# What is a conical node? How can you find one?

##### 1 Answer

It's a kind of angular node that you can see, for instance, in the

Last year, I also drew in the orbital and its node myself, so it may help you to look at this too:

For a

- The total number of
**nodes**is#n - 1 = 3 - 1 = bb(2)# . - The total number of
**radial**nodes is#n - l - 1 = 3 - 2 - 1 = bb(0)# . - The total number of
**angular**nodes is#l = bb(2)# .

Thus, the **two** conical nodes (top half and bottom half) it has are **both** angular, and the conical node is a kind of angular node.

The **angular wave function** that demonstrates this is:

#color(green)(Y_(l)^(m_l)(theta,phi)) = Y_(2)^(0)(theta,phi) = color(green)(sqrt(5/(16pi))(3cos^2theta - 1))#

Since a node is when the wave function

#0 = 3cos^2theta - 1#

#=> cos^2theta = 1/3#

#=> theta =# #arccos(pm1/sqrt3)#

#=> color(blue)(theta ~~ 54.74^@, 125.26^@)# ,

which are the degrees of descent from the top of the

Due to how there is no

- Tilt
#54.74^@# away from the top of the#z# axis, and revolve around the#z# axis, and you have one of the angular nodes. - Tilt
#125.26^@# away from the top of the#z# axis, and revolve around the#z# axis, and you have the other angular node.

Thus, due to the **two** solution angles, there is an exact copy of the top half of the conical node at the bottom, giving us **two angular conical nodes**.