# Determine the energy (in MeV) required to remove a single neutron in 56 26Fe?

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### 1 Answer

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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