Question

Nuclearissionoccursthroughmanydifferent pathways. For the ission of U-235 induced by a neutron, write a nuclear equation to form: a. bromine-87, lanthanum-146, and more neutrons. b. a nucleus with 56 protons, a second with a total of 94 neutrons and protons, and 2 additional neutrons.

Answer 1

Answer: a) $^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n$

b) $^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}$

Explanation:

A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.

a) The given reaction is $^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + x^{1}_{0}n$

Now,  as the mass on both reactant and product side must be equal:

$235+1=87+146+x$

$x=3$

Thus three neutrons are produced and nuclear equation will be: $^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n$

b) For the another fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{A}_{56}\textrm{Ba}+^{94}_{Z}\textrm{X}+2^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 2

A = 140

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = 56 + Z + 0

Z = 36

As Krypton has atomic number of 36,Thus the nuclear equation will be :

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}$