Physics · Nuclear Physics
A 57Fe nucleus decays from an excited state to the ground state by emitting a gamma ray. The energy of the emitted photon is 14.4 keV when the nucleus is held fixed in a lattice. If the nucleus is free to recoil, the energy of the emitted photon will be:
Verified Solution4 OptionsUpdated: October 2026👁️1 View
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- 1Higher by approximately 20 eV
- 2Higher by approximately 2 eV
- 3Lower by approximately 2 meV✓
- 4Lower by approximately 2 keV
Correct answer
Lower by approximately 2 meV
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