The binding energy per nucleon for ${O^{16}}$ and ${O^{17}}$ are $7.97 \, MeV$ and $7.75 \, MeV$ respectively. How much energy in $MeV$ is required to remove one neutron from ${O^{17}}$?

  • A
    $3.52$
  • B
    $3.64$
  • C
    $4.23$
  • D
    $7.86$

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Similar Questions

Nuclear fission is possible because the binding energy per nucleon ...

$A$ plot of the number of neutrons $(N)$ against the number of protons $(Z)$ for stable nuclei exhibits upward deviation from linearity for atomic number $Z > 20$. For an unstable nucleus having an $N/Z$ ratio less than $1$,the possible mode$(s)$ of decay is(are):
$(A)$ $\beta^{-}$-decay ($\beta$ emission)
$(B)$ Orbital or $K$-electron capture
$(C)$ Neutron emission
$(D)$ $\beta^{+}$-decay (positron emission)

The electrostatic energy of $Z$ protons uniformly distributed throughout a spherical nucleus of radius $R$ is given by $E = \frac{3}{5} \frac{Z(Z-1) e^2}{4 \pi \varepsilon_0 R}$. The measured masses of the neutron,${ }_1^1 H$,${ }_7^{15} N$,and ${ }_8^{15} O$ are $1.008665 \ u$,$1.007825 \ u$,$15.000109 \ u$,and $15.003065 \ u$,respectively. Given that the radii of both the ${ }_7^{15} N$ and ${ }_8^{15} O$ nuclei are the same,$1 \ u = 931.5 \ MeV/c^2$ ($c$ is the speed of light),and $e^2 / (4 \pi \varepsilon_0) = 1.44 \ MeV \ fm$. Assuming that the difference between the binding energies of ${ }_7^{15} N$ and ${ }_8^{15} O$ is purely due to the electrostatic energy,the radius of either of the nuclei is $(1 \ fm = 10^{-15} \ m)$: (in $fm$)

Which one of the following statements is correct?

Which element has the maximum binding energy per nucleon? Write down its value.

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