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U^(235) + n^(1)rarr fission product + ne...

`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

A

`12.75 xx 10^(8) kJ`

B

`18.60 xx 10^(9) kJ`

C

`8.21 xx 10^(7) kJ`

D

`6.55 xx 10^(6) kJ`

Text Solution

Verified by Experts

The correct Answer is:
C
`1g U-235 = (6.023 xx 10^(23))/(235)` atoms
`:.` Energy released `= 3.2 xx 10^(-11) xx (6.023 xx 10^(23))/(235) J`
`= 8.21 xx 10^(1) J = 8.2 xx 10^(7) kJ`
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The ._(92)^(235)U absorbs a slow neutron (thermal neutron) & undergoes a fission represented by (i) The energy release E per fission. (ii) The energy release when 1g of ._(92)^(236)U undergoes complete fission. Given : ._(92)^(235)U = 235.1175 am u (atom), ._(56)^(141)Ba = 140.9577 am u (atom), ._(36)^(92)Kr = 91.9264 am u (atom), ._(0)^(1)n = 1.00898 am u .

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)

Knowledge Check

  • The enegry released per fission of .^(235)u is nearly

    A
    `200 eV`
    B
    `20eV`
    C
    `2000 eV`
    D
    `200MeV`

  • How much energy is released when 2 mole of U^(235) is fission :

    A
    ` 10^(24) Me V`
    B
    ` 24 xx 10^(25) Me V`
    C
    ` 10 ^(24) J`
    D
    ` 10^(24) kWh`

  • Energy released by 1kg . U^(235) when it is fissoned

    A
    `8 xx 10^(10) kWH`
    B
    ` 5 xx 10^(30) eV`
    C
    `10^(10) ` Joule
    D
    `5 xx 10^(26) Me V`

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