Question

A uranium nucleus ${}_{92}{U}^{238}$ emits an $\alpha$ - particle and $\beta$-particle in succession. The atomic number and mass number of the final nucleus will be

• A. 90,233
• B. 90,238
• C. 91,234
• D. 93,238

Answer 1

Answer: (C)

When a radioactive nucleus emits an $\alpha$ -particle (a helium nucleus ${}_{2}H{e}^4$ ), its mass number decreases by $4$ while the atomic number decreases by $2$. When a radioactive nucleus emits a $\beta$ -particle (an electron, $e^{-}$ ) its mass number remains unchanged while the atomic number increases by $1$.
$\therefore {}_{92}{U}^{238}\stackrel{\alpha }{\to }{}_{90}T{h}^{234}\stackrel{\beta }{\to }{}_{91}P{a}^{234}$