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)

$8.36\,MeV$

Answer 2

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