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Fill in the blanks: (a) .(92)^(235)U +...

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)`

Text Solution

Verified by Experts

The correct Answer is:
(a) `2_(0)^(1)n` , (b) ` ._(36)^(82) Kr`
`._(92)^(235)U+_(0)^(1)nrarr ._(52)^(137)A+_(40)^(97)B+X ._(0)n^(1)`
(as the reaction is balanced with respect to nuclear charge, the missing particle must be neutral i.e. neutron)
Applying mass nuclear charge balancing
`34=-2+z`
`z=36" "` (This implies element is kr)
Applying mass number balancing
`82=0+A`
`A=82`
Final balanced reaction is
`._(34)^(82)Se rarr 2 ._(-1)^(0)e + _(36)^(82) Kr`
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