2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Nuclear Physics And Radioactivity

JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Radioactivity

  • question_answer
    A radioactive element \[_{90}{{X}^{238}}\]decay into\[_{83}{{Y}^{222}}\]. The number of \[\beta -\]particles emitted are  [BHU 1997; JIPMER 2001, 02]

    A)            4    

    B)            6

    C)            2    

    D)            1

    Correct Answer: D

    Solution :

                       Number of \[\alpha -\]particles emitted \[=\frac{238-222}{4}=4\] This decreases atomic number to \[90-4\times 2=82\] Since atomic number of \[_{83}{{Y}^{222}}\]is 83, this is possible if one \[\beta -\]particle is emitted.


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