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When a uranium isotope ""(92)^(235)U is ...

When a uranium isotope `""_(92)^(235)U` is bombared with neutron, it generates `""_(36)^(89)Kr` three neutrons and

A

`""_(56)^(144)Ba`

B

`""_(40)^(91)Zr`

C

`""_(36)^(101)Kr`

D

`""_(36)^(103)Kr`

Text Solution

Verified by Experts

The correct Answer is:
A
`U_(92)^(235) + ""_(0)^(1)n rarr Kr_(36)^(89) + 3n_(0)^(1) + X_(Z)^(A)`
`92+0= 36+Z`
`rArr Z= 56`
`235 +1= 89 +3+A`
So, `""_(56)^(144)Ba` is generated
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