
How many numbers of nodes are present in $3s$ orbital?
Answer
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Hint: To solve this problem, find out both the nodes and you can use the direct formula for the number of angular nodes which equals to the azimuthal quantum number while for the number of spherical nodes equals to $n-l-1$, where $n$ is the principal quantum number and $l$ is the azimuthal quantum number.
Complete step by step answer:
So, nodes are the points where the electron density is zero. There are two types of nodes for a given orbital. That are:
- Angular nodes: Angular nodes are also called a “nodal plane” and they are found in p, d, and f-orbitals. The s-orbital has no angular nodes.
- Radial or spherical nodes: Radial nodes are also called as the nodal regions. They are found in $2s$, $2p$, $3p$, $4p$, $4d$, $5d$and so on orbitals.
Here, we have to find out the number of nodes that is present in $3s$-orbital. So, we will be finding both the angular nodes and the radial nodes.
Since here, the orbital is s-orbital, so it has azimuthal quantum number (which is denoted by “$l$”) equals zero.
And we know, the formula for finding angular nodes is equal to its azimuthal quantum number. So, here as the azimuthal quantum is zero for the s-orbital, so it will have no angular nodes i.e. the angular node for $3s$-orbital will be equal to zero.
Now coming to the radial nodes, we know it is given by the formula, $n-l-1$, where $n$ is the principal quantum number and here it is given as $3$while $l$ is the azimuthal quantum number and for s-orbital, it is zero.
So, on putting these values in the formula we get,
The number of spherical or radial nodes in $3s$-orbital is $3-0-1=2$.
Therefore, on solving we get,
The number of spherical nodes $=2$.
The number of angular nodes $=0$.
Hence, the angular nodes and spherical nodes present in $3s$-orbital are $0$ and $2$.
Note: The total number of nodes of any orbital are given by the formula $n-1$, where n represents the principal quantum number. And we know that angular nodes are equal to the azimuthal quantum number. So, the number of radial nodes can also be calculated by differentiating the total number of nodes and angular nodes.
Complete step by step answer:
So, nodes are the points where the electron density is zero. There are two types of nodes for a given orbital. That are:
- Angular nodes: Angular nodes are also called a “nodal plane” and they are found in p, d, and f-orbitals. The s-orbital has no angular nodes.
- Radial or spherical nodes: Radial nodes are also called as the nodal regions. They are found in $2s$, $2p$, $3p$, $4p$, $4d$, $5d$and so on orbitals.
Here, we have to find out the number of nodes that is present in $3s$-orbital. So, we will be finding both the angular nodes and the radial nodes.
Since here, the orbital is s-orbital, so it has azimuthal quantum number (which is denoted by “$l$”) equals zero.
And we know, the formula for finding angular nodes is equal to its azimuthal quantum number. So, here as the azimuthal quantum is zero for the s-orbital, so it will have no angular nodes i.e. the angular node for $3s$-orbital will be equal to zero.
Now coming to the radial nodes, we know it is given by the formula, $n-l-1$, where $n$ is the principal quantum number and here it is given as $3$while $l$ is the azimuthal quantum number and for s-orbital, it is zero.
So, on putting these values in the formula we get,
The number of spherical or radial nodes in $3s$-orbital is $3-0-1=2$.
Therefore, on solving we get,
The number of spherical nodes $=2$.
The number of angular nodes $=0$.
Hence, the angular nodes and spherical nodes present in $3s$-orbital are $0$ and $2$.
Note: The total number of nodes of any orbital are given by the formula $n-1$, where n represents the principal quantum number. And we know that angular nodes are equal to the azimuthal quantum number. So, the number of radial nodes can also be calculated by differentiating the total number of nodes and angular nodes.
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