13.24 The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei C a 20 41  and  from the following data:

m( C a ) 20 40  = 39.962591 u

m( C a ) 20 41 = 40.962278 u

m A l ) 13 26 = 25.986895 u

m( A l ) 13 27 = 26.981541 u

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7 months ago

13.24 If a neutron n 0 1 )  is removed from C a 20 41 , the corresponding reaction can be written as:

C a 20 41 C a 20 40 + n 0 1

The separation energies are

For C a 20 41  : Separation energy = 8.363007 MeV

For A l 13 27  : Separation energy = 13.059 MeV

It is given that

m( C a ) 20 40  = 39.962591 u

m( C a ) 20 41 = 40.962278 u

m( n 0 1 ) = 1.008665 u

 

The mass defect of the reac

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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