Calculate the binding energy per nucleon of ${}_{17}^{35}{Cl}$ nucleus. Given that mass of ${}_{17}^{35}{Cl}$ nucleus = 34.98000 u, mass of proton = 1.007825 u, mass of neutron = 1.008665 u and 1 u is equivalent to 931 Mev.

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6.2 Mev

7.2 Mev

4.2 Mev

8.2 Mev

answer is D.

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Detailed Solution

$M_{cl} = 34.98000 amu ∆ m = [{ZM}_{p} + (A - Z) m_{n}] - M_{cl}$

= [17(1.007825)+18(1.008665)]–34.98

= [17.133025 + 18.15597]–34.98

$∆ m = 0.308995 amu BE = ∆ m x 931 Mev B . E = 0.308995 x 931 = 287.6743 MeV$

$\frac{B . E}{A} = \frac{287.6743}{35} = 8.2 MeV$

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