Structure of Atom Part 1

As, we know,
For 2s orbital, n=2,I=0
Angular node (l)=0
Radial nodes =n−1−1‌Rightarrow‌2−0−1=1
For 2p orbital, n=2 and I=1
Angular node (l)=1
Therefore, number of radial nodes =2−1−1=0
It has one angular and zero radial node.

For 3p orbital, n=3,I=1
Angular node (l)=1
It has one angular and one radial node.
For 3s orbital, n=3,I=0
No. of angular node (l)=0
No. of radial node Rightarrow‌n−I−1
=3−0−1=2=2
Hence, for 3 s-orbital has no angular node and 2 radial nodes.