The low lying energy levels due to the vibrational excitations of an even-even nucleus are shown in the figure below. The spin-parity $J^p$ of the level $E_1$ is
The question asks for the spin-parity $J^p$ of the level $E_1$ based on the energy level diagram provided.
Given the vibrational level scheme for an even-even nucleus, we observe energy levels at:
- $0^+$ at the ground state ($E_0 = 0 \,\text{MeV}$)
- $2^+$ at $E_0$ (indicating it is the first excited state due to quadrupole vibrations)
- $4^+$, $2^+$, and $0^+$ close to each other at higher energies (indicating further collective excitations)
In even-even nuclei, the first excited state is typically the $2^+$ state due to quadrupole vibrations. These states are collectively enhanced, contributing significantly to the energy spectrum.
Therefore, the energy level $E_1 = 0.56 \,\text{MeV}$ corresponds to the first excited state with spin-parity $J^p = 2^+$. Other options such as $1^+$, $1^-$, and $2^-$ do not typically represent the first excited state in such nuclei.
Thus, the correct answer is \(2^+\).
Important Questions from Miscellaneous
Which one of the following pairs of international organizations and their headquarters is incorrect?
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