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.(92)^(235)U+(0)^(1)n rarr (56)^(139)Ba ...

`._(92)^(235)U+_(0)^(1)n rarr _(56)^(139)Ba +_(36)^(94)Kr +3_(0)^(1)n+200MeV`
Total energy released (in MeV) after `5^(th)` stage of fission is:

A

48600

B

16200

C

24200

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
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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)

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

U^(235) + n^(1)rarr fission product + neutron + 3.2 xx 10^(-11)J . The energy released , when 1g of U^(235) finally undergoes fission , is

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 .

When slow neutrons are incident on a target containing underset(92)overset(235).U , a possible fission reaction is underset(92)overset(235).U+underset(0)overset(1).nrarr underset(56)overset(141).Ba+underset(36)overset(92).Kr+3_(0)^(1)n Estimate the amount of energy released using the following data Given, mass of underset(92)overset(235).U=235.04 amu , mass of underset(0)overset(1).n=1.0087 amu , mass of underset(56)overset(141).Ba=140.91 amu , mass of underset(36)overset(92).Kr=91.926 amu , and energy equivalent to 1 amu=931 MeV.

Complete the equations for the following nuclear processes: (a) ._(17)^(35) Cl + ._(0)^(1)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^(-))...

Consider one of fission reactions of ^(235)U by thermal neutrons ._(92)^(235)U +n rarr ._(38)^(94)Sr +._(54)^(140)Xe+2n . The fission fragments are however unstable and they undergo successive beta -decay until ._(38)^(94)Sr becomes ._(40)^(94)Zr and ._(54)^(140)Xe becomes ._(58)^(140)Ce . The energy released in this process is Given: m(.^(235)U) =235.439u,m(n)=1.00866 u, m(.^(94)Zr)=93.9064 u, m(.^(140)Ce) =139.9055 u,1u=931 MeV] .

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In each fission of U^235 , 200 MeV of energy is released. If a reactor produces 100MW power the rate of fission in it will be