Question

what type of nuclear decay is shown by the reaction below?
ce{_{92}^{238}u
ightarrow _{2}^{4}he + _{90}^{234}th}

a. beta
b. fusion
c. gamma
d. alpha

Explanation:

1. Recall the characteristics of each nuclear decay type:

• Alpha decay involves the emission of an alpha particle ($\ce{^4_2He}$), which has 2 protons and 2 neutrons.
• Beta decay involves the emission of a beta particle (electron or positron), and the mass number doesn't change by 4 or the atomic number by 2.
• Gamma decay is the emission of gamma rays (high - energy photons) and doesn't change the atomic or mass number of the nucleus.
• Fusion is the combination of two or more lighter nuclei to form a heavier nucleus, not a decay process.

1. Analyze the given reaction $\ce{^{238}_{92}U \to ^4_2He + ^{234}_{90}Th}$:

• The particle emitted is $\ce{^4_2He}$, which is an alpha particle. The mass number of the parent nucleus (238) decreases by 4 (238 - 234 = 4) and the atomic number decreases by 2 (92 - 90 = 2), which is consistent with alpha decay.

Answer:

D. Alpha