Question

uranium-238 undergoes alpha decay to form what daughter isotope and decay particle? option 1: $_{92}^{238}u+_{2}^{4}he$ option 2: $_{92}^{234}u+_{2}^{4}he$ option 3: $_{90}^{238}th+_{2}^{4}he$ option 4: $_{90}^{234}th+_{2}^{4}he$

Explanation:

Step1: Recall alpha - decay concept

In alpha - decay, an atomic nucleus emits an alpha particle ($_{2}^{4}\text{He}$). The mass number of the parent nucleus decreases by 4 and the atomic number decreases by 2.

Step2: Analyze Uranium - 238 decay

Uranium - 238 has a mass number ($A$) of 238 and an atomic number ($Z$) of 92. After alpha - decay, the mass number of the daughter nucleus is $238 - 4=234$ and the atomic number is $92 - 2 = 90$. The element with atomic number 90 is Thorium (Th). So the decay equation is $_{92}^{238}\text{U}
ightarrow_{90}^{234}\text{Th}+_{2}^{4}\text{He}$.

Answer:

Option 4. $_{90}^{234}\text{Th}+_{2}^{4}\text{He}$