Fill in the blanks (a) `._(92)^(235)U+._(0)^(1)n rarr ._(52)^(137)A+ ._(40)^(97) B + __________. (b) `._(34)^(82) Se rarr 2 _(-1)^(0)e+__________`.
Fill in the blanks
(a) `._(92)^(235)U+._(0)^(1)n rarr ._(52)^(137)A+ ._(40)^(97) B + __________.
(b) `._(34)^(82) Se rarr 2 _(-1)^(0)e+__________`.
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Correct Answer - (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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