JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2006

  • question_answer
        A nucleus decays by\[{{\beta }^{+}}-\] emission followed by a\[\gamma -\]emission. If the atomic and mass numbers of the parent nucleus are Z and A respectively, the corresponding numbers for the daughter nucleus are respectively

    A)  \[Z-1\]and\[A-1\]          

    B)  \[Z+1\] and A

    C)  \[Z-1\]and A     

    D)  \[Z+1\]and\[A-1\]

    Correct Answer: C

    Solution :

                    Key Idea: \[\beta -\]rays are fast moving electrons. When there is an excess of protons in the nucleus and it is not energetically possible to emit an\[\alpha -\]particle,\[{{\beta }^{+}}\]decay occurs. Resulting in reducing atomic numbers by 1. New atomic number\[=Z-1,\]mass number\[=A\]Gamma ray emission occurs with p4 emission. Since, gamma rays have no charge or mass their emission does not change the chemical composition of the atom.  Hence, atomic number\[=Z-1,\] mass number         = A Note: During\[{{\beta }^{+}}\]decay, a positron (a particle with the same mass as an electron but. With positive charge) and a neutrino are released.


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