The binding energy per nucleon of {O^{16}} is | vedclass

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(C) The nuclear reaction for removing a neutron from ${O^{17}}$ is: ${O^{17}} 	o {O^{16}} + n$.The total binding energy $(B.E.)$ of a nucleus is give

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$4.23$

Vedclass · Answer

(C) The nuclear reaction for removing a neutron from ${O^{17}}$ is: ${O^{17}} 	o {O^{16}} + n$.The total binding energy $(B.E.)$ of a nucleus is given by the product of the number of nucleons $(A)$ and the binding energy per nucleon $(B.E./A)$.Total $B.E.$ of ${O^{17}} = 17 	imes 7.75 \, MeV = 131.75 \, MeV$.Total $B.E.$ of ${O^{16}} = 16 	imes 7.97 \, MeV = 127.52 \, MeV$.The energy required to remove the neutron is the difference in the total binding energies:Energy required $= B.E.({O^{17}}) - B.E.({O^{16}})$.Energy required $= 131.75 \, MeV - 127.52 \, MeV = 4.23 \, MeV$.

The binding energy per nucleon of ${O^{16}}$ is $7.97 \, MeV$ and that of ${O^{17}}$ is $7.75 \, MeV$. The energy (in $MeV$) required to remove a neutron from ${O^{17}}$ is

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