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Question

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 correct answer is
$2^+$

The question asks for the spin-parity $J^p$ of the level $E_1$ based on the energy level diagram provided.

Energy Level Diagram

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 \(2^+\).

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