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Fill in the blank ""(92)U^(235) +""(0)...

Fill in the blank
`""_(92)U^(235) +""_(0)n^(1) to ? +""_(36)^(92)Kr + 3""_(0)^(1)n`

A

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

B

`""_(56)^(139)Ba`

C

`""_(56)^(139)Ba`

D

`""_(54)^(141)Ba`

Text Solution

Verified by Experts

The correct Answer is:
A
`""_(92)U^(235)+""_(0)n^(1) to ""_(Z)Ba^(A)+""_(36)Kr^(92)+3""_(0)n^(1)`
`92+0=Z + 36 + 0 therefore Z=56`
`235+1=A+92 + 3 therefore A=141`
Hence the missing nuclide is `""_(56)Ba^(141)`
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Identify the missingf product in the given reaction ""_(92)""^(235)Uto""_(0)""^(1)nto?+""_(36)""^(92)Kr+3""_(0)""^(1)n

Fill in the blanks: (a) ._(92)^(235)U + ._(0)^(1)n rarr ._(55)^(142)A + ._(37)^(92) B+.... (b) ._(34)^(82)Se rarr ... 2 ._(-1)e^(0)

When Uranium is bombarded with neutrons , it undergoes fission . The fission reaction can be written as : ""_(92) U^(235) + ""_(0) n^(1) to ""_(56) Ba^(141) + ""_(36) Kr^(92) + 3x + Q (energy) where three particles named x are produced and energy Q is released . What is the name of the particle x ?

""_(0)n^(1) + ""_(92)U^(235) to ""_(92)U^(236) to .............. + ""_(36)Kr^(89) + 3""_(0)n^(1)

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