If the mass defect of a nuclide is 3.32 times | vedclass

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Question

(B) The mass defect is given as $3.32 	imes 10^{-26} \ g$. First,convert the mass defect into atomic mass units $(amu)$ by dividing by the mass of on

Vedclass · Accepted Answer

$18.62$

Vedclass · Answer

(B) The mass defect is given as $3.32 	imes 10^{-26} \ g$. First,convert the mass defect into atomic mass units $(amu)$ by dividing by the mass of one $amu$ $(1.66 	imes 10^{-24} \ g)$: $	ext{Mass defect in } amu = \frac{3.32 	imes 10^{-26} \ g}{1.66 	imes 10^{-24} \ g/amu} = 0.02 \ amu$. The binding energy is calculated using the relation: $	ext{Binding energy} = 	ext{Mass defect in } amu 	imes 931 \ MeV/amu$. $	ext{Binding energy} = 0.02 	imes 931 \ MeV = 18.62 \ MeV$.

If the mass defect of a nuclide is $3.32 \times 10^{-26} \ g$,its binding energy in $MeV$ is:

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