Uranium-233 decays to thorium-229 by α decay, but the emissions have different energies and products: 83% emit an α particle with energy of 4.816 MeV and give 229Th in its groundstate; 15% emit an α particle of 4.773 MeV and give 229Th in ex-cited state I; and 2% emit a lower energy α particle and give 229Th in the higher excited state II. Excited state II emits a γ ray of 0.060 MeV to reach excited state I. (a) Find the γ ray energyand wavelength that would convert excited state I to the groundstate. (b) Find the energy of the α particle that would convert 233U to excited state II.
Uranium-233 decays to thorium-229 by α decay, but the emissions have different energies and products: 83% emit an α particle with energy of 4.816 MeV and give 229Th in its groundstate; 15% emit an α particle of 4.773 MeV and give 229Th in ex-cited state I; and 2% emit a lower energy α particle and give 229Th in the higher excited state II. Excited state II emits a γ ray of 0.060 MeV to reach excited state I. (a) Find the γ ray energyand wavelength that would convert excited state I to the groundstate. (b) Find the energy of the α particle that would convert 233U to excited state II.
The relationship between energy and wavelength of the light is expressed as,
......(1)
Where,
- h is Planck's constant.
- c is the speed of light.
- is the wavelength of the light.
(a)
The energy of the gamma-ray that converts excited state I to the ground state is calculated by the formula,
Where,
- E1 is the energy of the alpha-particle emitted by 83% of decay.
- E2 is the energy of the alpha-particle emitted by 15% of decay.
Substitute the energy of alpha-particles emitted by 83% and 15% of decay in the above formula.
Convert 0.043 MeV to eV.
Convert to J.
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