Question

Mass defect in a₁₁Na²³ atom is 0.21 amu. Then the binding energy per nucleon will be –

• A. 8.3 MeV
• B. 8.8 MeV
• C. 8.5 MeV
• D. 9.2 MeV

Answer 1

Answer: (C)

8.5 MeV

Answer 2

DE = 0.21 × 931 MeV

$\frac { \Delta \mathrm { E } } { \mathrm { A } }$= $\frac { 0.21 \times 931 } { 23 }$MeV