Show that the first excited state of 8-Be nucleus fits the experimental value of the rotational band $E(2^+) = 92 keV$ (this is the first excited state, which is 92 keV above the ground state). To do so model 8-Be to be made of 2 alpha particles $4.5 fm$ apart rotating about their center of mass.
The method I used is:
1) Get the moment of inertia of the system.
For two particles rotating each other (in the same plane) about its center of mass, one gets:
$$I = m_1 r_1^2 + m_2 r_2^2 = 2m_{\alpha} (d/2)^2 = 6.728 \times 10^{-56}kgm^2$$
where:
$$m_{\alpha} = 6.645 \times 10^{-27} kg$$
$$d = 4.5 \times 10^{-15}m$$
I've checked this value and it's correct.
2) Verify the experimental value for the rotational band by applying the rotational energy formula:
$$E = J (J + 1) \frac{\hbar^2}{2I}$$
OK let's first get what's called "characteristic rotational energy": $\frac{\hbar^2}{2I}$
$$\frac{\hbar^2}{2I} = 8.256 \times 10^{-14} J= 0.516 MeV$$
where:
$$\hbar = 1.054 \times 10^{-34}Js$$
$$1 eV = 1.6 \times 10^{-19}J$$
$2^+$ band has $J=2$ associated with it. So it's just a matter of plugging numbers in:
$$E = J (J + 1) \frac{\hbar^2}{2I} = 3096 keV$$
Which is way off $92keV$.
Where is my method gone wrong? Maybe the original question provided an incorrect experimental result.
PS/ For anyone interested in rotational energy related to nucleus this is a good video to check: https://www.youtube.com/watch?v=rwdBnwznt3s
