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Question: The orbital angular moment of a \[p\]- electron is given as A) \(\sqrt 3 \;\dfrac{h}{{2\pi }}\) ...

The orbital angular moment of a pp- electron is given as
A) 3  h2π\sqrt 3 \;\dfrac{h}{{2\pi }}
B)   32  hπ\;\sqrt {\dfrac{3}{2}} \;\dfrac{h}{\pi }
C) 6  h2π\sqrt 6 \;\sqrt {\dfrac{h}{{2\pi }}}
D) h2  π\dfrac{h}{{\sqrt 2 \;\pi }}

Explanation

Solution

We need to know that the orbital angular moment of any electron can be calculated. As it depends on the quantum number of the orbital in which electrons are present. In atoms, electrons revolve around the nucleus. During the revolution, the electron attains an angular moment. This moment is dependent on the mass of the electron, velocity of the electron and radius of that orbit. During this moment, the electron experiences attraction from the nucleus and repulsion from inner electrons. Net of attraction and repulsion determines whether the electron is stable in its orbit.

Formula used: The orbital angular moment of the electron is expressed as
L=l(l+1)  h2πL = \sqrt {l(l + 1)} \;\dfrac{h}{{2\pi }}
Here, ll is the azimuthal quantum number of electrons. It varies from 0  to 30\;{\text{to 3}}.
hh is Planck’s constant and its value is 6.626  ×1034  m2 kg s - 16.626\; \times {\text{1}}{{\text{0}}^{ - 34}}\;{{\text{m}}^{\text{2}}}{\text{ kg }}{{\text{s}}^{{\text{ - 1}}}}.
π\pi is Pi and its value is 3.143.14.

Complete step by step answer:
The orbital angular moment of app electron can be calculated by the azimuthal quantum number of pp orbitals. The symbol of azimuthal quantum number is represented as ll. The ll value of pp orbital is 11.
The orbital angular moment of pp- electron is calculated as
L=l(l+1)  h2πL = \sqrt {l(l + 1)} \;\dfrac{h}{{2\pi }}
Here, the value of ll is 11, because it is app orbital electron.
L=1(1+1)  h2πL = \sqrt {1(1 + 1)} \;\dfrac{h}{{2\pi }}
L=2  h2πL = \sqrt 2 \;\dfrac{h}{{2\pi }}
L=2  h2  ×2  πL = \sqrt 2 \;\dfrac{h}{{\sqrt 2 \; \times \sqrt 2 \;\pi }}
L=  h  ×2  πL = \sqrt {\not{2}} \;\dfrac{h}{{\sqrt {\not{2}} \; \times \sqrt 2 \;\pi }}
L=  h 2  πL = \;\dfrac{h}{{{\text{ }}\sqrt 2 \;\pi }}
From the derivation, we have found that the orbital angular moment of pp- electrons is h2  π\dfrac{h}{{\sqrt 2 \;\pi }}

Therefore, option D is correct.

Note: There are four quantum numbers for each electron in an atom. They are principal quantum number, azimuthal quantum number, magnetic quantum number and spin quantum number. These quantum numbers are used to describe the orbital energy level, orientation and spin of an electron inside the atom. By using principle quantum numbers, the angular moment of that energy level can be determined. By using azimuthal quantum number, the orbital angular momentum can be calculated.