= Solution
{c}
The remaining six positions give the other <isotope>'s sequence
$$
\boxed{\begin{array}{c|rrrrrr}
\text{peak}&d&f&g&h&j&k\\\hline
F&3&4&5&6&7&8\\
\nu\ (\mathrm{GHz})&10.99&9.74&8.17&6.29&4.09&1.58
\end{array}}
$$
The adjacent differences divided by $4,5,6,7,8$ are $0.3125,0.3140,0.3133,0.3143,0.3138\,\mathrm{GHz}$, respectively. Thus $a_{\nu,153}\simeq-0.314\,\mathrm{GHz}$ and the same endpoint argument gives $\boxed{I_{153}=5/2}$.
For the same electronic <energy level>, the magnetic-dipole <hyperfine structure> constant is proportional to $\mu_I/I$, to leading order in the small <isotope>-dependent change of the electronic <wavefunction>. Since the two <nuclear spins> are equal,
$$
\frac{\mu_{151}}{\mu_{153}}\simeq\frac{a_{151}}{a_{153}}.
$$
Using the full spans averages over the approximate <Landé interval rule>: $\sum_{F=4}^8F=30$, so
$$
|a_{\nu,151}|\simeq\frac{21.19}{30},\qquad
|a_{\nu,153}|\simeq\frac{10.99-1.58}{30}\quad\mathrm{GHz},
\qquad\boxed{\frac{\mu_{151}}{\mu_{153}}\simeq\frac{21.19}{9.41}\simeq2.25.}
$$
The constants have the same sign, so the <nuclear magnetic dipole moments> have the same relative sign. Their absolute sign cannot be obtained from their ratio without the electronic field's sign. Finite nuclear-size corrections to the <hyperfine interaction>, often called the <hyperfine anomaly>, and electric-quadrupole corrections are neglected at this accuracy.
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