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The product p of the nuclear reaction ...

The product p of the nuclear reaction
`._(92)^(235)U + ._(0)^(1) n rarr p + ._(36)^(92)Kr + 3 ._(0)^(1)n` is

A

`""_(56)^(141)Sr`

B

`""_(56)^(141)La`

C

`""_(56)^(141)Ba`

D

`""_(56)^(141)Cs`

Text Solution

Verified by Experts

The correct Answer is:
C
U – 235 nucleus when hit by a neutron undergoes the reaction.
`""_(92)U^(235) + ""_(0)n^1 to ""_(56)Ba^(141) + ""_(36)Kr^(92) + 3 ""_(0)^(1)n`
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