Question

The isotope ${}_{92}{\text{U}}^{238}$ decays successively to form ${}_{90}{\text{Th}}^{234}$, ${}_{91}{\text{Pa}}^{234}$, ${}_{92}{\text{U}}^{234}$, ${}_{90}{\text{Th}}^{230}$ and ${}_{88}{\text{Ra}}^{226}$. What are the radiation emitted in these five steps?

• A. $\alpha ,{\beta }^{-},{\beta }^{-},\alpha ,\alpha$
• B. $\alpha ,{\beta }^{+},\alpha ,\alpha ,\alpha$
• C. $\alpha ,{\beta }^{-},{\beta }^{+},{\beta }^{-},\alpha$
• D. $\alpha ,\alpha ,\alpha ,{\beta }^{-},{\beta }^{-}$

Answer 1

The correct option is A $\alpha ,{\beta }^{-},{\beta }^{-},\alpha ,\alpha$

The given successive decays can be represented by the following reactions,

$\left(1\right)$ ${}_{92}{\text{U}}^{238}\stackrel{\alpha }{\to }{}_{90}{\text{Th}}^{234}+{}_2{\text{He}}^4$

$\left(2\right)$ ${}_{90}{\text{Th}}^{234}\stackrel{{\beta }^{-}}{\to }{}_{91}{\text{Pa}}^{234}+{}_{-1}{e}^0$

$\left(3\right)$ ${}_{91}{\text{Pa}}^{234}\stackrel{{\beta }^{-}}{\to }{}_{92}{\text{U}}^{234}+{}_{-1}{e}^0$

$\left(4\right)$ ${}_{92}{\text{U}}^{234}\stackrel{\alpha }{\to }{}_{90}{\text{Th}}^{230}+{}_2{\text{He}}^4$

$\left(5\right)$ ${}_{90}{\text{Th}}^{230}\stackrel{\alpha }{\to }{}_{88}{\text{Ra}}^{226}+{}_2{\text{He}}^4$

Therefore, the radiations emitted in these five steps are $\alpha ,{\beta }^{-},{\beta }^{-},\alpha ,\alpha$

Hence, option $\left(A\right)$ is correct.

Why this Question? This will help in understanding the successive decay of nuclide due to $\alpha$ and $\beta$ decay.