devarshi-dt-logo

Question:

The number of neutrons emitted when 23592U undergoes controlled nuclear fission to 14254Xe and 9038Sr is:

0

1

3

2

Solution:

92U235→54Xe142+38Sr90+xn

Now, by balancing weight on both sides of the reaction, we get:
235=90+142+x
x=3

Hence, option D is correct.