Question

Consider the nuclear reaction $ {{X}^{200}}\to {{A}^{110}}+{{B}^{80}} $ 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. $10\, MeV$
• E. $1480\, MeV$

Answer 1

Answer: (A)

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