Question

When a uranium isotope 92
235 w is bombarded with
a neutron, it generates 89 Kr. three neutrons
36
and :
144
56 Ba.
when a uranium
Isotope 2354 is bombarro,
(2)
91 Zr
40
101 kr
(3)
with. aneutrron
it
36 kr
103
4)
36 kr​

Answer 1

Correct Question:

When a uranium isotope $\rm {^{235}_{92}U}$ is bombarded with a neutron, it generates $\rm {^{89}_{36}Kr}$ , three neutrons and:

(1) $\rm {^{91}_{40}Zr}$

(2) $\rm {^{101}_{36}Kr}$

(3) $\rm {^{103}_{36}Kr}$

(4) $\rm {^{144}_{56}Ba}$

Answer:

$\boxed{\mathfrak{(4) {^{144}_{56}Ba} }}$

Explanation:

According to question:

$\rm \implies {U^{235}_{92}} +{n^{1}_{0}} \longrightarrow {Kr^{89}_{36}} + 3{n^{1}_{0}} + {X^{A}_{Z}}$

Total atomic mass & atomic number before and after bombardment should be equal. so,

$\rm \implies 92 + 0 = 3 6 +3 \times 0 + Z \\ \\ \rm \implies Z = 92 - 36 \\ \\ \rm \implies Z = 56$

$\rm \implies 235 + 1 = 89 + 3 \times 1 + A \\ \\ \rm \implies 236 = 92 + A \\ \\ \rm \implies A = 236 - 92 \\ \\ \rm \implies A = 144$

So, $\rm {^{144}_{56}Ba}$ is generated.

Answer 2

Explanation:

$\sf{U_{235}^{92} + n_{0}^{1} → Kr_{36}^{89} + 3n_{0}^{1} + X_{Z}^{A}}$

Atomic number is the number of protons that a chemical element has in its centre.

Atomic mass is the sum of protons and neutrons in a single atom.

92 + 0 = 36 +3(0) + Z

92 = 36 + 0 + Z

92 = 36 + Z

Z = 92 - 36

Z = 56

235 + 1 = 89 + 3(1) + A

236 = 89 + 3 + A

236 = 92 + A

A = 236 - 92

A = 144

Hence, $\sf{Ba_{56}^{144}}$ is produced.