If the binding energy of $N^{14}$ is $7.5 \text{ MeV}$ per nucleon and that of $N^{15}$ is $7.7 \text{ MeV}$ per nucleon, then the energy required to remove a neutron from $N^{15}$ is (in $\text{ MeV}$)

  • A
    $5.25$
  • B
    $0.2$
  • C
    $10.5$
  • D
    $0.4$

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What is the order of binding energy per nucleon?

$A$ nucleus of mass $M$ emits a $\gamma$-ray photon of frequency $\nu$. The loss of internal energy by the nucleus is:

In the nuclear reaction $_1^2H + _1^3H \to _2^4He + _0^1n$, if the binding energies of $_1^2H$, $_1^3H$, and $_2^4He$ are respectively $a$, $b$, and $c$ (in $MeV$), then the energy (in $MeV$) released in this reaction is:

The binding energy per nucleon for $^6C^{12}$ nucleus is ......... $MeV$.
(Nuclear mass of $^6C^{12} = 12.00000 \text{ a.m.u.}$
Mass of hydrogen nucleus $= 1.007825 \text{ a.m.u.}$
Mass of neutron $= 1.008665 \text{ a.m.u.}$)

The mass equivalent of $931\, MeV$ energy is

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