Question

Determine the mass defect and the binding energy in MeV/nucleon for 10146 Pd (atomic mass = 100.908287 amu). Proton = 1.00783 amu, neutron = 1.00866 amu, and 1 amu = 931.5 MeV?

Answer 1

The given nuclei is .

Hence, number of protons,

Number of neutrons, .

Now, mass defect is defined as mass of the nuclei subtracted from the sum of mass of the protons and neutrons.

Hence,

$\Delta m = \left (n_p \times m_p + n_n \times m_n \right ) - m_{Pd}$

Where

Hence,

$\Delta m = \left (n_p \times m_p + n_n \times m_n \right ) - m_{Pd} \\ \Rightarrow \Delta m = (46 \times 1.00783 \ amu + 55 \times 1.00866 \ amu) - 100.908287 \ amu \\ \Rightarrow \Delta m = (46.36018 \ amu + 55.4763 \ amu) - 100.908287 \ amu \\ \Rightarrow \Delta m = 0.928193 \ amu$

Hence, the binding energy of the nucleus can be calculated by converting the amu units to MeV unit.

$E = 0.928193 \ amu \times \frac{931.5 \ MeV}{1 \ amu} \approx 864.6 \ MeV$

Now, binding energy per nucleon can be calculated by dividing the energy above by the number of nucleons i.e. 46+55=101.

Hence,

$E_{per \ nucleon} = \frac{ 864.6 \ MeV}{101 \ nucleon} \approx {\color{Red} 8.560 \ MeV/nucleon}$

Hence, the binding energy per nucleon of is about 8.560 MeV/nucleon.