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

Option: 1

{^{144}_{56}}Ba


Option: 2

{^{91}_{40}}Zr


Option: 3

{^{101}_{36}}Kr


Option: 4

{^{103}_{36}}Kr


Answers (1)

best_answer

Let the unknown element is { }_{A} \mathrm{X}^{Z}

{ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow {}_{36} \mathrm{Ba}^{89}+{ }_{A} \mathrm{X}^{Z}+3\ { }_{0} \mathrm{n}^{1}

The sum of the atomic number on L.H.S=92

The sum of the atomic number on R.H.S=92=36+A

On comparing 

A=56

SImilarly

The sum of the mass number on L.H.S=235+1

The sum of the atomic number on R.H.S= 89+3+Z

On comparing 

235+1=89+3+Z

So Z=144

So the known element is {^{144}_{56}}Ba

Posted by

Ritika Jonwal

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