Atoms & Nuclei

JEE Main level, single correct, medium

207 of 239

Question 207 of 239

Find the Binding energy per nucleon for Sn50120Sn50120. Mass of proton mp=1.00783U{m}_{p}=1.00783U, mass of neutron mn=1.00867U{m}_{n}=1.00867U and mass of tin nucleus msn=119.902199U{m}_{sn}=119.902199U. (take 1U=931MeV1U=931MeV)

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Answer: (D) 8.5MeV8.5MeV

Binding energy =(ΔM)C2=(AM)931=(\Delta M){C}^{2}=(AM)931
put the value of
∆M∆M
BE=8.5MeVBE=8.5MeV

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