The number of micro states corresponding to the atomic term symbol 4F is
28
The number of microstates corresponding to a specific atomic term symbol ${}^{(2S+1)}\text{L}$ can be calculated using a simple formula. The term symbol provides information about the total spin angular momentum ($S$) and the total orbital angular momentum ($L$) of an atomic state.
The given atomic term symbol is ${}^{4}\text{F}$. Let's break down what this symbol represents:
| L Value | Letter Code |
|---|---|
| 0 | S |
| 1 | P |
| 2 | D |
| 3 | F |
| 4 | G |
| ... | ... |
The total number of microstates associated with a term symbol ${}^{(2S+1)}\text{L}$ is given by the product of the spin multiplicity and the number of possible orientations for the orbital angular momentum. The number of orientations for orbital angular momentum $L$ is $2L+1$.
The formula for the number of microstates is:
$\text{Number of Microstates} = (2S+1)(2L+1)$
Now, we substitute the values derived from the term symbol ${}^{4}\text{F}$:
Using the formula:
$\text{Number of Microstates} = (4)(7) = 28$
Therefore, the number of microstates corresponding to the atomic term symbol ${}^{4}\text{F}$ is 28.
The possible terms arising from a p1d1 configuration are
The ground state term symbol and the gj value for Pr3+ ion, respectively are (Atomic number of Pr is 59)
For the d3 electron configuration, the ground state term symbol is
The term symbol for the ground dinitrogen cation radical (\(N^+_2\)) is
For the Eu3+ ion (At No: 63),
A. the calculated and the observed magnetic moments are in agreement with each other.
B. the higher energy states 7F1 and 7F2 are populated and increase the observed magnetic moment.
C. the 4f orbital is more than half-filled.
D. the ground state term symbol is 7F0.
Of the above, the correct statements are