Nuclear Rotational Spectrum Energy Levels

The energy levels of rotational spectra in nuclei, especially even-even nuclei like $^{238}_{\ 92}\text{U}$, often follow a pattern related to the rotational quantum number '$J$'. The ground state has '$J=0$', the first excited state has '$J=2$', and the second excited state has '$J=4$'.

The energy '$E_J$' of a rotational state is approximately proportional to '$J(J+1)$'. The formula is often expressed as:

$E_J = \frac{\hbar^2}{2I} J(J+1)$

Where '$I$' is the moment of inertia and '$\hbar$' is the reduced Planck constant.

Calculating Excited State Energies

Let '$K = \frac{\hbar^2}{2I}$'. We can express the energies relative to the ground state ($E_0 = 0$):

• First excited state ($J=2$): $E_2 = K \times 2(2+1) = 6K$. We are given $E_2 = 45$ keV.
• Second excited state ($J=4$): $E_4 = K \times 4(4+1) = 20K$. This is the energy we need to find.

From the first excited state energy, we can find '$K$':

$6K = 45 \text{ keV}$

$K = \frac{45 \text{ keV}}{6}$

Now, substitute this value of '$K$' into the expression for the second excited state energy:

$E_4 = 20K = 20 \times \left(\frac{45 \text{ keV}}{6}\right)$

$E_4 = \frac{20 \times 45}{6} \text{ keV} = \frac{900}{6} \text{ keV}$

$E_4 = 150 \text{ keV}$

Therefore, the energy of the second excited state of the $^{238}_{\ 92}\text{U}$ nucleus is 150 keV.