Question

The fact that BE/A is greatest for A near 60 implies that the range of the nuclear force is about the diameter of such nuclides.

(a) Calculate the diameter of an A = 60 nucleus.
________ fm
(b) Calculate BE/A for ⁵⁶Fe and ¹⁰²Ru. The first is one of the most tightly bound nuclides, while the second is larger and less tightly bound.
⁵⁶Fe
________ MeV
¹⁰²Ru
________ MeV

Answer 1

a.)

Average diameter of nucleus is given by,

d = 2*r0*A^(1/3)

here, r0 = 1.20*10^-15 m

A = 60

then, d = 2*(1.20*10^-15)*(60)^(1/3)

d = 9.396*10^-15 m

b.)

1.

Mass defect is given by:

dm = Mass of Protons in iron + mass of neutrons in iron + Mass of electrons in iron - atomic Mass of iron

Given that

Atomic mass of iron = 55.9349375 u

Mass of 1 proton = 1.007276467 u

Mass of 1 electron = 9.11*10^-31 kg = 0.000549 amu

Mass of 1 neutron = 1.008664916 u

Number of protons in iron = 26

Number of neutrons in iron = 56 - 26 = 30

Number of electrons in iron = 26

So,

dm = 26*mp + 30*mn + 26*me - M_Fe

dm = 26*1.007276467 + 30*1.008664916 +26*0.000549 - 55.9349375

dm = mass defect = 0.528472122 amu

Binding energy is given by:

E = dm*c^2

c^2 = 931.5 Mev/u

So,

E = (0.528472122 u)*(931.5 MeV/u) = 0.528472122*931.5

E = Binding energy = 492.271782 MeV = 492.3 MeV

Binding energy per nucleon will be:

BE per nucleon = BE/total number of nucleons

Total number of nucleons = 56

BE per nucleon = BE/total number of nucleons = 492.271782/56

Binding energy per nucleon of 56Fe = 8.79 MeV

2.

Mass defect is given by:

dm = Mass of Protons in Ruthenium + mass of neutrons in Ruthenium + Mass of electrons in Ruthenium - atomic Mass of Ruthenium

Given that

Atomic mass of Ruthenium = 101.904349312 u

Mass of 1 proton = 1.007276467 u

Mass of 1 electron = 9.11*10^-31 kg = 0.000549 amu

Mass of 1 neutron = 1.008664916 u

Number of protons in Ruthenium = 44

Number of neutrons in Ruthenium = 102 - 44 = 58

Number of electrons in Ruthenium = 44

So,

dm = 44*mp + 58*mn + 44*me - M_Ru

dm = 44*1.007276467 + 58*1.008664916 +44*0.000549 - 101.904349312

dm = mass defect = 0.942536364 amu

Binding energy is given by:

E = dm*c^2

c^2 = 931.5 Mev/u

So,

E = (0.942536364 u)*(931.5 MeV/u) = 0.942536364*931.5

E = Binding energy = 877.972623066 MeV = 877.98 MeV

Binding energy per nucleon will be:

BE per nucleon = BE/total number of nucleons

Total number of nucleons = 102

BE per nucleon = BE/total number of nucleons = 877.972623066/102

Binding energy per nucleon of 102Ru = 8.61 MeV

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