Originally Posted by **pranimaboity2050** In a many electron atom the orbital spin and total angular momenta are denoted by **L**,**S** and **J** respectively.If **L**=2, **S**=1, and **J**=2, find the angle between **L**,**S**
using vector atomic model.
Attempt: **J**=**L**+**S** **J**.**J**=|**L**|^2 +|**S**|^2 + 2|**L**|.|**S**| cosθ.
So cosθ= [|**J**|^2 -|**L**|^2-|**S**|^2]/ [2|**L**|.|**S**| ]
Now |**L**|=([L (L+1)]^1/2 ) hbar
|**S**|=([S(S+1)]^1/2 ) hbar ,|**J**|=([J(J+1)]^1/2 ) hbar
So cosθ=[J(J+1) - L (L+1)-S (S+1)]/[2([L (L+1)]^1/2 )×([S (S+1)]^1/2 ) ]
Now cosθ= 6-6-2/2×(6×2)^1/2 =-0.2886
θ=106.78º.Have I done this correctly? I couldn't find a suitable worked example to be sure by myself. |

Your question is ill posed. There are a number of possible answers to your question.

Your overall method is okay but the value of l is not fixed when we give a value for |

**L**|. In your example, |

**L**| = 2 so we have three possible values for l: l = 0, 1, 2. These are the kinds of values you use in your cosine equation: we are not using J, L, and S...we are using j, l, and s. We can generate a list of possible values for cos(t) (I am using t = theta). The j(j + 1) equation is better to use as we might have l = 0 and/or s = 0.

For j = 1, l = 1, and s = 1 we have cos(t) = 1/2. For j = 2, l = 2, and s = 1 we have that cos(t) = -1/(2 sqrt(3)). As these two angles are not the same there is more than one answer to your question.

(You can also do this using mj = ml + ms. Given L = 2 implies l = 0, 1, 2 implies ml = -2, -1, 0, 1, and 2 and similar for J and S.)

-Dan

Edit: Apparently the codecogs website has changed something. You can barely make out the equation. It is the same as your J(J + 1) equation using the cosine rule, but with lowercase letters in place of the capitals.