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The mass of a $^7_3Li$ nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of $^7_3Li$ nucleus is nearly
A
23 MeV
B
46 MeV
C
5.6 MeV
D
3.9 MeV
Detailed Solution
Mass defect: $\Delta m = 0.042$ u
Binding energy: $BE = \Delta m\times931$ MeV = $0.042\times931 = 39.1$ MeV
Number of nucleons in $^7_3Li$: A = 7
Binding energy per nucleon = $\frac{39.1}{7}$
$\approx 5.6$ MeV
Binding energy: $BE = \Delta m\times931$ MeV = $0.042\times931 = 39.1$ MeV
Number of nucleons in $^7_3Li$: A = 7
Binding energy per nucleon = $\frac{39.1}{7}$
$\approx 5.6$ MeV