The nuclear binding energy for Ar (39.962384 | vedclass

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(A) For $_{18}Ar^{40}$:Number of protons $= 18$Number of neutrons $= 40 - 18 = 22$Mass defect $(\Delta m) = [18 	imes m_{p} + 22 	imes m_{n}] - 	ex

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$343.81$

Vedclass · Answer

(A) For $_{18}Ar^{40}$:Number of protons $= 18$Number of neutrons $= 40 - 18 = 22$Mass defect $(\Delta m) = [18 	imes m_{p} + 22 	imes m_{n}] - 	ext{mass of } Ar$$\Delta m = [18 	imes 1.007825 + 22 	imes 1.008665] - 39.962384$$\Delta m = [18.14085 + 22.19063] - 39.962384$$\Delta m = 40.33148 - 39.962384 = 0.369096 \, amu$Binding energy $= \Delta m 	imes 931.5 \, MeV/amu$Binding energy $= 0.369096 	imes 931.5 \approx 343.81 \, MeV$.

The nuclear binding energy for $Ar$ $(39.962384 \, amu)$ is ......... $MeV.$ (Given: mass of proton $= 1.007825 \, amu$,mass of neutron $= 1.008665 \, amu$)

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