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NEET PHYSICSMedium

A nucleus with 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. The Z of the resulting nucleus is:

A

74

B

76

C

78

D

82

Step-by-Step Solution

  1. Identify the Decay Rules:
  • Alpha (α\alpha) decay: The nucleus emits a helium nucleus (24He{}_{2}^{4}\mathrm{He}). The atomic number (ZZ) decreases by 2 .
  • Beta-minus (β−\beta^-) decay: The nucleus emits an electron (−10e{}_{-1}^{0}e). The atomic number (ZZ) increases by 1 .
  • Beta-plus (β+\beta^+) decay: The nucleus emits a positron (+10e{}_{+1}^{0}e). The atomic number (ZZ) decreases by 1 .
  1. Count the Particles:
  • Number of α\alpha particles (nαn_\alpha) = 8.
  • Number of β−\beta^- particles (nβ−n_{\beta^-}) = 4.
  • Number of β+\beta^+ particles (nβ+n_{\beta^+}) = 2.
  1. Calculate Final Atomic Number:
  • Zfinal=Zinitial+nα(−2)+nβ−(+1)+nβ+(−1)Z_{\text{final}} = Z_{\text{initial}} + n_\alpha(-2) + n_{\beta^-}(+1) + n_{\beta^+}(-1)
  • Zfinal=92+8(−2)+4(1)+2(−1)Z_{\text{final}} = 92 + 8(-2) + 4(1) + 2(-1)
  • Zfinal=92−16+4−2Z_{\text{final}} = 92 - 16 + 4 - 2
  • Zfinal=78Z_{\text{final}} = 78.
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