 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2a/section-a/16/b/solution

For

B ≫ Δ_{o}

, the corresponding asymptotic expansion is

225 B^{2} + 18 B Δ_{o} + Δ_{o}^{2} = 15 B + \frac{3}{5} Δ_{o} + \frac{8 Δ _{o}^{2}}{375 B} + \dots .

Hence

E_{0} E_{1} E_{2} E_{3} = - \frac{3}{5} Δ_{o} - \frac{4 Δ _{o}^{2}}{375 B} + \dots, = \frac{1}{5} Δ_{o}, = 15 B + \frac{4 Δ _{o}^{2}}{375 B} + \dots, = \frac{6}{5} Δ_{o} .

At zero field,

E_{0}, E_{1}, E_{3}

converge to the

^{3} F

E_{2}

^{3} P

15 B

. The

^{3} T_{1 g} (P)

branch therefore has zero initial slope; its high-field slope is not zero.

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