Question

(b) In the nuclear reaction
n +92U235
> 54Xea + 6Sr94 + 2n
determine the values of a and b.​

Answer 1

Answer:

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The neutron will be written as

0

n

1

so conserving the atomic number in both sides of the equation we get 0+92=54+b+2×0 so b=48

and on conserving the atomic mass in both sides we get 1+235=a+94+2×1 or a=140

Finally we get a=140 and b=48