Question

Calculate the neutron separation energy from the following data:
$ m (^{40}_{20} Ca) = 39.962591\,u $ ,
$ m (^{41}_{20} Ca) = 40.962278\,u $ ,
$ m_n = 1.00865, \,1u = 931.5\,Me V/c^2 $

• A. 7.57 MeV
• B. 8.36 MeV
• C. 9.12 MeV
• D. 9.56 MeV

Answer 1

Answer: (B)

Mass defect $= m (\,{}^{40}Ca_{20}) + m_a - m (\,{}^{41}Ca_{20})$
$= 39.962591 + 1.00865 - 40.962278$
$= 8.963 \times 10^{-3}$
$= 8.963 \times 10^{-3} \times 931.5$
$= 8.3490 \,MeV$