Question

question 12
which is the balanced equation for the fission of u - 235?
a. $_{92}^{235}u+_{0}^{1}n
ightarrow_{35}^{87}br+_{57}^{146}la+3_{0}^{1}n$
b. $_{92}^{235}u
ightarrow_{56}^{138}ba+_{36}^{95}kr+_{0}^{1}n$
c. $_{92}^{235}u+_{0}^{1}n
ightarrow_{52}^{137}te+_{40}^{97}zr$
d. $_{92}^{235}u
ightarrow_{55}^{144}cs+_{37}^{90}rb+_{0}^{1}n$

Explanation:

Step1: Check mass number balance

For nuclear fission reactions, the sum of mass numbers on the left - hand side (LHS) must equal the sum of mass numbers on the right - hand side (RHS).
For option A:
LHS: \(235 + 1=236\)
RHS: \(87+146 + 3\times1=236\)
For option B:
LHS: \(235\)
RHS: \(138 + 95+1=234
eq235\)
For option C:
LHS: \(235 + 1=236\)
RHS: \(137+97=234
eq236\)
For option D:
LHS: \(235\)
RHS: \(144+90 + 1=235\)

Step2: Check atomic number balance

The sum of atomic numbers on the LHS must equal the sum of atomic numbers on the RHS.
For option A:
LHS: \(92+0 = 92\)
RHS: \(35+57+3\times0=92\)
For option D:
LHS: \(92\)
RHS: \(55+37+0=92\)

Step3: Consider the nature of fission

Nuclear fission of \(^{235}_{92}U\) is a neutron - induced process. So, there must be a neutron (\(^{1}_{0}n\)) on the LHS. Option D does not have a neutron on the LHS, while option A has a neutron on the LHS.

Answer:

A. \(^{235}_{92}U+^{1}_{0}n
ightarrow^{87}_{35}Br + ^{146}_{57}La+3^{1}_{0}n\)