The orbital angular momentum of an electron in a $ s $ orbital is given as:
(A) $ 1 $
(B) zero
(C) $ \dfrac{{\sqrt {2h} }}{{2\pi }} $
(D) All of these
Answer
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Hint: The orbital angular momentum of an electron of an object about a fixed origin is called the angular momentum of the centre of mass about the origin. The net angular momentum of an object is the sum total of spin and the orbital angular momentum.
Complete step by step answer
In chemistry, quantum numbers are a set of numbers that gives the complete idea of an electron position and energy of an atom. There are four types of quantum numbers, these are Principal quantum numbers, Azimuthal quantum numbers, magnetic quantum numbers and spin quantum numbers.
The principal quantum number is denoted by $ 'n' $ and its value is given by $ n = 1,2,3.... $ here $ 'n' $ is an integer. The Azimuthal quantum number is denoted by $ 'l' $ and is given as $ l = 0 - (n - 1) $ where $ 'n' $ is principal quantum number. The shape of the orbital is governed by the azimuthal quantum number. The magnetic quantum number is denoted by $ 'm' $ and for a given value of $ 'l' $ the value of $ 'm' $ is given by $ ( - l\,to + l) $ . The spin quantum number has only two values i.e. $ + \dfrac{1}{2}\,and\, - {\raise0.7ex\hbox{ $ 1 $ } \!\mathord{\left/
{\vphantom {1 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }} $ and is represented by $ {m_s} $ .
The orbital angular momentum is given by the formula $ = \dfrac{h}{{2\pi }}\sqrt {l(l + 1} $ here $ 'l' $ is azimuthal quantum number. For a $ s $ orbital the value of $ l $ will be $ 0 $ . Hence putting the value of $ l $ In the formula we get the orbital angular momentum as:
$ = \dfrac{h}{{2\pi }}\sqrt {l(l + 1} $
$ = \dfrac{h}{{2\pi }}\sqrt {0(0 + 1} ) = 0 $
Hence we can say that the orbital angular momentum for an electron in s orbital will be zero.
Therefore, option (B) is correct.
Note
The value of the Principal quantum number will always be an integer from $ n = 1,2,3.... $ and for a given value of $ n $ the value of Azimuthal quantum number $ (l) $ can vary from zero to $ n - 1 $ . The azimuthal quantum number is also called the orbital angular quantum number.
Complete step by step answer
In chemistry, quantum numbers are a set of numbers that gives the complete idea of an electron position and energy of an atom. There are four types of quantum numbers, these are Principal quantum numbers, Azimuthal quantum numbers, magnetic quantum numbers and spin quantum numbers.
The principal quantum number is denoted by $ 'n' $ and its value is given by $ n = 1,2,3.... $ here $ 'n' $ is an integer. The Azimuthal quantum number is denoted by $ 'l' $ and is given as $ l = 0 - (n - 1) $ where $ 'n' $ is principal quantum number. The shape of the orbital is governed by the azimuthal quantum number. The magnetic quantum number is denoted by $ 'm' $ and for a given value of $ 'l' $ the value of $ 'm' $ is given by $ ( - l\,to + l) $ . The spin quantum number has only two values i.e. $ + \dfrac{1}{2}\,and\, - {\raise0.7ex\hbox{ $ 1 $ } \!\mathord{\left/
{\vphantom {1 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }} $ and is represented by $ {m_s} $ .
The orbital angular momentum is given by the formula $ = \dfrac{h}{{2\pi }}\sqrt {l(l + 1} $ here $ 'l' $ is azimuthal quantum number. For a $ s $ orbital the value of $ l $ will be $ 0 $ . Hence putting the value of $ l $ In the formula we get the orbital angular momentum as:
$ = \dfrac{h}{{2\pi }}\sqrt {l(l + 1} $
$ = \dfrac{h}{{2\pi }}\sqrt {0(0 + 1} ) = 0 $
Hence we can say that the orbital angular momentum for an electron in s orbital will be zero.
Therefore, option (B) is correct.
Note
The value of the Principal quantum number will always be an integer from $ n = 1,2,3.... $ and for a given value of $ n $ the value of Azimuthal quantum number $ (l) $ can vary from zero to $ n - 1 $ . The azimuthal quantum number is also called the orbital angular quantum number.
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