3. The 6s¹6p¹ electron configuration, as discussed by Mason, gives rise to ³P₀, ³P₁, ³P₂, and ¹P₁ terms. Show that the sum of the degeneracies of each J state is equal to the total number of microstates for the 6s¹6p¹ configuration.

Number of microstates = 

The degeneracy of a J state = 2J+1, so:

³P₀: J = 0, 2(0) + 1 = 1 state

³P₁: J = 1, 2(1) + 1 = 3 states

³P₂: J = 2, 2(2) + 1 = 5 states

¹P₁: J = 1, 2(1) + 1 = 3 states

Total number of states = 1 + 3 + 5 + 3 = 12 equals the number of microstates.