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 ${}_{20}^{41}{Ca} and {}_{13}^{27}{Al}$ from the following data:
$m ({}_{20}^{40}{Ca}) = 39.962591 u m ({}_{20}^{41}{Ca}) = 40.962278 u m ({}_{13}^{26}{Al}) = 25.986895 u m ({}_{13}^{27}{Al}) = 26.981541 u .$

Solution

The enrgy equation, when neutron is seperated from

{}_{20}^{41}{Ca}

is given by,

{}_{20}^{41}{Ca} + Q_{1} \to {}_{20}^{40}{Ca} + {}_{0}^{1}n

Mass defect is given by,

∆ m = m ({}_{20}^{40}{Ca}) + m ({}_{0}^{1}n) - m ({}_{20}^{41}{Ca})

= 39.962591 + 1.008665 - 40.962278 = 0.008978 u

But,

1 u \equiv 931.5 MeV

Hence,

0.008978 u \equiv 0.008978 \times 931.5 = 8.363 MeV

is the neutron seperation energy.

The equations for the neutron separation in second case can be written as,

{}_{13}^{27}{Al} + Q_{2} \to {}_{13}^{26}{Al} + {}_{0}^{1}n

Mass defect,

∆ m = m ({}_{13}^{26}{Al}) + m ({}_{0}^{1}n) - m ({}_{13}^{27}{Al})

But,

1 u \equiv 931.5 MeV

Hence,

0.014019 u \equiv 0.014019 \times 931.5 = 13.06 MeV .

is the neutron seperation energy.

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