Chapter 25, Problem 57GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# A technique to date geological samples uses rubidium-87, a long-lived radioactive isotope of rubidium (t½ = 4.8 × 1010 years). Rubidium-87 decays by β emission to strontium-87. If rubidium-87 is part of a rock or mineral, then strontium-87 will remain trapped within the crystalline structure of the rock. The age of the rock dates back to the time when the rock solidified. Chemical analysis of the rock gives the amounts of 87Rb and 87Sr. From these data, the fraction of 87Rb that remains can be calculated. Suppose analysis of a stony meteorite determined that 1.8 mmol of 87Rb and 1.6 mmol of 87Sr (the portion of 87Sr formed by decomposition of 87Rb) were present. Estimate the age of the meteorite. (Hint: The amount of 87Rb at t0 is moles 87Rb + moles 87Sr.)

Interpretation Introduction

Interpretation:

Age of Rubidium-87 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

A long lived radioactive isotope of Rubidium-87(t1/2=Â 4.8Ã—1010years)

Analysis of a stony meteorite determined that 1.8mmol of Rubidium-87 and 1.6 mmol of Strontium-87 were present.

The following reaction is R3787bâ†’S3887rÂ +Â e-10.

By the conservation of mass then the initial amount of Rubidium-87 should be 1.8mol + 1.6mol = 3.4mol.

Thus the ratio of Rubidium left over the original amount should be 1.83.4=0.529, thus less than one half life has eclipsed.

tage=Â halfÂ lifeÂ Â 0.693Â Ã—Â ln(1/y)

=â€‰4.8Ã—10100

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