A stationary uranium-238 nucleus (\(^{238}_{92}U\)) decays into a nucleus of thorium-234 by emitting an alpha-particle - OCR - A-Level Physics A - Question 1 - 2017 - Paper 1

First, we calculate the mass of the alpha particle:

1. Mass of uranium nucleus: ( m_U = 4.0 \times 10^{-25} , \text{kg} )
2. Mass of thorium nucleus: ( m_{Th} = m_U - m_{alpha} ) where ( m_{alpha} ) is the mass of the alpha particle (approximately ( 6.7 \times 10^{-27} , \text{kg} )).
3. Calculate momentum of the alpha particle using ( p = mv ): [ p = 6.7 \times 10^{-27} , \text{kg} \times 2.4 \times 10^{5} , m/s \approx 1.608 \times 10^{-21} , \text{kg m/s} ]
4. Calculate the kinetic energy using ( KE = \frac{p^2}{2m} ): [ KE = \frac{(1.608 \times 10^{-21})^2}{2 \times 6.7 \times 10^{-27}} \approx 1.161 \times 10^{-13} , J ]
5. Convert to MeV: ( 1 , ext{J} = 6.242 \times 10^{12} , ext{MeV} ): [ KE \approx 1.161 \times 10^{-13} \times 6.242 \times 10^{12} \approx 0.73 , ext{MeV} ]

The decay chain of uranium-238 to lead-206 involves a change in the atomic number from 92 to 82 and a change in the mass number from 238 to 206. The changes can be calculated as follows:

1. Change in atomic number (Z): ( 92 - 82 = 10 )
2. Change in mass number (A): ( 238 - 206 = 32 )
3. Each alpha decay reduces Z by 2 and A by 4, meaning:
   • For Z: Each emission of an alpha particle decreases atomic number by 2, so the number of alpha decays required is ( \frac{10}{2} = 5 ).
   • For A: Each emission of an alpha particle decreases mass number by 4, so the number of alpha decays is ( \frac{32}{4} = 8 ).
4. Total particles emitted: 5 alpha particles + 6 beta particles (since each beta emission only decreases Z by 1, resulting in total changes): ( 5 + 6 = 14 ).
5. Conclusion: The total of 14 particles is emitted during the decay chain.