Question

19.36:

Part A

Ca-40 (atomic mass = 39.96259 amu)

Express your answer using five decimal places.

Mass defect =		amu

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Part B

Express your answer using four significant figures.

Binding energy per nucleon =		MeV/nucleon

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Part C

V-51 (atomic mass = 50.94396 amu)

Express your answer using five decimal places.

Mass defect =		amu

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Part D

Express your answer using four significant figures.

Binding energy per nucleon =		MeV/nucleon

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Part E

Ag-107 (atomic mass = 106.905092 amu)

Express your answer using five decimal places.

Mass defect =		amu

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Part F

Express your answer using four significant figures.

Binding energy per nucleon =		MeV/nucleon

Answer 1

Part A

Mass defect =( Combined mass of protons and neutrons in the atom )- (actual mass of the atom)

No. of protons in Ca-40 =20

Mass of 1 proton = 1.00728 amu

Mass of protons in Ca-40 = 20 * 1.00728 amu = 20.1456 amu

No. of neutrons in Ca-40 =20

Mass of 1 neutron = 1.00867 amu

Mass of neutrons in Ca-40 = 20 * 1.00867 amu = 20.1734 amu

Combined mass of protons and neutrons in Ca-40 = 20.1456 amu + 20.1734 amu

= 40.319 amu

Actual atomic mass of Ca-40 =39.96259 amu

Therefore,

Mass defect = ( Combined mass of protons and neutrons in the Ca-40)- (actual atomic mass of the Ca-40)

= 40.319 amu - 39.96259 amu

= 0.35641 amu

Part B

Mass defect per nucleus, m = 0.35461 amu / nucleus

Therefore, Mass defect per nucleon, m^' = [0.35461 amu / 40 nucleons] =0.00886525 amu /nucleon

Mass defect per nucleon in kg, m' =( 0.00886525 amu) (1.6606 x 10^-27 kg/amu)

= 1.47216 * 10^-29 kg / nucleon

Converting mass to energy using Einsten's equation ,E =mc², where c = 2.9979 x 10⁸ m/s

E =mc²

=  (1.47216 * 10^-29 kg / nucleon )*(2.9979 x 10⁸ m / s)²

= 1.3230897 * 10^-12 J / nucleon

{1 MeV = 1.602*10^-13 J ; 1J =1 / 1.602*10^-13 MeV ]

Therefore,

1.3230897 * 10^-12 J / nucleon = [ 1.3230897 * 10^-12 J * (1 MeV /   1.602*10^-13 J)] / nucleon

=8.259 MeV/ nucleon

Binding energy per nucleon = 8.259 MeV/ nucleon

Part C

Mass defect =( Combined mass of protons and neutrons in the atom )- (actual mass of the atom)

No. of protons in V-51 =23

Mass of 1 proton = 1.00728 amu

Mass of protons in V-51 = 23 * 1.00728 amu = 23.16744  amu

No. of neutrons in V-51 =28

Mass of 1 neutron = 1.00867 amu

Mass of neutrons in V-51 = 28 * 1.00867 amu = 28.24276 amu

Combined mass of protons and neutrons in V-51 = 23.16744  amu+ 28.24276 amu

=51.4102 amu

Actual atomic mass of V-51 = 50.94396 amu

Therefore,

Mass defect = ( Combined mass of protons and neutrons in the Ca-40)- (actual atomic mass of the Ca-40)

= 51.4102 amu -  50.94396 amu

= 0.46624  amu

Part D

Mass defect per nucleus, m = 0.46624  amu / nucleus

Therefore, Mass defect per nucleon, m^' =[ 0.46624  amu/ 51 nucleons] =0.00914196 amu /nucleon

Mass defect per nucleon in kg, m' =(0.00914196 amu /nucleon ) (1.6606 x 10^-27 kg/amu)

= 1.5181 * 10^-29 kg / nucleon

Converting mass to energy using Einsten's equation ,E =mc², where c = 2.9979 x 10⁸ m/s

E =mc²

=  (1.5181 * 10^-29 kg / nucleon )*(2.9979 x 10⁸ m / s)²

= 1.364377 * 10^-12 J / nucleon

{1 MeV = 1.602*10^-13 J ; 1J =1 / 1.602*10^-13 MeV ]

Therefore,

1.364377 * 10^-12 J / nucleon = [ 1.364377 * 10^-12 J * (1 MeV /   1.602*10^-13 J)] / nucleon

=8.517 MeV/ nucleon

Binding energy per nucleon = 8.517 MeV/ nucleon