Find the binding energy per nucleon for ^{120 | vedclass

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(A) The number of protons $Z = 50$ and the number of neutrons $N = A - Z = 120 - 50 = 70$.The expected mass of the nucleus is $M_{expected} = Z m_{p}

Vedclass · Accepted Answer

$8.5$

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(A) The number of protons $Z = 50$ and the number of neutrons $N = A - Z = 120 - 50 = 70$.The expected mass of the nucleus is $M_{expected} = Z m_{p} + N m_{n}$.$M_{expected} = 50(1.00783) + 70(1.00867) = 50.3915 + 70.6069 = 120.9984 \, U$.The mass defect $\Delta m = M_{expected} - m_{Sn} = 120.9984 - 119.902199 = 1.096201 \, U$.The binding energy $B.E. = \Delta m 	imes 931 \, MeV/U = 1.096201 	imes 931 \approx 1020.56 \, MeV$.The binding energy per nucleon is $\frac{B.E.}{A} = \frac{1020.56}{120} \approx 8.5 \, MeV$.

Find the binding energy per nucleon for $^{120}_{50}Sn$. Given: mass of proton $m_{p} = 1.00783 \, U$,mass of neutron $m_{n} = 1.00867 \, U$,and mass of tin nucleus $m_{Sn} = 119.902199 \, U$. (Take $1 \, U = 931 \, MeV$) (in $, MeV$)

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