Determine the energy (in MeV) required to remove a single neutron in 56 26Fe?
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Fe-56 has the next stable isotope at mass number = 54, i.e. Fe-54.
Driving two neutrons out of the Fe-56 nucleus produces a Fe-54 nucleus.
Hence, we can say, 2*E_n = (M_Fe56 – M_Fe54 – 2*M_n)*931 MeV
Where E_n is the energy required to drive out a neutron, and M_ terms are mass of that particular species in amu.
M_Fe56 : =55.934937u
M_Fe54 : =53.939612u
M_n = 1.00866492u
Therefore, E_n = (55.934937-53.939612-2*1.00866492)*931/2 MeV
= -0.022*931/2 MeV
= -10.2433 MeV
The negative sign indicating that energy has to be input in order to drive the neutron out of the nuclear binding forces.
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