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Complete the equation for the following ...

Complete the equation for the following fission process `._92 U^235 ._0 n^1 rarr ._38 Sr^90 +....`.

A

`""_(57)X^(142)+3_0n^1`

B

`""_(54)X^(145)+3_0n^1`

C

`""_(54)X^(143)+3_0n^1`

D

`""_(54)X^(142)+""_0n^1`

Text Solution

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The correct Answer is:
C
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Complete the equations for the following nuclear processes: (a) ._(17)^(35) Cl + ._(1)^(0)n rarr... + ._(2)^(4)He (b) ._(92)^(235)U + ._(0)^(1) n rarr ...+ ._(54)^(137)Xe + 2 _(0)^(1)n (c) ._(13)^(27) Al + ._(2)^(4) He rarr ... + ._(0)^(1) n (d) ...(n,p) ._(16)^(35) S (e) ._(94)^(239) Pu (alpha, beta^(-))...

Complete the following nuclear equation by suppling the symbol for the other product of the fission : ._(92)^(235)U+_(0)^(1)n rarr _(38)^(94)Sr +………+2_(0)^(1)n

The ._(92)U^(235) absorbs a slow neturon (thermal neutron) & undergoes a fission represented by ._(92)U^(235)+._(0)n^(1)rarr._(92)U^(236)rarr._(56)Ba^(141)+_(36)Kr^(92)+3_(0)n^(1)+E . Calculate: The energy released when 1 g of ._(92)U^(235) undergoes complete fission in N if m=[N] then find (m-2)/(5) . [N] greatest integer Given ._(92)U^(235)=235.1175"amu (atom)" , ._(56)Ba^(141)=140.9577 "amu (atom)" , ._(36)r^(92)=91.9263 "amu(atom)" , ._(0)n^(1)=1.00898 "amu", 1 "amu"=931 MeV//C^(2)

Which of the following are fission reactions? _0^1 n+ _92^235 U rarr _92^235 U rarr _56^144 Ba + _36^89 Kr + 3_0^1n _0^1 n+ _92^235 U rarr _92^235 U rarr _51^133 Sb + _41^99 Nb + 4_0^1n _0^1n + _92^235 U rarr _54^140 Xe + _ 38^94 Sr + 2_0^1 n _1^2 H+ _1^2 H rarr _2^3 He + _0^1n

Which of the following elements is an isodiaphere of ._(92)^(235)U ?

U-235 is decayed by bombardment by neutron as according to the equation: ._(92)U^(235) + ._(0)n^(1) rarr ._(42)Mo^(98) + ._(54)Xe^(136) + x ._(-1)e^(0) + y ._(0)n^(1) Calculate the value of x and y and the energy released per uranium atom fragmented (neglect the mass of electron). Given masses (amu) U-235 = 235.044 , Xe = 135.907, Mo = 97.90, e = 5.5 xx 10^(-4), n = 1.0086 .

Percentage of mass lost during the fission of ._(92)U^(235) approximately is

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