Question

The energy required to separate the typical middle mass nucleus ${}_{50}^{120}Sn$ into its constituent nucleons is :
(Mass of ${}_{50}^{120}Sn=$ $119.902199amu$; mass
of proton$=1.007825amu$ and mass of neutron$=1.008665amu$)

• A. $921MeV$
• B. $821MeV$
• C. $1021MeV$
• D. $1121MeV$

Answer 1

The correct option is C $1021MeV$

$Z=50,A-Z=120-50=70$

$\Delta m=Z.m_p+(A-Z)m_n-M$

$\Delta m=[50\times 1.007825+70\times 1.008665-119.902199]$
$=1.095601MeV$

$E=1.095601\times 931.478MeV$
$=1020.53MeV$ $\approx$ $1021MeV$