The mass of a proton is 1.00728 amu andthat of a neutron is 1.00867 amu. What is the binding energy pernucleon (in J) of a Co nucleus? (The mass of a cobalt-60 nucleus is59.9338 amu.) a. 3.039 * 10^-12 b. 2.487 * 10^-12 c. 7.009 * 10^-14 d. 1.368 * 10^-12 e. 9.432 * 10^-13

The binding energy per nucleon = 1.368*10^-12 (option D)

Explanation:

Step 1: Data given

The mass of a proton is 1.00728 amu

The mass of a neutron is 1.00867 amu

The mass of a cobalt-60 nucleus is59.9338 amu

Step 2: Calculate binding energy

The mass defect = the difference between the mass of a nucleus and the total mass of its constituent particles.

Cobalt60 has 27 protons and 33 neutrons.

The mass of 27 protons = 27*1.00728 u = 27.19656 u

The mass of 33 neutrons = 33*1.00867 u = 33.28611 u

Total mass of protons + neutrons = 27.19656 u + 33.28611 u = 60.48267 u

Mass of a cobalt60 nucleus = 59.9338 amu

Mass defect = Δm = 0.54887 u

ΔE = c²*Δm

ΔE = (3.00 * 10^8 m/s) ² * (0.54887 amu)) * (1.00 g / 6.02 * 10^23 amu) * (1kg/1000g)

Step 3: Calculate binding energy per nucleon

ΔE = 8.21 * 10^-11 J

8.21 * 10^-11 J / 59.9338 = 1.368 * 10^-12

The binding energy per nucleon = 1.368*10^-12 (option D)