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JEE Main 2003
Exam
JEE Main
Year
2003
Subject
Physics
Chapter
Nuclear Physics

JEE Main 2003 · Physics · Nuclear Physics

A nucleus with Z=92 emits the following in a sequence: α, β^ - , β^...

A nucleus with Z=92\mathrm{Z}=92 emits the following in a sequence: α,β−,β−α,α,α,α,α,β−,β−,α,β+,β+,α\alpha, \beta^{-}, \beta^{-} \alpha, \alpha, \alpha, \alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \beta^{+}, \beta^{+}, \alpha. Then Z\mathrm{Z} of the resulting nucleus is

Options

  1. A7676
  2. B7878
  3. C8282
  4. D7474

Correct answer

B. 7878

Step-by-step solution

No. of α\alpha particles emitted =8,=8, \quad No. of β−\beta^{-}particles emitted =4,=4, \quad No.of β+\beta^{+}particles emitted =2=2 z=92−2×8+4−2=78z=92-2 \times 8+4-2=78
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