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 given by: ${O^{17}} 	o {O^{16}} + {n^1}$.The total binding energy $(B.E.)$ of a nu

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$4.23$

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(C) The nuclear reaction for removing a neutron from ${O^{17}}$ is given by: ${O^{17}} 	o {O^{16}} + {n^1}$.The total binding energy $(B.E.)$ of a nucleus is calculated as: $B.E. = (	ext{Number of nucleons}) 	imes (	ext{Binding energy per nucleon})$.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 between the total binding energies of the parent and daughter nuclei:$	ext{Energy required} = B.E.({O^{17}}) - B.E.({O^{16}})$$	ext{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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