the samarium - 147 nuclide radioactively decays by alpha emission. write a balanced nuclear chemical…

Explanation:

Step1: Recall alpha - particle composition

An alpha - particle is a helium nucleus, represented as $_{2}^{4}\text{He}$.

Step2: Determine the atomic and mass numbers of samarium - 147

Samarium has an atomic number of 62, so samarium - 147 is written as $_{62}^{147}\text{Sm}$.

Step3: Apply conservation of mass and atomic numbers

When ${62}^{147}\text{Sm}$ emits an alpha - particle (${2}^{4}\text{He}$), the new element's atomic number is $62 - 2=60$ (neodymium, Nd) and the mass number is $147 - 4 = 143$. The balanced nuclear chemical equation is ${62}^{147}\text{Sm}\rightarrow{60}^{143}\text{Nd}+_{2}^{4}\text{He}$.

Answer:

${62}^{147}\text{Sm}\rightarrow{60}^{143}\text{Nd}+_{2}^{4}\text{He}$