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Chemical Kinetics and Nuclear Chemistry question

2009 · Shift 2 · Q15
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Chemical Kinetics and Nuclear Chemistry question

2009 · Shift 2 · Q15

JEE AdvancedChemistryChemical Kinetics and Nuclear ChemistryNumerical+3 / −1
The total number of α\alphaα and β\betaβ particles emitted in the nuclear reaction 92238U→82214Pb{}_{92}^{238}U \to _{82}^{214}Pb92238​U→82214​Pb is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 8

  1. Let the number of α\alphaα particles emitted be xxx and the number of β−\beta^-β− particles emitted be yyy.

  2. Use conservation of mass number:

    • Each α\alphaα particle reduces mass number by 444.
    • β\betaβ emission does not change mass number.

    So, 238−4x=214238 - 4x = 214238−4x=214 4x=244x = 244x=24 x=6x = 6x=6

  3. Use conservation of atomic number:

    • Each α\alphaα particle reduces atomic number by 222.
    • Each β−\beta^-β− particle increases atomic number by 111.

    Starting from uranium, after 6α6\alpha6α emissions: 92−2(6)=8092 - 2(6) = 8092−2(6)=80

    Final atomic number is 828282, so we need: 80+y=8280 + y = 8280+y=82 y=2y = 2y=2

  4. Therefore,

    • Number of α\alphaα particles =6= 6=6
    • Number of β\betaβ particles =2= 2=2

    Total number emitted: 6+2=86 + 2 = 86+2=8

Therefore, the required total number of α\alphaα and β\betaβ particles emitted is 888.

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