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Nucleus $\mathrm{A}$ is having mass number 220 and its binding energy per nucleon is $\mathrm{5.6\, \mathrm{MeV}}$ . It splits in two fragments $\mathrm{' B ' \, and \, ' C'}$ of mass numbers 105 and 115. The binding energy of nucleons in $\mathrm{'B' and \, 'C'\,\, is\, \, 6.4\, \mathrm{MeV}}$ per nucleon. The energy Q released per fission will be :
Option: 1
$\mathrm{0.8\, \mathrm{MeV}}$

Option: 2
$\mathrm{275\, \mathrm{MeV}}$

Option: 3
$\mathrm{220\, \mathrm{MeV}}$

Option: 4
$\mathrm{176\, \mathrm{MeV}}$

Answers (1)

best_answer

$\mathrm{A_{A}}= 220$
$\mathrm{\left ( BE \right )_{A}= BE}\; \text{per nucleons}\mathrm{=5.6\,MeV }$
For atom A

$\mathrm{\text{Total energy of A's nucleus}= -220\left ( BE \right ) _{A} }$
                                           $\mathrm{\left ( TE \right )_{A} = -220\times 5.6\, Me V}$

$\mathrm{Q-value= \left [ \left ( TE_{A} \right )-\left ( TE_{B} +TE_{C}\right ) \right ] }$
                       $\mathrm{= \left [ \left ( -220\times 5.6 \right )-\left ( -220\times 6.4 \right )\right ] }$
                       $\mathrm{= 176\, Me V}$

The correct option is (4)

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