Question

Binding energy of a Nitrogen nucleus $\left[{}_7^{14}N\right]$, given $m\left[{}_7^{14}N\right]=14.00307u$

• A. 206.5 MeV
• B. 78 MeV
• C. 104.7 MeV
• D. 85 MeV

Answer 1

Answer: (C)

$BE=\left[Z{m}_p+(A-Z)m_n-m_X\right]\times 931.5$
$Be=[7.05481+7.06069-14.0030]\times 931.5$
$=104.7MeV$