Question

Calculate the binding energy per nucleon for ${}_{10}^{20}Ne{,}_{26}^{56}Fe$ and ${}_{92}^{238}U$. Given that mass of neutron is 1.008665 amu, mass of proton is 1.007825 amu, mass of ${}_{10}^{20}Ne$ is 19.9924 amu, mass of ${}_{26}^{56}Fe$ is 55.93492 amu, ${}_{92}^{238}U$ is 238.050783 amu

Answer 1

For ${}_{10}^{20}Ne$ : mass defect ,$\Delta m=[10m_p+(20-10\left)m_n\right]+M_{Ne}$

$\Delta m=[10\times 1.007825+10\times 1.008665]-19.9924=0.1725amu$

Binding energy per nucleon $BE=\frac{\Delta m{c}^2}{A}=\frac{0.1725c^2}{20}=0.0086c^{2}amu=0.0086\times 931.5=8.03MeV$ as $(1amu=931.5MeV/c^2)$

For ${}_{26}^{56}Fe$ : mass defect ,$\Delta m=[26m_p+(56-26\left)m_n\right]+M_{Fe}$

$\Delta m=[26\times 1.007825+30\times 1.008665]-55.93492=0.5285amu$

Binding energy per nucleon $BE=\frac{\Delta m{c}^2}{A}=\frac{0.5285c^2}{56}=0.0086c^{2}amu=0.0094\times 931.5=8.76MeV$ as $(1amu=931.5MeV/c^2)$

For ${}_{92}^{238}U$ : mass defect ,$\Delta m=[92m_p+(238-92\left)m_n\right]+M_{Ne}$

$\Delta m=[92\times 1.007825+146\times 1.008665]-238.050783=1.934amu$

Binding energy per nucleon $BE=\frac{\Delta m{c}^2}{A}=\frac{1.934c^2}{238}=0.0086c^{2}amu=0.0081\times 931.5=7.57MeV$ as $(1amu=931.5MeV/c^2)$