= Solution
{c}
The $4f^7$ shell carries a large <spin> (its ground <atomic term> under <Hund's rules> has $L=0,S=7/2$), and the $6s$ <electron> penetrates the inner shells. Its nonzero <probability density> at the <atomic nucleus> supplies a potentially large <Fermi contact interaction> with the <nuclear magnetic dipole moment>. The $6s$ shell here is singly occupied, so the obligatory paired-$6s^2$ cancellation is absent. Both stable <europium> <isotopes> have nonzero <nuclear spin>, as the spectrum below also establishes.
The <term symbol> ${}^8D_{11/2}$ specifies $S=7/2$, $L=2$, $J=11/2$. Its large $J$, combined with appreciable <nuclear spin>, permits a wide range of total <angular momentum> $F$ and hence a substantial total <hyperfine structure> span. For $I=5/2$, $F=3,\ldots,8$ and the span is $30|a|$. The penetrating $6s$ contribution and the available range of $F$ explain the large <hyperfine structure>; the <term symbol> alone does not determine its constant, which also depends on electronic recoupling, radial <wavefunctions> and <configuration interaction>.
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