2. Solved Papers
  3. RAJASTHAN PMT

RAJASTHAN PMT Rajasthan - PMT Solved Paper-2004

  • question_answer
    A nucleus of mass number 232 and atomic number 90 after many disintegrations of \[\alpha \] and \[\] radiations, decays into other nucleus whose mass number is 220 and atomic number is 86. The numbers of \[\alpha \] and \[\beta \] radiations will be:

    A)  4, 0                       

    B)  3, 6

    C)  3, 2       

    D)         2, 1

    Correct Answer: C

    Solution :

                Nuclear reaction can be shown as \[_{90}{{X}^{232}}\xrightarrow{{{n}_{2}}H{{e}^{4}}}\Upsilon {{\xrightarrow{n1{{\beta }^{0}}}}_{86}}{{M}^{220}}\] Since, atomic mass and atomic number remain conserved in nuclear reaction, so, \[90=2n+(-n)+86\] \[\Rightarrow \]               \[2n=n=4\]                         ?(i) and        \[232=4n+0+220\] \[\Rightarrow \]               \[4n=12\] \[\therefore \]  \[n=3\] Substituting the value of n in equation (i), we get \[2\times 3-n=4\] \[\Rightarrow \]                               \[n=6-4=2\] Hence, 3\[\text{-}\]particles and 2p-particles will be disintegrated.


You need to login to perform this action.
You will be redirected in 3 sec spinner