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J & K CET Medical J & K - CET Medical Solved Paper-2013

  • question_answer
    The radioactive decay of thorium (A = 232, Z = 90) releases six alpha and four beta particles. The atomic number and mass number of the final product is

    A)  Z = 80, A = 207

    B)  Z = 82, A = 208

    C)  Z = 92, A = 209

    D)  Z=90, A=207

    Correct Answer: B

    Solution :

     Decrease in mass number due to\[6a=6\times 4=24\]. Decrease in change number due to \[6\alpha =6\times 2=12\] Increase in change number due to four\[\beta =4\] \[\therefore \]Net mass number\[=232-24=208\] Net atomic number\[=90-12+4=82\]


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