In the decay sequence: U 92238 - x _ 1 → Th 90234 - x _ 2 → Pa 91234 - x _ 3 → Z 234 - x _ 4 → Th 90234 x _ 1 , x _ 2 , x _ 3 and x _ 4 are the particles/ radiation emitted by the respective isotopes. The correct

Correct answer: A. Z is an isotope of uranium. I.e., x _ 1 = α, x _ 2 = β, x _ 3 = β and x _ 4 = α .

JEE Advanced 2019Paper 1

Exam: JEE Advanced
Year: 2019
Paper: Paper 1
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Advanced 2019 · Paper 1 · Chemistry · Chemical Kinetics

In the decay sequence: U 92238 - x _ 1 → Th 90234 - x _ 2 → Pa...

In the decay sequence:

U 92238 \overset{- {x}_{1}}{\to} Th 90234 \overset{- {x}_{2}}{\to} Pa 91234 \overset{- {x}_{3}}{\to} Z 234 \overset{- {x}_{4}}{\to} Th 90234

{x}_{1}, {x}_{2}, {x}_{3}

and

{x}_{4}

are the particles/ radiation emitted by the respective isotopes. The correct option(s) is/are: In the decay sequence: U 92238 - x _ 1 → Th 90234 - x _ 2 → Pa 91234 - x _ 3 → Z 234 - x _ 4 → Th 90234 x _ 1 , x _ 2 ,

In the decay sequence: U 92238 - x _ 1 → Th 90234 - x _ 2 → Pa 91234 - x _ 3 → Z 234 - x _ 4 → Th 90234 x _ 1 , x _ 2 ,

Options

A $Z$ is an isotope of uranium.
B ${x}_{2}$ is ${\beta}^{-}$ .
C ${x}_{1}$ will deflect towards the negatively charged plate.
D ${x}_{3}$ is $\gamma$ -ray.

Correct answer

A. $Z$ is an isotope of uranium.

Step-by-step solution

I.e.,

{x}_{\text{1}} = \alpha,

{\text{x}}_{\text{2}} = \text{\beta},

{\text{x}}_{\text{3}} = \text{\beta}

and

{\text{x}}_{\text{4}} = \alpha

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