Question

Calculate the mass defect and binding energy per nucleon for

$27Co^{59}$. [The mass of $Co^{59} = 58.95amu$, mass of

hydrogen atom = 1.008142 amu and mass of neutron = 1.008982 amu].

• A. 8.77 MeV
• B. 8.25 MeV
• C. 9.01 MeV
• D. 8.00 MeV

Answer 1

Answer: (A)

8.77 MeV

Answer 2

Number of protons in $27Co^{59} = 27$

∴ Number of neutrons = 59 – 27 = 32

$\Delta m = (1.008142 \times 27 + 1.008982 \times 32) - 58.95 = 0.556438amu$The binding energy $(E_{B})$ per nucleon

$= \frac{\Delta m \times 931}{\text{Mass of cobalt}} = \frac{0.556438 \times 931}{59}MeV = 8.77MeV$