   Chapter 19, Problem 88IP

Chapter
Section
Textbook Problem

# Radioactive cobalt-60 is used to study defects in vitamin B12 absorption because cobalt is the metallic atom at the center of the vitamin B12 molecule. The nuclear synthesis of this cobalt isotope involves a three-step process. The overall reaction is iron-58 reacting with two neutrons to produce cobalt-60 along with the emission of another particle. What particle is emitted in this nuclear synthesis? What is the binding energy in J per nucleon for the cobalt-60 nucleus (atomic masses: 60Co = 59.9338 u; 1H = 1.00782 u)? What is the de Broglie wavelength of the emitted particle if it has a velocity equal to 0.90c, where c is the speed of light?

Interpretation Introduction

Interpretation: Particle emitted in the given nuclear synthesis is to be stated. Binding energy in J/nucleon for cobalt- 60 is to be calculated. De Broglie wavelength of the emitted particle is to be stated.

Concept introduction: A process through which an unstable nuclide looses its energy due to excess of protons or neutrons is known as radioactive decay. Decay constant is the quantity that expresses the rate of decrease of number of atoms of a radioactive element per second. Half life of radioactive sample is defined as the time required for the number of nuclides to reach half of the original value.

To determine: Particle emitted in the given nuclear synthesis, binding energy in J/nucleon for cobalt- 60 and De Broglie wavelength of the emitted particle.

Explanation

Explanation

The nuclear reaction for the synthesis of cobalt- 60 is,

2658Fe+201n2760Co+10e .

The synthesis involves the emission of beta particle.

The atomic mass of 60Co=59.9338amu

The mass of a neutron is 1.0087amu .

The mass of 11H proton is 1.00782amu .

Number of protons in 60Co=27 .

Number of neutrons in 60Co=33 .

The mass defect is calculated by the formula,

Δm=Atomicmass of 60Co[Numberofprotons×massof11HprotonNumberofneutrons×massofneutron]

Substitute the value of the atomic mass of 60Co , the number of protons and mass of the 11H proton and that of the neutron in the above equation.

Δm=59.9338[(27×1.00782)+(33×1.00866)]Δm=0.56312amu/nucleus

The conversion of amu/nucleus to Kg/nucleus is done as,

1amu=1.66×1027Kg

Therefore, the conversion of 0.56444amu/nucleus into kg/nucleus is,

0.56312amu/nucleus=(0.56312×1.66×1027)kg/nucleus=9.347×10-28Kg/nucleus

Therefore, the mass defect (Δm) of 60Co is 9.347×10-28Kg/nucleus .

The binding energy per nucleon is calculated by Einstein’s mass energy equation, that is,

ΔE=ΔmC2

Where,

• ΔE is the change in energy.
• Δm is the change in mass.
• C is the velocity of light (3×108) .

Substitute the values of Δm and C in the equation.

ΔE=Δmc2ΔE=(9.347×1028)(3×108)2ΔE=8.4123×1011J/nucleus

Therefore, the energy released per nucleus is 8.4123×1011J/nucleus .

The binding energy per nucleon is calculated by the formula,

Binding energy per nucleon=Bindingenergy per nucleusNumberofprotons+Numberofneutrons

Substitute the value of the binding energy per nucleus, the number of protons and the number of neutrons in the above equation

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