Uranium- 233 decays to thorium- 229 by \(\alpha\) decay, but the emissions have different energies and products: \(83 \%\) emit an \(\alpha\) particle with an energy of \(4.816 \mathrm{MeV}\) and give \({ }^{229} \mathrm{Th}\) in its ground state; \(15 \%\) emit an \(\alpha\) particle of \(4.773 \mathrm{MeV}\) and give \({ }^{229} \mathrm{Th}\) in excited state I; and \(2 \%\) emit a lower energy \(\alpha\) particle and give \({ }^{229} \mathrm{Th}\) in the higher excited state II. Excited state II emits a \(\gamma\) ray of \(0.060 \mathrm{MeV}\) to reach excited state I. (a) Find the \(\gamma\) -ray energy and wavelength that would convert excited state I to the ground state. (b) Find the energy of the \(\alpha\) particle that would raise \({ }^{233} \mathrm{U}\) to excited state II.

Alpha decay is a type of radioactive decay where an unstable atom ejects an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, essentially a helium-4 nucleus. This process reduces the atomic number by 2 and the mass number by 4. For example, in the case of Uranium-233 decaying into Thorium-229, Uranium undergoes an alpha decay resulting in the emission of an alpha particle and transforming into Thorium.

Gamma rays are high-energy photons emitted during radioactive decay, often from an excited state of the nucleus moving to a lower energy level. To calculate the wavelength of a gamma ray, we use the energy-wavelength relationship given by the equation:
\ \( \lambda = \frac{hc}{E} \)\.
Here, \( \lambda \) is the wavelength, \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \) eV·s), \( c \) is the speed of light (\( 3 \times 10^8 \) m/s), and \( E \) is the energy of the gamma ray.
For example, if the gamma ray energy is 0.043 MeV, the calculation would be:
\ \( \lambda = \frac{4.135667696 \times 10^{-15} \times 3 \times 10^8}{0.043 \times 10^6} = 2.89 \times 10^{-11} \) meters\.

Energy calculations involve determining the energy change when a nucleus undergoes decay or transitions between states. For instance, if excited state I of Thorium-229 emits a gamma-ray of 0.060 MeV to reach a lower energy state, this energy release must match the difference in the initial and final states.
For calculating the energy of alpha particles, consider the energy difference required. Given the energy of alpha emission at excited state I is 4.773 MeV and ground state is 4.816 MeV, the gamma-ray energy that converts excited state I to the ground state would be:
\ \( 4.816 \text{ MeV} - 4.773 \text{ MeV} = 0.043 \text{ MeV} \)\.

Nuclear excited states occur when a nucleus has higher energy than its ground state. These states are often reached and then decay by emitting radiation to lower energy levels, such as gamma rays.
In the case of Thorium-229, excited state I and II are higher energy forms of the nucleus compared to the ground state. Excited state II can emit a gamma-ray of 0.060 MeV to de-excite to excited state I.
To find the energy of alpha particles contributing to these states, subtract the energy emitted when transitioning between states. For example, converting Uranium-233 to excited state II involves alpha emission with an energy calculation as follows:
Given that the energy of alpha particle emission for state I is 4.773 MeV, and excited state II emits an additional 0.060 MeV, the alpha particle energy for state II would be:
\ \( 4.773 \text{ MeV} - 0.060 \text{ MeV} = 4.713 \text{ MeV} \)\.