   Chapter 19, Problem 39E

Chapter
Section
Textbook Problem

# A rock contains 0.688 mg 206Pb for every 1.000 mg 238U present. Assuming that no lead was originally present, that all the 206Pb formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between 238U and 206Pb is negligible, calculate the age of the rock. (For 238U, t1/2 = 4.5 × 109 years.)

Interpretation Introduction

Interpretation: A rock containing 0.688mg 206Pb for every 1.000mg 238U is given.

Assuming no lead was originally present, and all the 206Pb formed over the years remained in the rock and nuclides formed in intermediate stages of decay is negligible, the age of the rock is to be calculated.

Concept introduction: A process through which an unstable nuclide loses its energy due to excess of protons or neutrons is known as radioactive decay.

To determine: The age of the rock.

Explanation

Explanation

Given

The value of 238U is 4.5×109years .

The decay constant is calculated by the formula,

k=0.693t1/2

Where

• t1/2 is the half life of nuclide.
• k is the decay constant.

Substitute the value of half life in the above formula.

k=0.6934.5×109years1=1.5×10-10years-1_ .

The decay constant is 1.5×10-10years-1 .

Explanation

Given

The amount of 206Pb for every 1.000mg 238U is 0.688mg .

The conversion of mg into g is done as,

1mg=103g

Therefore, the conversion of 0.688mg into g is,

0.688mg=0.688×103g

Molar mass of 206Pb is 206g/mole .

In 206g of 206Pb , atoms of 206Pb present =6.022×1023

Therefore, in 0.688×103g of 238U , atoms of 206Pb present =6.022×1023×0.688×103206=2.01×1018atoms_

Explanation

Given

The amount of 238U is 1mg .

The conversion of mg into g is done as,

1mg=103g

Molar mass of 238U is 238g/mole .

In 238g of 238U , atoms of 238U present =6

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