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NCERT Class XII Chapter
Nuclei
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Question : 24 of 31
Marks: +1, -0
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 2041Ca\mathrm{^{41}_{20}Ca} and 1327Al\mathrm{^{27}_{13}Al} from the following data :
m (2040Ca)\left(\mathrm{^{40}_{20}Ca}\right) = 39.962591u, m (2041Ca)\left(\mathrm{^{41}_{20}Ca}\right) = 40.962278u,
m (1326Al)\left(\mathrm{^{26}_{13}Al}\right) = 25.986895u, m (1327Al)\left(\mathrm{^{27}_{13}Al}\right) = 26.981541u, mnm_n = 1.008665 u
Solution:  
Neutron separation of 2040Ca\mathrm{^{40}_{20}Ca} can be obtained as
E = Energy equivalent of total mass afterward – Energy equivalent of nucleus before
E = m(40Ca)+mn−m(2041Ca)20c2\underset{20}{ m(\mathrm{^{40}Ca})+m_n-m(\mathrm{^{41}_{20}Ca}) } c^2
E = {39.962591 + 1.008665 – 40.962278} 931.5 MeV
E = 0.008978 × 931.5 MeV = 8.363 MeV
Similarly, neutron separation energy of 1327Al\mathrm{^{27}_{13}Al} can be calculated as
E = [Energy equivalent of 26Al\mathrm{^{26}Al} + Energy equivalent of mass of neutron – Energy equivalent of nucleus 27Al\mathrm{^{27}Al} before]
E = [m26(Al)+mn−m27(Al)]c2\left[ m^{26}(\mathrm{Al})+m_n-m^{27}(\mathrm{Al}) \right] c^2
E = [25.986895 + 1.008665 – 26.981541] 931.5 MeV
E = 0.014079 × 931.5 MeV = 13.058 MeV
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