Obtain the binding energy (in $MeV$) of a nitrogen nucleus $\left(^{14}_{7} N \right)$ given $m\left(^{14}_{7} N \right)=14.00307 \; u$. (Given: $m_{H} = 1.007825 \; u$,$m_{n} = 1.008665 \; u$) (in $; MeV$)

  • A
    $142.66$
  • B
    $104.66$
  • C
    $204.43$
  • D
    $84.15$

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Calculate the binding energy of a nitrogen nucleus $\left[{ }_7^{14} N\right]$,given that the mass of the nucleus $m\left[{ }_7^{14} N\right] = 14.00307 \ u$. (Take mass of proton $m_p = 1.00783 \ u$ and mass of neutron $m_n = 1.00867 \ u$) (in $MeV$)

If the binding energy per nucleon in $Li^7$ and $He^4$ nuclei are respectively $5.60 \ MeV$ and $7.06 \ MeV$, then the energy of the reaction $Li^7 + p \to 2He^4$ is ......... $MeV$.

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$(a)$ The Equivalence of Mass and Energy
$(b)$ Nuclear Energy
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