Atoms & Nuclei

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Question 238 of 239

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

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Answer: (B) 7878

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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