In the decay sequence:
${}_{92}{}^{238}\mathrm{U}\stackrel{-{\mathrm{x}}_1}{\to }{}_{90}{}^{234}\mathrm{T}\mathrm{h}\stackrel{-{\mathrm{x}}_2}{\to }{}_{91}{}^{234}\mathrm{P}\mathrm{a}\stackrel{-{\mathrm{x}}_3}{\to }{}_{\mathrm{ }}{}^{234}\mathrm{Z}\stackrel{-{\mathrm{x}}_4}{\to }{}_{90}{}^{234}\mathrm{T}\mathrm{h}$
${\mathrm{x}}_1,\mathrm{ }{\mathrm{x}}_2,\mathrm{ }{\mathrm{x}}_3$ and ${\mathrm{x}}_4$ are the particles/ radiation emitted by the respective isotopes. The correct option(s) is/are:

• Correct answer
  A

  $\mathrm{Z}$ is an isotope of uranium.
• Correct answer
  B

  ${\mathrm{x}}_2$ is ${\mathrm{\beta }}^{-}$.
• Correct answer
  C

  ${\mathrm{x}}_1$ will deflect towards the negatively charged plate.
• D

  ${\mathrm{x}}_3$ is $\mathrm{\gamma }$-ray.

Solution

I.e., ${\mathrm{x}}_{\text{1}}=\mathrm{\alpha },$ ${\text{x}}_{\text{2}}=\text{β},$ ${\text{x}}_{\text{3}}=\text{β}$ and ${\text{x}}_{\text{4}}=\mathrm{\alpha }$.

Hence, the correct answer is options: 'A', 'B', 'C'.