The binding energy per nucleon of { }_5 B ^{1 | vedclass

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Question

(A) The reaction for removing a neutron from ${ }_5 B ^{11}$ is given by: ${ }_5 B ^{11} \longrightarrow { }_5 B ^{10} + { }_0 n ^1$ The total binding

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(A) The reaction for removing a neutron from ${ }_5 B ^{11}$ is given by: ${ }_5 B ^{11} \longrightarrow { }_5 B ^{10} + { }_0 n ^1$ The total binding energy of a nucleus is calculated as: $BE = (	ext{Binding energy per nucleon}) 	imes (	ext{Mass number } A)$ For ${ }_5 B ^{11}$: $BE({ }_5 B ^{11}) = 7.5 \,MeV 	imes 11 = 82.5 \,MeV$ For ${ }_5 B ^{10}$: $BE({ }_5 B ^{10}) = 8.0 \,MeV 	imes 10 = 80.0 \,MeV$ The energy required to remove the neutron is the difference in binding energies: $E = BE({ }_5 B ^{11}) - BE({ }_5 B ^{10})$ $E = 82.5 \,MeV - 80.0 \,MeV = 2.5 \,MeV$

The binding energy per nucleon of ${ }_5 B ^{10}$ is $8.0 \,MeV$ and that of ${ }_5 B ^{11}$ is $7.5 \,MeV$. The energy required to remove a neutron from ${ }_5 B ^{11}$ is .......... $MeV$.

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