
For a 3s−orbital,
\[\psi \left( {3s} \right) = \dfrac{1}{{9\sqrt 3 }}{\left( {\dfrac{1}{{{a_o}}}} \right)^{3/2}}\left( {6 - 5\sigma + {\sigma ^2}} \right){e^{ - \sigma /2}}\]
where, \[\sigma = \dfrac{{2r.Z}}{{3{a_o}}}\]What is the maximum radial distance of the node from the nucleus?
Answer
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Hint:An orbit is referred to as the simple planar representation of an electron. On the other hand, an orbital is referred to as the dimensional motion or movement of an electron around the nucleus in the three-dimensional motion. An orbital is actually a space or a region where an electron is most likely to be found.
Complete answer:
The names of the orbital i.e. s, p, d, or f stand for the names given to the groups of lines which are originally noted in the spectra of alkali metals. The line groups are known as sharp, principal, diffuse, or fundamental, respectively.
A node in an orbital refers to a plane of zero electron density. There are two types of nodes: (i) radial nodes which is equal to the total number of nodes (n-1) – the number of angular nodes (I) = (n-1) - l and (ii) angular nodes which is equal to the orbital angular momentum quantum number i.e. l.
We know that value of wave function ($\Psi $) and its square (${\Psi ^2}$) is zero at nodes i.e.
$\Psi (3s) = 0$ which means:
\[
\dfrac{1}{{9\sqrt 3 }}{\left( {\dfrac{1}{{{a_o}}}} \right)^{3/2}}\left( {6 - 5\sigma + {\sigma ^2}} \right){e^{ - \sigma /2}} = 0 \\
\Rightarrow 6 - 5\sigma + {\sigma ^2} = 0 \\
\Rightarrow 6 - 3\sigma - 2\sigma + {\sigma ^2} = 0 \\
\Rightarrow 3(2 - \sigma ) - \sigma (2 - \sigma ) = 0 \\
\Rightarrow (3 - \sigma )(2 - \sigma ) = 0 \\
\Rightarrow \sigma = 2,3
\]
In the question we are given \[\sigma = \dfrac{{2r.Z}}{{3{a_o}}}\], where r is the maximum radial distance of a node from the nucleus. This means:
\[
\dfrac{{2r.Z}}{{3{a_o}}} = 2{\text{ }}or{\text{ }}\dfrac{{2r.Z}}{{3{a_o}}} = 3 \\
\Rightarrow r = \dfrac{{3{a_o}}}{Z}{\text{ }}or{\text{ }}r = \dfrac{{9{a_o}}}{{2Z}} \\
\]
Note:
A wave function node exists at the points where wave function is zero and alters signs. The electron possesses zero probability of being positioned at a node. Owing to the separation of variables for an electronic orbital, the wave function is zero if any one of the component functions is zero.
Complete answer:
The names of the orbital i.e. s, p, d, or f stand for the names given to the groups of lines which are originally noted in the spectra of alkali metals. The line groups are known as sharp, principal, diffuse, or fundamental, respectively.
A node in an orbital refers to a plane of zero electron density. There are two types of nodes: (i) radial nodes which is equal to the total number of nodes (n-1) – the number of angular nodes (I) = (n-1) - l and (ii) angular nodes which is equal to the orbital angular momentum quantum number i.e. l.
We know that value of wave function ($\Psi $) and its square (${\Psi ^2}$) is zero at nodes i.e.
$\Psi (3s) = 0$ which means:
\[
\dfrac{1}{{9\sqrt 3 }}{\left( {\dfrac{1}{{{a_o}}}} \right)^{3/2}}\left( {6 - 5\sigma + {\sigma ^2}} \right){e^{ - \sigma /2}} = 0 \\
\Rightarrow 6 - 5\sigma + {\sigma ^2} = 0 \\
\Rightarrow 6 - 3\sigma - 2\sigma + {\sigma ^2} = 0 \\
\Rightarrow 3(2 - \sigma ) - \sigma (2 - \sigma ) = 0 \\
\Rightarrow (3 - \sigma )(2 - \sigma ) = 0 \\
\Rightarrow \sigma = 2,3
\]
In the question we are given \[\sigma = \dfrac{{2r.Z}}{{3{a_o}}}\], where r is the maximum radial distance of a node from the nucleus. This means:
\[
\dfrac{{2r.Z}}{{3{a_o}}} = 2{\text{ }}or{\text{ }}\dfrac{{2r.Z}}{{3{a_o}}} = 3 \\
\Rightarrow r = \dfrac{{3{a_o}}}{Z}{\text{ }}or{\text{ }}r = \dfrac{{9{a_o}}}{{2Z}} \\
\]
Note:
A wave function node exists at the points where wave function is zero and alters signs. The electron possesses zero probability of being positioned at a node. Owing to the separation of variables for an electronic orbital, the wave function is zero if any one of the component functions is zero.
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