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)

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.