# Moment of inertia of the nucleus

1. Nov 16, 2007

### genloz

1. The problem statement, all variables and given/known data
The unstable Tungsten isotope:
$$^{174}_{74}W$$
has various excited states. The spin parity
$$J^{P}$$
and energy levels of some of these states are given below in the table. Show that this series of states is generated by rotational motion in the nucleus and calculate moment of inertia of the nucleus in the 2+, 6+ and 10+ excited states. Comment on your results and compre them with the moment of inertia the nucleus would have were it a rigidly rotating sphere of radius R with an appropriate value of R.

$$J^{P} E(KeV)$$
0+ 0
2+ 112
4+ 355
6+ 704
8+ 1137
10+ 1635
12+ 2186

2. Relevant equations
Moment of inertia of a rigid sphere:
$$\frac{2}{5}MR^{2}$$
$$1.3A^{1/3} fm$$

3. The attempt at a solution
Moment of inertia of a rigid sphere:
$$=\frac{2}{5}MR^{2}$$
$$=\frac{2}{5}*(74*938.272 +74*0.511 +100*939.566)*(1.3*174^{1/3})^{2}$$
$$= as above MeV fm^2$$

So firstly I'm really unsure about the units of the above equation, and secondly I have no idea how to start to show that the series of states is generated by rotational motion...

I guess once I find out how to determine the moment of inertia of the nucleus then the excited states themselves would increase the size of the nucleus by a particular proportion equivalent to what's shown in the table?

A few hints would be very useful! Thanks!