   Chapter 25, Problem 47PS

Chapter
Section
Textbook Problem

Radioactive cobalt-60 is used extensively in nuclear medicine as a γ-ray source. It is made by a neutron capture reaction from cobalt-59 and is a β emitter; β emission is accompanied by strong γ radiation. The half-life of cobalt-60 is 5.27 years.(a) How long will it take for a cobalt-60 source to decrease to one eighth of its original activity?(b) What fraction of the activity of a cobalt-60 source remains after 1.0 year?

(a)

Interpretation Introduction

Interpretation:

Time required for Cobalt-60 to decrease to one eighth of its original activity has to be calculated.

Concept Introduction:

Radiocarbon dating: The dynamic equilibrium exists in all living organism by exhaling or inhaling, maintain the same ratio of 12C and 14C that takes in. when the living organism dies, the intake of 14C stops and it’s the ratio no longer exhibits equilibrium; undergoes radioactive decay.

Half-life period: The time required to reduce to half of its initial value.

Formula used to calculate half-life:

t1/2=0.693kwhere,kis rate constant.(or)ln[N]t- ln[N]0= -kt

Explanation

Given info:

Half-life of cobalt-60(t1/2)=5.27years.

The decreasing original activity is A/A0=1/8.

Calculation:

Formula:

The decay rate constant is calculated as follows,

k=0.693t1/2=0.6935.27yr=0.131y1

Consider, Initial amount as 1

Final amount as (1/8)th=0

(b)

Interpretation Introduction

Interpretation:

The fraction of the activity of a Cobalt-60 source remains after one year has to be calculated.

Concept Introduction:

Radiocarbon dating: The dynamic equilibrium exists in all living organism by exhaling or inhaling, maintain the same ratio of 12C and 14C that takes in. when the living organism dies, the intake of 14C stops and it’s the ratio no longer exhibits equilibrium; undergoes radioactive decay.

Half-life period: The time required to reduce to half of its initial value.

Formula used to calculate half-life:

t1/2=0.693kwhere,kis rate constant.(or)ln[N]t- ln[N]0= -kt

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