Question

A hypothetical nuclear fission takes place as following:

${}_{92}^{236}X\to {}_{56}^{140}Y+{}_{36}^{94}Y+\text{neutrons}$

How many neutrons will get emitted in the process?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

${}_{92}^{236}X\to {}_{56}^{140}Y+{}_{36}^{94}Y+\text{neutrons}$

Z in LHS = 92
Z in RHS = 56 + 36 = 92

A in LHS = 236
A in RHS = 234
A required in RHS = 236 - 234 = 2

We know a neutron is ${}_0^1\text{n}$ which has a Z of 1. So, two neutrons are emitted in the process.