The calculated mass of {}_{20}Ca^{40} is 40.3 | vedclass

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Question

(A) The binding energy $(BE)$ released is related to the mass defect $(\Delta m)$ by the equation: $BE = \Delta m 	imes 931 \ MeV$. Given $BE = 306.3

Vedclass · Accepted Answer

$39.998$

Vedclass · Answer

(A) The binding energy $(BE)$ released is related to the mass defect $(\Delta m)$ by the equation: $BE = \Delta m 	imes 931 \ MeV$. Given $BE = 306.3 \ MeV$,we calculate the mass defect: $\Delta m = \frac{306.3}{931} = 0.3290 \ u$. The mass defect is defined as: $\Delta m = 	ext{Calculated mass} - 	ext{Isotopic mass}$. Therefore,$	ext{Isotopic mass} = 	ext{Calculated mass} - \Delta m$. $	ext{Isotopic mass} = 40.328 - 0.3290 = 39.998 \ u$.

The calculated mass of ${}_{20}Ca^{40}$ is $40.328 \ u$. It releases $306.3 \ MeV$ energy in a nuclear process. Its isotopic mass is

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