Binding energy of a certain nucleus is 18 tim | vedclass

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(D) The mass defect $\Delta m$ is related to the binding energy $BE$ by the Einstein mass-energy equivalence relation: $BE = \Delta m c^2$.Given $BE =

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$20$

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(D) The mass defect $\Delta m$ is related to the binding energy $BE$ by the Einstein mass-energy equivalence relation: $BE = \Delta m c^2$.Given $BE = 18 	imes 10^8 \ J$ and the speed of light $c = 3 	imes 10^8 \ m/s$.Substituting the values: $18 	imes 10^8 = \Delta m 	imes (3 	imes 10^8)^2$.$18 	imes 10^8 = \Delta m 	imes 9 	imes 10^{16}$.$\Delta m = \frac{18 	imes 10^8}{9 	imes 10^{16}} = 2 	imes 10^{-8} \ kg$.Converting to micrograms: $2 	imes 10^{-8} \ kg = 2 	imes 10^{-8} 	imes 10^9 \ \mu g = 20 \ \mu g$.

Binding energy of a certain nucleus is $18 \times 10^8 \ J$. How much is the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus (in $\mu g$)?

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