# An electron is in the hydrogen atom with #n = 3#. #|L| = sqrt6 ℏ# . Which is a possible angle between #vecL# and the #z# axis?

##### 1 Answer

The only angle that satisfies the criteria is 65.9°.

#### Explanation:

In the **Schrödinger equation** for the hydrogen atom, the **squared orbital angular momentum operator**,

#bar(ul(|color(white)(a/a) ul(hat(L)^2)Y_(l)^(m_l)(theta,phi) = ul(l(l+1)ℏ^2)Y_(l)^(m_l)(theta,phi) color(white)(a/a)|))" "#

or

#color(blue)(bar(ul(|color(white)(a/a) ul(hat(L)) harr ul(sqrt(l(l+1))ℏ) color(white)(a/a)|)))" "#

where

#l# is the**angular momentum quantum number**.#Y_(l)^(m_l)(theta,phi)# is the spherical harmonic wave function for the orbital of a given#l# and#m_l# .#m_l# is the**magnetic quantum number**, and the projection of#l# along the#z# axis.

The

#bar(ul(|color(white)(a/a)ul(hatL_z)Y_(l)^(m_l)(theta,phi) = ul(m_lℏ)Y_(l)^(m_l)(theta,phi)color(white)(a/a)|))" "#

or

#color(blue)(bar(ul(|color(white)(a/a)ul(hatL_z) harr ul(m_lℏ) harr |L|cosθcolor(white)(a/a)|)))" "#

where

#m_l# is the**magnetic quantum number**and#θ# is the angle that#vecL# makes with the#z# -axis, or the angle of#hatL# with respect to#m_l# .

Thus

Some possibilities for

It appears that our electron is a

and

The possible angles for a

#2.4495 color(white)(mml)2color(white)(mml)0.8165color(white)(mll)35.26#

#2.4495 color(white)(mml)1color(white)(mml)0.4082color(white)(mll)65.90#

#2.4495 color(white)(mml)0color(white)(mml)0color(white)(mmmml)90#

#2.4495 color(white)(mm)"-1"color(white)(mm)"-0.4082"color(white)(m)114.09#

#2.4495 color(white)(mm)"-2"color(white)(mm)"-0.8165"color(white)(m)144.74#

The only angle that agrees with those given in the question is 65.90°.