Find the binding energy (BE) per nucleon of { | vedclass

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(B) The number of protons $Z = 26$ and the number of neutrons $N = 56 - 26 = 30$.The mass defect $\Delta m$ is given by:$\Delta m = [Z m_{p} + N m_{n}

Vedclass · Accepted Answer

$8.52$

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(B) The number of protons $Z = 26$ and the number of neutrons $N = 56 - 26 = 30$.The mass defect $\Delta m$ is given by:$\Delta m = [Z m_{p} + N m_{n} - m({^{56}Fe})]$$\Delta m = [26 	imes 1.00727 + 30 	imes 1.00866 - 55.936] \ u$$\Delta m = [26.18902 + 30.2598 - 55.936] \ u = 0.51282 \ u$.The total binding energy $BE$ is:$BE = \Delta m 	imes 931.5 \ MeV/u$$BE = 0.51282 	imes 931.5 \approx 477.7 \ MeV$.The binding energy per nucleon is:$BE/A = \frac{477.7}{56} \approx 8.53 \ MeV$.Rounding to the nearest provided option,the answer is $8.52 \ MeV$.

Find the binding energy $(BE)$ per nucleon of ${^{56}Fe}$,where $m({^{56}Fe}) = 55.936 \ u$,$m_{n} = 1.00866 \ u$,and $m_{p} = 1.00727 \ u$ (in $MeV$).

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