Question

complete each nuclear fission reaction.

$_{92}^{235}u+_{0}^{1}n\to_{36}^{90}kr+_{56}^{a}ba+3_{0}^{1}n$

a:

$_{94}^{239}pu+_{0}^{1}n\to_{c}^{b}ba+_{38}^{91}sr+3_{0}^{1}n$

b:

c:

done

Explanation:

Step1: For the first reaction (finding A)

In nuclear reactions, mass - number is conserved.
The mass - number of the left - hand side: \(235 + 1=236\)
The mass - number of the right - hand side: \(90+A + 3\times1\)
Set up the equation \(236=90 + A+3\)
\(A=236-(90 + 3)=143\)

Step2: For the second reaction (finding B and C)

• Mass - number conservation (finding B):

The mass - number of the left - hand side: \(239+1 = 240\)
The mass - number of the right - hand side: \(B + 91+3\times1\)
Set up the equation \(240=B + 91+3\)
\(B=240-(91 + 3)=146\)

• Atomic - number conservation (finding C):

The atomic - number of the left - hand side: \(94\)
The atomic - number of the right - hand side: \(C + 38+0\) (since the neutron has atomic - number \(0\))
Set up the equation \(94=C + 38\)
\(C=94 - 38=56\)

Answer:

A: \(143\)
B: \(146\)
C: \(56\)