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

In the Schrödinger equation for the hydrogen atom, the squared orbital angular momentum operator, #hatL^2#, corresponds to the eigenvalue, #l(l+1)ℏ^2#, according to this formula:

#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 #z#-component of the angular momentum similarly corresponds to the eigenvalue #m_lℏ# according to:

#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

#θ = cos^"-1"(m_l/|L|)#

Some possibilities for #n = 3# are

#bb(lcolor(white)(mmm)|L|//ℏ)#
#0color(white)(ml)sqrt(0(1)) = 0#
#1color(white)(ml)sqrt(1(2)) = sqrt2#
#2color(white)(ml)sqrt(2(3)) = sqrt6#

It appears that our electron is a #"d"# electron (#l = 2#).

and

#|L| = sqrt6ℏ = 2.4495ℏ#

The possible angles for a #"d"# electron are

#bb(color(white)(m)|L|"/"ℏcolor(white)(mm)m_l//ℏcolor(white)(ml)cosθcolor(white)(mmm)θ)#

#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°.