From atomic structure we know an electron can only have has an intrinsic spin vector of s = 1/2. From quantum mechanics the magnitude of the spin vector is ׀s׀ is,

where h (h bar) is Planck’s constant divided by 2π. Quantum mechanics also restricts the value of the component of s along the z-axis of the electron to m_{Z} to ±1/2h (spin up ↑ and spin down ↓). What about the other two Cartesian components of the spin vector (m_{x} and m_{y}), can we say anything about them? Due to Heisenberg’s Uncertainty Principle if we know my exactly then we can know nothing about m_{x} and m_{y} hence our spin vector points somewhere on a cone of uncertainty as in below Figre. Calculate the half-angle of the cone of uncertainty θ of the electron spin angular momentum vectors.

The 2p electrons have an orbital angular momentum ** l** = lh. From quantum mechanics the magnitude of the orbital angular momentum vector is ׀

**׀.**

*l***׀ l ׀ = √ l (l + 1) h = √2 h**

Quantum mechanics also restricts the value of the component along the z-axis to one of three possible values mz equal to -1- lh, 0 h, or – 1h. Again because we know the value of the component along the z-axis exactly, we know nothing about the values along the x-axis and y-axis. The orbital angular momentum vector points somewhere on one of three “cones of uncertainty” as in below Figure. Calculate the half-angles of the cones of uncertainty El of the electron orbital angular momentum vectors.

**Solution**

The angles of the two half cones for the electron spin are measured from the positive z direction, so the sn,= -1/2 cone half-angle is 0 = 180°-125.3° = 54.7° from the negative z-axis. The electron orbital angular momentum vector point in one of three orientations for the p atomic orbitals.

The m_{1} = -1 cone half-angle is θ = 1800-135° = 45° from the negative z-axis.

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