Question

Find the ground state term symbol for Pr\(^{3+}\) (Pr = 59).

• A. \( ^3H_4 \)
• B. \( ^3H_6 \)
• C. \( ^3H_5 \)
• D. \( ^4F_{5/2} \)

Hint:

For rare earth ions like Pr\(^{3+}\), the term symbol is derived from the \( 4f^n \) configuration and the values of \( S \) and \( L \).

Answer 1

The Correct Option is C

Step 1: Electronic Configuration of Pr\(^{3+}\).
Praseodymium (Pr) has an atomic number of 59, and the electronic configuration is: \[ \text{Pr:} [Xe] 4f^3 6s^2 \] For \( \text{Pr}^{3+} \), three electrons are removed from the 4f shell, leaving a configuration of \( 4f^3 \).

Step 2: Term Symbol Calculation.
For \( 4f^3 \), the total spin quantum number (\( S \)) is 3/2, and the total orbital quantum number (\( L \)) is 3 (which corresponds to the letter H). The multiplicity is given by \( 2S+1 \), which results in \( 2(3/2) + 1 = 4 \). Hence, the term symbol is \( ^3H_5 \).

Step 3: Conclusion.
Thus, the ground state term symbol for Pr\(^{3+}\) is \( ^3H_5 \), corresponding to option (3).