Question

The stability of the iron-57 nucleus is directly proportional to its binding energy.

See Periodic Table (1 point) x Feedback Part 1 Given the exact mass of a 57 Fe atom is 9.4543502x10 26 kg, calculate the total nuclear binding energy in joules per mole 57 Fe nuclei x 1015 J/mol 5.12 (1 point) See Hint Part 2 Feedback Calculate the binding energy in joules per mole nucleons. x 10 13 x8.9 J/mol nucleon

Please show work!!

Answer 1

⁵⁷Fe has ,

no. of proton = 26. & Mass of proton = 1.672623×10^-27 kg.

no. of neutrons = mass no. - no. of protons = (57-26) = 31 and mass of neutron = 1.674929×10^-27 kg.

So, Fe should have mass = ( 26×1.672623×10^-27kg)+(31×1.674929×10- 27 kg) = 9.5410997×10^-26 kg.

But the actual mass of Fe = 9.4543502×10^-26 kg.

So,
m.
(9.5410997×10^-26 - 9.4543502×10^-26 ) kg.

=0.08665977 × 10^-26 kg.

So binding energy per atom =c² = 0.08665977×10^-26×10³ g ×3×10⁸ ms^{​​​​​-1}​​​

m

= 0.25997931×10^-18×10³ gms^{​​​​-1}

= 0.25997931×10^-15 J

Now we have to calculate this energy per mole of Fe.

bindingenergy per mole will be

= 0.26×10^-15×6.022×10²³J/mol

=1.5656×10⁸J/mol.

One atom of ⁵⁷Fe atom has no. of nucleon= 57

Binding energy per atom of Fe =0.26×10^-15J

So, binding energy per nucleon will be = (0.26×10^-15)/57 J/nucleon.

= 4.561×10^-18 J/nucleon.