A nucleus with mass number 242 and binding en | vedclass

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Question

(B) The initial binding energy of the nucleus is calculated as: $E_{i} = 242 	imes 7.6\,MeV = 1839.2\,MeV$.The final state consists of two fragments,

Vedclass · Accepted Answer

$121$

Vedclass · Answer

(B) The initial binding energy of the nucleus is calculated as: $E_{i} = 242 	imes 7.6\,MeV = 1839.2\,MeV$.The final state consists of two fragments,each with a mass number of $121$. The binding energy per nucleon for each fragment is $8.1\,MeV$.The total final binding energy is: $E_{f} = (121 	imes 8.1\,MeV) + (121 	imes 8.1\,MeV) = 242 	imes 8.1\,MeV = 1960.2\,MeV$.The total gain in binding energy is the difference between the final and initial binding energies:$\Delta E = E_{f} - E_{i} = 242 	imes (8.1 - 7.6)\,MeV$.$\Delta E = 242 	imes 0.5\,MeV = 121\,MeV$.

$A$ nucleus with mass number $242$ and binding energy per nucleon as $7.6\,MeV$ breaks into two fragments,each with mass number $121$. If each fragment nucleus has a binding energy per nucleon of $8.1\,MeV$,the total gain in binding energy is $........MeV$.

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