Question

the no of nodes in s orbital of any energy level is equal to​

Answer 1

Answer:

Angular node in s orbital is zero

Explanation:

Angular node in s orbital is zero but no. of radial node in s orbital is equal to n-1 where n is principle quantum number.

Answer 2

Answer:

Explanation:

Question:

The number of nodes in s orbital of any energy level is equal to​

A) $n^$

B) $2n^{2}$

C) $n-1$
D) $n-2$

Step by step explanation:

Hint: To solve this problem, identify both nodes. You can use the direct formula for the number of angular nodes, which equals the azimuthal quantum number, or the direct formula for the number of spherical nodes, which equals $n-1-1$,

where $n$ is the principal quantum number and $1$ is the azimuthal quantum number.

Therefore, nodes are the locations where there is no electron density. For a specific orbital,

there are two sorts of nodes.

those are:

• Angular nodes: The p, d, and f-orbitals include angular nodes, which are also known as "nodal planes."  No angular nodes exist in the $s$ -orbital.

• Spherical or radial nodes: These structures are also referred to as nodal regions. They are present in orbitals with $2s, 2p, 3p, 4p, 4d, 5d,$ and and so on orbitals.

Here, we need to figure out how many nodes are part of the $3s$ orbit. So, we'll be looking for both angular and radial nodes.

Since here, the orbital is $s$ -orbital, so it has azimuthal quantum number (which is denoted by “$1$”) 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, $n-1-1$,

where $n$ is the principal quantum number and here it is given as 3 while 1 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.

Answer: Option (c)

The number of nodes in s orbital of any energy level is equal to​ $n-1$.

#SPJ3