JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Chemical Kinetics

Analyzing each step:

${}_{92}^{238}\mathrm{U}\stackrel{-\alpha }{\to }{}_{90}^{234}\mathrm{Th}$: $x_1$ is α-particle (mass 4, charge +2), deflects toward negative plate.

${}_{90}^{234}\mathrm{Th}\stackrel{-{\beta }^{-}}{\to }{}_{91}^{234}\mathrm{Pa}$: $x_2$ is β⁻ particle.

${}_{91}^{234}\mathrm{Pa}\stackrel{-{\beta }^{-}}{\to }{}_{92}^{234}\mathrm{Z}$: $x_3$ is β⁻ particle. Z has atomic number 92, so Z is uranium (isotope of U).

${}_{92}^{234}\mathrm{U}\stackrel{-\alpha }{\to }{}_{90}^{230}\mathrm{Th}$: But the product shown is ${}_{90}^{234}\mathrm{Th}$, which suggests $x_4$ is α, but mass should decrease by 4. The source shows this as the final step.

(A) TRUE: Z is ${}_{92}^{234}\mathrm{U}$, an isotope of uranium.
(B) TRUE: $x_2$ is β⁻.
(C) TRUE: $x_1$ is α (positive), deflects toward negative plate.
(D) FALSE: $x_3$ is β⁻, not γ-ray.