Consider the nuclear reaction X200→ A110+B80 binding energy per nucleon for X, text A and B are 7.4 MeV, 8.2 MeV and 8.1 MeV respectively, then the energy released in the reaction is

Question

Consider the nuclear reaction X200→ A110+B80 binding energy per nucleon for X, text A and B are 7.4 MeV, 8.2 MeV and 8.1 MeV respectively, then the energy released in the reaction is (A) 70 MeV (B) 200 MeV (C) 190 MeV (D) 10 MeV

Tardigrade Admin · Accepted Answer

For X, energy =200× 7.4=1480 text MeV For A, energy =110× 8.2=902 text MeV For B, energy =80× 8.1=648 text MeV Therefore, energy released =(902+648)-1480 =1550-1480=70 text MeV

Q. Consider the nuclear reaction $X^{200} \to A^{110} + B^{80}$ binding energy per nucleon for $X, A$ and $B$ are $7.4 M e V, 8.2 M e V$ and $8.1 M e V$ respectively, then the energy released in the reaction is

$70 M e V$

$200 M e V$

$190 M e V$

$10 M e V$

$1480 M e V$

For $X,$ energy $= 200 \times 7.4 = 1480 M e V$
For A, energy $= 110 \times 8.2 = 902 M e V$
For B, energy $= 80 \times 8.1 = 648 M e V$
Therefore, energy released
$= (902 + 648) - 1480$
$= 1550 - 1480 = 70 M e V$

Q. Consider the nuclear reaction X200→A110+B80 binding energy per nucleon for X,A and B are 7.4MeV,8.2MeV and 8.1MeV respectively, then the energy released in the reaction is

70MeV

200MeV

190MeV

10MeV

1480MeV

For X, energy =200×7.4=1480MeV For A, energy =110×8.2=902MeV For B, energy =80×8.1=648MeV Therefore, energy released =(902+648)−1480 =1550−1480=70MeV