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NEET PHYSICSNUCLEIMedium

Question

In the nuclear emission stated above: 82290XαYe+ZβPeQ{}_{82}^{290}X \xrightarrow{\alpha} Y \xrightarrow{e^+} Z \xrightarrow{\beta^-} P \xrightarrow{e^-} Q, the mass number and atomic number of the product Q respectively, are:

A

286, 80

B

288, 82

C

286, 81

D

280, 81

Step-by-Step Solution

  1. Initial Nucleus (XX): Mass Number (AA) = 290, Atomic Number (ZZ) = 82.
  2. Step 1: Alpha Decay (α\alpha): Emission of a Helium nucleus (24He{}_{2}^{4}He).
  • A=2904=286A' = 290 - 4 = 286
  • Z=822=80Z' = 82 - 2 = 80
  • Nucleus YY is 80286Y{}_{80}^{286}Y.
  1. Step 2: Positron Emission (e+e^+): Emission of a positron (+10e{}_{+1}^{0}e). ZZ decreases by 1, AA remains unchanged.
  • A=286A'' = 286
  • Z=801=79Z'' = 80 - 1 = 79
  • Nucleus ZZ is 79286Z{}_{79}^{286}Z.
  1. Step 3: Beta Minus Decay (β\beta^-): Emission of an electron (10e{}_{-1}^{0}e). ZZ increases by 1, AA remains unchanged.
  • A=286A''' = 286
  • Z=79+1=80Z''' = 79 + 1 = 80
  • Nucleus PP is 80286P{}_{80}^{286}P.
  1. Step 4: Electron Emission (ee^-): Another β\beta^- decay. ZZ increases by 1.
  • Afinal=286A_{final} = 286
  • Zfinal=80+1=81Z_{final} = 80 + 1 = 81
  • Nucleus QQ is 81286Q{}_{81}^{286}Q .

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from NUCLEI. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSNUCLEInuclearemissionstatedxrightarrowalphaxrightarrowe

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