Question

A hypothetical nuclear fission takes place as following:
${}_{92}^{236}X{\to }_{56}^{140}Y{+}_{36}^{94}Y+neutron\left(s\right)$
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

${}_{92}^{236}X{\to }_{56}^{140}Y{+}_{36}^{94}Y+neutron/s$
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 ${}_0^{1}n$.
So, two neutrons will solve the purpose.