JEE Challenger
More from Structure of Atom
ChemMedAdv 2019 • Paper 1

Nuclear Decay Sequence and Identification of Emitted Radiation

In the decay sequence,

92238U→−x190234Th→−x291234Pa→−x3Z→−x490230Th{}^{238}_{92}\text{U} \xrightarrow{-x_1} {}^{234}_{90}\text{Th} \xrightarrow{-x_2} {}^{234}_{91}\text{Pa} \xrightarrow{-x_3} \text{Z} \xrightarrow{-x_4} {}^{230}_{90}\text{Th}

x1,x2,x3x_1, x_2, x_3 and x4x_4 are particles/radiation emitted by the respective isotopes. The correct option(s) is(are)

Options

A

x1x_1 will deflect towards negatively charged plate

Correct
B

x2x_2 is β−\beta^-

Correct
C

x3x_3 is γ\gamma-ray

D

Z\text{Z} is an isotope of uranium

Correct

Topics & Concepts

Structure of AtomQuantum NumbersElectronic ConfigurationPeriodic Trends

Step-by-Step Solution

We analyze each step of the radioactive decay sequence by applying the conservation of mass number (AA) and atomic number (ZZ).

The given decay series is:
92238U→−x190234Th→−x291234Pa→−x3234Z→−x490230Th{}^{238}_{92}\text{U} \xrightarrow{-x_1} {}^{234}_{90}\text{Th} \xrightarrow{-x_2} {}^{234}_{91}\text{Pa} \xrightarrow{-x_3} {}^{234}\text{Z} \xrightarrow{-x_4} {}^{230}_{90}\text{Th}

Step 1: Identification of x1x_1

92238U→−x190234Th{}^{238}_{92}\text{U} \xrightarrow{-x_1} {}^{234}_{90}\text{Th}

  • Change in mass number: ΔA=238−234=4\Delta A = 238 - 234 = 4
  • Change in atomic number: ΔZ=92−90=2\Delta Z = 92 - 90 = 2

This corresponds to the emission of an α\alpha-particle (24He2+{}^{4}_{2}\text{He}^{2+}):
x1=α-particlex_1 = \alpha\text{-particle}

Since α\alpha-particles carry a net positive charge (+2e+2e), they deflect towards the negatively charged plate in an electric field.
  ⟹  Option (A) is correct.\implies \textbf{Option (A) is correct.}

Step 2: Identification of x2x_2

90234Th→−x291234Pa{}^{234}_{90}\text{Th} \xrightarrow{-x_2} {}^{234}_{91}\text{Pa}

  • Change in mass number: ΔA=234−234=0\Delta A = 234 - 234 = 0
  • Change in atomic number: ΔZ=91−90=+1\Delta Z = 91 - 90 = +1

An increase of 11 in atomic number with no change in mass number signifies β−\beta^--decay (−10e{}^{0}_{-1}\text{e} or β−\beta^-):
x2=β−x_2 = \beta^-
  ⟹  Option (B) is correct.\implies \textbf{Option (B) is correct.}

Step 3: Identification of x3x_3 and Nuclide Z\text{Z}

Let the atomic number of nuclide Z\text{Z} be ZZZ_{\text{Z}}.

In Step 4:
ZZ234Z→−x490230Th{}^{234}_{Z_{\text{Z}}}\text{Z} \xrightarrow{-x_4} {}^{230}_{90}\text{Th}

  • Change in mass number: ΔA=234−230=4\Delta A = 234 - 230 = 4
  • Emission x4x_4 must be an α\alpha-particle (24He{}^{4}_{2}\text{He}), meaning ΔZ=2\Delta Z = 2.
  • Thus: ZZ−2=90  ⟹  ZZ=92Z_{\text{Z}} - 2 = 90 \implies Z_{\text{Z}} = 92.

Since nuclide Z\text{Z} has atomic number Z=92Z = 92, it is an isotope of uranium (92234U{}^{234}_{92}\text{U}):
  ⟹  Option (D) is correct.\implies \textbf{Option (D) is correct.}

Now analyzing Step 3:
91234Pa→−x392234Z{}^{234}_{91}\text{Pa} \xrightarrow{-x_3} {}^{234}_{92}\text{Z}

  • Change in mass number: ΔA=0\Delta A = 0
  • Change in atomic number: ΔZ=92−91=+1\Delta Z = 92 - 91 = +1

This represents a β−\beta^--decay, so x3x_3 is a β−\beta^- particle, not a γ\gamma-ray.
  ⟹  Option (C) is incorrect.\implies \textbf{Option (C) is incorrect.}

Conclusion

The correct options are A, B, and D.

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