Calculate the neutron separation energy from the following data: m (4020 Ca) = 39.962591 u , m (4120 Ca) = 40.962278 u , mn = 1.00865, 1u = 931.5 Me V/c2

Question

Calculate the neutron separation energy from the following data: m (4020 Ca) = 39.962591 u , m (4120 Ca) = 40.962278 u , mn = 1.00865, 1u = 931.5 Me V/c2  (A) 7.57 MeV (B) 8.36 MeV (C) 9.12 MeV (D) 9.56 MeV

Tardigrade Admin · Accepted Answer

Mass defect = m ( 40Ca20) + ma - m ( 41Ca20) = 39.962591 + 1.00865 - 40.962278 = 8.963 × 10-3 = 8.963 × 10-3 × 931.5 = 8.3490 MeV

Q. Calculate the neutron separation energy from the following data:
$m (_{20}^{40} C a) = 39.962591 u$ ,
$m (_{20}^{41} C a) = 40.962278 u$ ,
$m_{n} = 1.00865, 1 u = 931.5 M e V / c^{2}$

7.57 MeV

8.36 MeV

9.12 MeV

9.56 MeV

Mass defect $= m (^{40} C a_{20}) + m_{a} - m (^{41} C a_{20})$
$= 39.962591 + 1.00865 - 40.962278$
$= 8.963 \times 1 0^{- 3}$
$= 8.963 \times 1 0^{- 3} \times 931.5$
$= 8.3490 M e V$

Q. Calculate the neutron separation energy from the following data: m(2040​Ca)=39.962591u , m(2041​Ca)=40.962278u , mn​=1.00865,1u=931.5MeV/c2