Question

complete the following nuclear fission reaction

$_{92}^{235}u+_{0}^{1}n\to+_{56}^{142}ba+_{36}^{91}kr+3\underline{\quad\quad}$

a $_{1}^{1}h$

b $_{0}^{1}n$

c $_{-1}^{0}e$

d $_{+1}^{0}e$

Explanation:

Step1: Calculate mass number

Mass number of left - hand side: \(235 + 1=236\)
Mass number of right - hand side: \(142+91 + 3x\)
Set up equation: \(236=142 + 91+3x\), \(3x=236-(142 + 91)=3\), \(x = 1\)

Step2: Calculate atomic number

Atomic number of left - hand side: \(92+0 = 92\)
Atomic number of right - hand side: \(56+36+3y\)
Set up equation: \(92=56 + 36+3y\), \(3y=92-(56 + 36)=0\), \(y = 0\)

Answer:

B. \(_{0}^{1}n\)