Uranium-233decays to thorium-229by decay, but the emissions have different energies and products: 83%emit an particle with energy of 4.816MeV and give 229Th in its ground state; 15%emit an particle of4.773MeV and give ²²⁹Th in excited state I; and2%emit a lower energy particle and give 229Th in the higher excited state II. Excited state II emits a ray of 0.060MeV 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 particle that would convert²³³U to excited state II.

The distance between “two crests” is the wavelength of a wave.

Representation of wave

In mathematics, the Greek symbol lambda () is used to indicate wavelength.

A. Determine the $\gamma$-ray energy and wavelength required to convert the excited state.

We need to use this formula:

….. (i)

• $\gamma$= Wavelength
• h = Planck’s constant
• c = Speed of light
• E = Energy

The binding energies are generally represented in millions of electron volts, or mega-electron volts (MeV), for the derivation:

$$\begin{gathered} 1MeV={10}^{6}eV \\ =1.602x{10}^{-13}J \end{gathered}$$

$$\begin{gathered} Energy(\gamma -ray)=(4.816-4.773)MeV \\ =0.043MeV \end{gathered}$$

The known values are substituted.

$\lambda =\frac{\left(6.626x{10}^{-34}J.s\right)\left(\left(2.99792x{10}^8\frac{m}{s}\right.\right)}{0.043MeV}\times \left(\frac{1MeV}{1.602\times {10}^{-13}}\right)$

Therefore, the gamma-ray energy is 0.043 MeV and wavelength $\lambda =2.8836\times {10}^{-11}m$.

B. The energy of a particle that would convert ²³³U to excited state II,

$$\begin{gathered} Energy=4.816MeV-\left(0.043MeV-0.060MeV\right) \\ =4.713MeV \end{gathered}$$

Therefore,the energy of the particle =4.173 MeV.