Question

What is the binding energy of ${}^{29}S{i}_{14}$ whose atomic $14$
mass is $28.976495u$
Mass of proton $=1.007276u$
Mass of neutron $=1.008664u$
$($ Neglect the electron mass $)$
(Assume $1u=931.5MeV$ )

• A. 237.84 MeV
• B. 421.72 MeV
• C. 387.21 MeV
• D. 116.35 MeV

Answer 1

Answer: (A)

Given, mass of ${}_{14}{}^{29}Si=28.976495u$
Now, mass of $14$ protons
$=14\times$ mass of $1$ proton
$=14\times 1.007276=14.101864u$
Also, mass of $29-14=15$ neutrons
$=15\times$ mass of $1$ neutron
$=15\times 1.008664=15.12996$
Total mass of constituents of $Si$ atom
$=14.101864+15.12996=29.231824$
So, mass defect $=$ mass of constituents $-$ mass of atom
$=29.231824-28.976495=0.255329u$
Binding energy $=$ Energy equivalent of mass defect
$=\Delta M\times 931.5MeV$
$=0.255329\times 931.5=237.84MeV$