Question

The proton mass is 1.007276 amu , the neutron mass is 1.008665 amu , and the electron mass is 5.486×10⁻⁴ amu .

What is the expected mass of a potassium-39 nucleus, based on the total mass of its protons and neutrons?

The actual measured mass of a potassium-39 atom is 38.963707 amu . What is the mass defect, Δm, for a potassium-39 nucleus?

What is the binding energy for a potassium-39 nucleus?

What is the binding energy, ΔE, for a potassium-39 nucleus in MeV/nucleon? Recall that 1 MeV=1.60×10−13 J.

Answer 1

atomic number of K = 19.

number of protons = 19

number of neutrons = mass number - atomic number

number of neutrons = 39 - 19 = 20

mass of one proton = 1.007276 amu

mass of 19 protons = 19 x 1.007276 amu = 19.138244 amu

mass of one neutrons = 1.008665 amu

mass of 20 neutrons = 20 x 1.008665 amu = 20.1733 amu

Part-A

expected mass of Potassium = 19.138244 amu+ 20.1733 amu= 39.311544 amu

actual mass of Potassium = 38.963707 amu

mass defect = 39.311544 amu - 38.963707 amu = 0.347837 amu

mass defect = 0.347837 amu

1 amu = 1.66 x10^-27 Kg

0.347837 amu

                       = 0.347837 amu x 1.66 x10-27 = 0.577409 x10^-27 Kg

mass defect = 5.77x 10^-28 Kg

Part-C

Velocity of light = C = 3.0 x 10^8 m/sec

Binding energy = mc^2

Binding energy = 5.77x10^-28 x ( 3.0x10^8)^2

Binding energy = 51.93 x10^-12 J

Binding energy =5.193 x10^-11 J/mole

1.60x10^-13 J = 1 MeV

5.193 x10^-11J = ?

                       = 5.193 x 10^-11 / 1.60x10^-13 = 3.2456 x10^2 MeV

Binding energy = 3.25 x 10^2 Mev = 325 Mev

Binding energy per nucleon = 325 / 39 = 8.33 MeV/neucleon