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PhysicsNEET 2024Radioactivity and decay law

Nuclei

82290X→ α Y→ e+ Z→ e− P→ e− Q^{290}_{82}X \xrightarrow{\ \alpha\ } Y \xrightarrow{\ e^{+}\ } Z \xrightarrow{\ e^{-}\ } P \xrightarrow{\ e^{-}\ } Q

In the nuclear emission stated above, the mass number and atomic number of the product QQ, respectively, are:

  1. A.286, 80
  2. B.288, 82
  3. C.286, 81✓
  4. D.280, 81

Solution

An α\alpha-emission lowers AA by 4 and ZZ by 2: (286,80)(286, 80). Each β\beta-emission leaves AA unchanged; a positron lowers ZZ by 1 and an electron raises it by 1. Net over e+,e−,e−e^+, e^-, e^-: Z=80−1+1+1=81Z = 80 - 1 + 1 + 1 = 81. So QQ is (286,81)(286, 81).

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