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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

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 

Grade:12

1 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 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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