The atomic mass of { }_7 N ^{15} is 15.000108 | vedclass

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Question

(B) The reaction for removing a proton from ${ }_8 O ^{16}$ is given by: ${ }_8 O ^{16} + 	ext{Energy} \longrightarrow { }_7 N ^{15} + { }_1 H ^1$.Th

Vedclass · Accepted Answer

$12.13$

Vedclass · Answer

(B) The reaction for removing a proton from ${ }_8 O ^{16}$ is given by: ${ }_8 O ^{16} + 	ext{Energy} \longrightarrow { }_7 N ^{15} + { }_1 H ^1$.The mass defect $\Delta m$ is calculated as: $\Delta m = (M_{N} + M_{H}) - M_{O}$.Substituting the given values: $\Delta m = (15.000108 + 1.007825) - 15.994915 	ext{ a.m.u.}$.$\Delta m = 16.007933 - 15.994915 = 0.013018 	ext{ a.m.u.}$.The energy required is $E = \Delta m 	imes 931.5 	ext{ MeV/a.m.u.}$.$E = 0.013018 	imes 931.5 \approx 12.126 	ext{ MeV}$,which is approximately $12.13 	ext{ MeV}$.

The atomic mass of ${ }_7 N ^{15}$ is $15.000108 \text{ a.m.u.}$ and that of ${ }_8 O ^{16}$ is $15.994915 \text{ a.m.u.}$ If the mass of a proton is $1.007825 \text{ a.m.u.}$,then the minimum energy required to remove the least tightly bound proton is ......... $MeV$.

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