1-00 kg of -235-U undergoes fission process- If energy released per event is 200 MeV- then the total energy released is5-12times 10-24- MeV6-02times 10-23- MeV5-12times 10-26- MeV6-02times 10-26- MeV

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Toppr Admin · Accepted Answer

M-1-xA0-kgNo- of moles -n-1000235-4-2553No- of atoms of U-nNA-2-562-xD7-1024 atomsEnergy released per atom -200-xA0-MeVTotal energy released -200-xD7-2-562-xD7-1024-5-12-xD7-1026MeV-xA0-

$1.00 k g$ of $^{235} U$ undergoes fission process. If energy released per event is $200 M e V$ , then the total energy released is
$5.12 \times 10^{24} M e V$
$6.02 \times 10^{23} M e V$
$5.12 \times 10^{26} M e V$
$6.02 \times 10^{26} M e V$

$M = 1 k g$
No. of moles $= n =$ $\frac{1000}{235} = 4.2553$
No. of atoms of $U =$ $n N_{A} = 2.562 \times 10^{24}$ atoms
Energy released per atom $= 200 M e V$
Total energy released $= 200 \times 2.562 \times 10^{24} = 5.12 \times 10^{26} M e V$

1.00kg of 235U undergoes fission process. If energy released per event is 200MeV, then the total energy released is5.12×1024MeV6.02×1023MeV5.12×1026MeV6.02×1026MeV

M=1 kgNo. of moles =n=1000235=4.2553No. of atoms of U=nNA=2.562×1024 atomsEnergy released per atom =200 MeVTotal energy released =200×2.562×1024=5.12×1026MeV

$1.00 k g$ of $^{235} U$ undergoes fission process. If energy released per event is $200 M e V$ , then the total energy released is
$5.12 \times 10^{24} M e V$
$6.02 \times 10^{23} M e V$
$5.12 \times 10^{26} M e V$
$6.02 \times 10^{26} M e V$

$M = 1 k g$
No. of moles $= n =$ $\frac{1000}{235} = 4.2553$
No. of atoms of $U =$ $n N_{A} = 2.562 \times 10^{24}$ atoms
Energy released per atom $= 200 M e V$
Total energy released $= 200 \times 2.562 \times 10^{24} = 5.12 \times 10^{26} M e V$