(IIT JEE 2005)
a) 2, 0
b) 0, 2
c) 1, 2
d) 2, 11
For a given orbital with principal quantum number, n and azimuthal quantum number, l
Number of radial nodes = (n - l - 1)
For 3s orbital: n = 3 and l = 0
Therefore, number of radial nodes = 3 - 0 - 1 = 2
For 2p orbital: n = 2 and l = 1
Therefore, number of radial nodes = 2 - 1 - 1 = 0
Correct option is "b".
For a given orbital, there are two types of nodes i.e. 1) Angular nodes (also known as nodal planes) 2) Radial nodes (also known as nodal regions).
The number of angular nodes = l
The number of radial nodes = (n - l - 1)
Total number of nodes = n - 1
n = Principal quantum number
l = Azimuthal quantum number
(Eamcet - 2011-E)
The number of radial nodes = (n - l - 1) = 4 - 0 - 1 = 3
Note: Do not confuse. No need of "m" value to get the number.
The orbital corresponding to valence electron and number of radial nodes for the given elements are shown below:
|Element||Orbital corresponding to valence electron||Number of radial nodes|
|Cu||4s||4 - 0 - 1 = 3|
|Li||2s||2 - 0 - 1 = 1|
|K||4s||4 - 0 - 1 = 3|
|Na||3s||3 - 0 - 1 = 2|
Only 3s orbital in Na contain 2 radial nodes.
The correct option is "d".
1) How many nodal planes are present in 2p, 3p and 4p orbitals?
2) Calculate the number of i) angular nodes ii) radial nodes and iii) total number of nodes for 4f orbital.
3) Which node in the atomic orbital passes through the nucleus?
4) Which atomic orbital has zero nodes?
Author: Aditya vardhan Vutturi Google+ profile