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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.
Posted by llltkl on May 21, 2013 at 7:20 PM (Answer #1)
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