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Grade 12IIT JEE Entrance Exam

question no 25 in jee 2010 paper I some have given the answerr as 4 and some have given as 3 . can you please explain

The number of neutorons emitted when 235/92 U undergoes controlled nuclear fission to 142/54 Xe and 90/38 Sr is

Profile image of veluri hasita
16 Years agoGrade 12
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1 Answer

Profile image of Badiuddin askIITians.ismu Expert
16 Years ago

Dear Veluri

Its a simple and straight forward question

clearly the emmited electron is 3

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Badiuddin

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