Question

For the given series reaction in `n^(th)` step, find out the number of protons & energy.
`_(92).^(238)U rarr Ba+Kr+3_(0)n^(1)+"Energy" (E)`

Answer 1

Correct Answer - `3^(n), 3^(n-1) E`
Nuclear fission is a series reaction `3, 9, 27`……. Neutron and `E, 3E, 9E…….` energy are emitted. Hence answer is `3^(n), 3^(n-1) E`.