the uranium - 238 nuclide radioactively decays by alpha emission. write a balanced nuclear chemical equation…

Explanation:

Step1: Recall alpha - particle composition

An alpha - particle is a helium nucleus with a mass number of 4 and an atomic number of 2, represented as $_{2}^{4}\text{He}$.

Step2: Write the initial nuclide

Uranium - 238 has a mass number of 238 and an atomic number of 92, written as $_{92}^{238}\text{U}$.

Step3: Determine the daughter nuclide

In alpha - decay, the mass number of the parent nuclide decreases by 4 and the atomic number decreases by 2. So the mass number of the daughter nuclide is $238 - 4=234$ and the atomic number is $92 - 2 = 90$. The element with atomic number 90 is thorium (Th).

Step4: Write the balanced nuclear equation

${92}^{238}\text{U}\rightarrow{90}^{234}\text{Th}+_{2}^{4}\text{He}$

Answer:

${92}^{238}\text{U}\rightarrow{90}^{234}\text{Th}+_{2}^{4}\text{He}$