Energy splitting in hyperfine structure

[Kara386]

Homework Statement

I've been given the worked answer to a very similar question, but there's a step I don't understand so I can't apply it. My question asks:
An energy level with $J=\frac{11}{2}$ has six hyperfine sub-levels with these relative energies:
0, 2.51, 4.71, 6.59, 8.16, 9.41 (GHz)
What is the value of I? Show the spacing obeys the interval rule and determine the value of F for each sub-level.

Homework Equations

The Attempt at a Solution

So based on this example question I have, start from the fact that $F = |I-J|...(I+J)$, integer steps, and label the lowest sub-level as $F_0 = |I-\frac{11}{2}|$. That means I have energy levels from $F_0$ to $F_0 +5$. Then I can make a table:
$\begin{array}{|c|c|} \hline F & \Delta E_F -\Delta E_{F-1}\\ \hline F_0 +5 & 1.25 \\ \hline F_0 + 4 & 1.57 \\ \hline F_0 + 3 & 1.88 \\ \hline F_0 + 2 & 2.2 \\ \hline F_0 + 1 & 2.51 \\ \hline \end{array}$
Found by referring back to the listed energy levels, so $F_0 + 5 = 9.41 - 8.16$ and so on. Then take ratios of energy splittings, and here's what I don't understand. From
$\frac{F_0+5}{F_0 + 4} = \frac{1.25}{1.57}$, a value for $F_0 +5$ is deduced.

In the example it states
$\frac{F_0 +2}{F_0 +1} = \frac{202}{151}$ therefore $F_0+1 = 3$. How did they work that out? They have $F_0 = |I - \frac{3}{2}|$ and the energy level $F_0$ is at $151$MeV. Don't understand that step. Thanks for any help!

 

[mfb]

In the example it states
$\frac{F_0 +2}{F_0 +1} = \frac{202}{151}$ therefore $F_0+1 = 3$. How did they work that out?

By finding a small fraction that fits, in this case $\frac{202}{151} \approx \frac 4 3$. You can find a similar fraction if you divide 1.57 by 1.25, and note a pattern if you continue with 1.88/1.57 and so on.

151 MeV is way too large for the hyperfine structure.

 

[Kara386]

By finding a small fraction that fits, in this case $\frac{202}{151} \approx \frac 4 3$. You can find a similar fraction if you divide 1.57 by 1.25, and note a pattern if you continue with 1.88/1.57 and so on.

151 MeV is way too large for the hyperfine structure.

Apologies, I meant MHz! 151MeV would be pretty huge. I don't think I can edit it anymore.

 

[Kara386]

By finding a small fraction that fits, in this case $\frac{202}{151} \approx \frac 4 3$. You can find a similar fraction if you divide 1.57 by 1.25, and note a pattern if you continue with 1.88/1.57 and so on.

151 MeV is way too large for the hyperfine structure.

Well for my question $\frac{F_0+5}{F_0+4} = \frac{1.25}{1.57} \approx 0.796 \approx \frac{4}{5}$ which gives $F_0 = -1$ or $1$, so I think I'm doing something wrong. And looking at the next fraction,
$\frac{F_0+4}{F_0+3} = \frac{1.57}{1.88} \approx \frac{5}{6}$ again gives two different answers for $F_0$ of 1 and 2. What am I doing wrong? Are the fractions I've approximated to incorrect?

 

[mfb]

It has been a long time since I last worked with hyperfine structures, but the series of fractions is so clear that it must be the right approach.