ = }}\dfrac{{\text{h}}}{{{{2\pi }}}}{\text{)}}\,\]as \[2{h^*}\].
C.\[\dfrac{{{h^*}}}{{\sqrt 2 }}\] is not the correct answer. Because, while substituting the values in the corresponding equation of Angular Momentum in an orbital\[ = \sqrt {1(1 + 1)} .\dfrac{h}{{2\pi }}\], we will not get the angular momentum of \[3\]p-orbitals in terms of \[{{\text{h}}^{\text{*}}}{\text{(}}{{\text{h}}^{\text{*}}}{\text{ = }}\dfrac{{\text{h}}}{{{{2\pi }}}}{\text{)}}\,\] as \[\dfrac{{{h^*}}}{{\sqrt 2 }}\].
D.\[{h^*}\] is not the correct answer. Because, while substituting the values in the corresponding equation of Angular Momentum in an orbital\[ = \sqrt {1(1 + 1)} .\dfrac{h}{{2\pi }}\], we will not get the angular momentum of \[3\]p-orbitals in terms of \[{{\text{h}}^{\text{*}}}{\text{(}}{{\text{h}}^{\text{*}}}{\text{ = }}\dfrac{{\text{h}}}{{{{2\pi }}}}{\text{)}}\,\]as\[{h^*}\]
We all know that Angular momentum is the quantity of rotation of a body that is the product of its moment of inertia and its angular velocity. The product of the mass, velocity or speed of rotation, and the radius of the circle determines the angular momentum of an object moving in a circle with radius 'r.' Angular momentum is a vector that points in the same direction as angular velocity. So by the accurate definitions of Angular momentum, the angular momentum of \[3\]p-orbitals in terms of \[{{\text{h}}^{\text{*}}}{\text{(}}{{\text{h}}^{\text{*}}}{\text{ = }}\dfrac{{\text{h}}}{{{{2\pi }}}}{\text{)}}\,{\text{is}}\] \[\sqrt 2 h^*\].

Note:
The kilogram squared per second (kg-\[{{\text{m}}^{\text{2}}}\]/sec) is the SI unit for angular momentum. The rotational counterpart of linear momentum is angular momentum (also known as moment of momentum or rotational momentum). Since it is a conserved quantity—the total angular momentum of a closed system stays constant—it is a significant quantity.